# Lambert series in analytic number theory

Jordan Bell
[email protected]
(February 22, 2023)
###### Abstract

Tour of 19th and early 20th century analytic number theory.

## 1 Introduction

Let $d(n)$ denote the number of positive divisors of $n$. For $|z|<1$,

 $\sum_{n=1}^{\infty}d(n)z^{n}=\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}.$

## 2 Euler

The first use of the term “Lambert series” was by Euler to describe the roots of an equation.

Euler writes in E25 [28] about the particular value of a Lambert series.

## 3 Lambert

Bullynck [7, pp. 157–158]: “As he recorded in his scientific diary, the Monatsbuch, Lambert started thinking about the divisors of integers in June 1756. An essay by G.W. Krafft (1701–1754) in the St. Petersburg Novi Commentarii seems to have triggered Lambert’s interest [Bopp 1916, p. 17, 40].”

Bullynck [7, p. 163]:

Lambert did more than deliver the factor table. He also addressed the absence of any coherent theory of prime numbers and divisors. Filling such a lacuna could be important for the discovery of new and more primality criteria and factoring tests. For Lambert the absence of such a theory was also an occasion to apply the principles laid out in his philosophical work. A fragmentary theory, or one with gaps, needed philosophical and mathematical efforts to mature.

To this aim [prime recognition] and others I have looked into the theory of prime numbers, but only found certain isolated pieces, which did not seem possible to make easily into a connected and well formed system. Euclid has few, Fermat some mostly unproven theorems, Euler individual fragments, that anyway are farther away from the first beginnings, and leave gaps between them and the beginnings. [Lambert 1770, p. 20]

Bullynck [7, pp. 164–165]:

In 1770, Lambert presented two sketches of what would be needed for something like a theory of numbers. The first dealt mainly with factoring methods [Lambert 1765-1772, II, pp. 1–41], while the second gave a more axiomatic treatment [Lambert 1770, pp. 20–48]. In the first essay, Lambert explained how, for composite number with small factors, Eratosthenes’ sieve could be used and optimised. For larger factors, Lambert explained that approximation from above, starting by division by numbers that are close to the square root of the tested number $p$, was more advantageous. For both methods, Lambert advised the use of tables. The second essay had more theoretical bearings. Lambert rephrased Euclid’s theorems for use in factoring, included the greatest common divisor algorithm, and put the idea of relatively prime numbers to good use. He also noted that binary notation, because of the frequent symmetries, could be helpful. Finally,Lambert also recognized Fermat’s little theorem as a good, though not infallible criterion for primality, “but the negative example is very rar” [Lambert 1770, p. 43].

Monatsbuch, September 1764, “Singula haec in Capp. ult. Ontol. occurunt”, and Anm. 5, Anm. 25, 1764, Anm. 12 1765, Anm. 19, 1765 [2].

Lambert [53, pp. 506–511, §875]

Youschkevitch [87]

Lorey [59, p. 23]

Löwenhaupt [60, p. 32]

## 4 Krafft

Krafft [50, pp. 244–245]

## 5 Servois

Servois [72] and [73, p. 166]

## 6 Lacroix

Lacroix [51, pp. 465–466, §1195]

## 7 Klügel

Klügel [46, pp. 52–53, s.v. “Theiler einer Zahl”, §12]:

Ist $N=\alpha^{m}\beta^{n}\gamma^{p}\cdots$, wo $\alpha,\beta,\gamma$, Primzahlen sind; so erhellet auch leicht, daßalle Theiler von $N$, die Einheit und die Zahl selbst mit engeschlossen, durch die Glieder des Products

 $(1+\alpha+\alpha^{2}+\cdots+\alpha^{m})(1+\beta+\beta^{2}+\cdots+\beta^{n})(1+% \gamma+\gamma^{2}+\cdots+\gamma^{p})\cdots$

argestelle werden. Die Anzahl der Glieder dieses Products, d. i. die Anzahl aller Theiler von $N$, ist offenbar $=(m+1)(n+1)(p+1)\cdots$. Für das obige Beispiel $=4\cdot 3\cdot 2=24$, wo die Einheit mit engeschlossen ist.
In der aus der Entwickelung von

 $\frac{x}{1-x}+\frac{x^{2}}{1-x^{2}}+\frac{x^{3}}{1-x^{3}}+\cdots+\frac{x^{n}}{% 1-x^{n}}+\cdots$

entspringenden Reihe:

 $x+2x^{2}+2x^{3}+3x^{4}+2x^{5}+4x^{6}+2x^{7}+\cdots$

welche Lambert in seiner Architektonik S. 507. mittheilt, enthalt jeder Coefficient so viele Einheiten, als der Exponent der entsprechendenden Potenz von $x$ Theiler hat.

Stern [77]

## 9 Clausen

Clausen [21] states that

 $\sum_{n=1}^{\infty}\frac{x^{n}}{1-x^{n}}=\sum_{n=1}^{\infty}x^{n^{2}}\left(% \frac{1+x^{n}}{1-x^{n}}\right),$

and that the right-hand series converges quickly for small $x$. Clausen does prove this expansion, and a proof is later given by Scherk [68]. Scherk’s argument uses the fact

 $1+2t+2t^{2}+2t^{3}+2t^{4}+\cdots=(1+t+t^{2}+t^{3}+t^{4}+\cdots)+t(1+t+t^{2}+t^% {3}+t^{4}+\cdots)=\frac{1+t}{1-t}.$

We write

 $\sum_{n=1}^{\infty}\frac{x^{n}}{1-x^{n}}=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty% }x^{nm}.$

The series is

 $\begin{array}[]{rrrrrrr}x&+x^{2}&+x^{3}&+x^{4}&+x^{5}&+x^{6}&+\mathrm{etc.}\\ +x^{2}&+x^{4}&+x^{6}&+x^{8}&+x^{10}&+x^{12}&+\mathrm{etc.}\\ +x^{3}&+x^{6}&+x^{9}&x^{12}&+x^{15}&+x^{18}&+\mathrm{etc.}\\ +x^{4}&+x^{8}&+x^{12}&+x^{16}&+x^{20}&+x^{24}&+\mathrm{etc.}\\ +x^{5}&+x^{10}&+x^{15}&+x^{20}&+x^{25}&+x^{30}&+\mathrm{etc.}\\ +x^{6}&+x^{12}&+x^{18}&+x^{24}&+x^{30}&+x^{36}&+\mathrm{etc.}\\ +\mathrm{etc.}&&&&&&\end{array}$

We sum the terms in the first row and column: the sum of these is

 $x+2x^{2}+2x^{3}+2x^{4}+\mathrm{etc.}=x\left(\frac{1+x}{1-x}\right).$

Then, from what remains we sum the terms in the second row and column: the sum of these is

 $x^{4}+2x^{6}+2x^{8}+2x^{10}+\mathrm{etc.}=x^{4}\left(\frac{1+x^{2}}{1-x^{2}}% \right).$

Then, from what remains, we sum the terms in the third row and column: the sum of these is

 $x^{9}+2x^{12}+2x^{15}+2x^{18}+\mathrm{etc.}=x^{9}\left(\frac{1+x^{3}}{1-x^{3}}% \right),$

etc.

## 10 Eisenstein

Eisenstein [27] states that for $|z|<1$,

 $\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}=\frac{1}{(1-x)(1-x^{2})(1-x^{3})% \cdots}\sum_{n=1}^{\infty}(-1)^{n+1}\frac{nz^{n(n+1)/2}}{(1-x)\cdots(1-x^{n})}.$

For $t=\frac{1}{z}$, Eisenstein states that

 $\frac{z}{1-z}+\frac{z^{2}}{1-z^{2}}+\frac{z^{3}}{1-z^{3}}+\frac{z^{4}}{1-z^{4}% }+\textrm{etc.}$

is equal to

 $\cfrac{1}{t-1-\cfrac{(t-1)^{2}}{t^{2}-1-\cfrac{t(t-1)^{2}}{t^{3}-1-\cfrac{t(t^% {2}-1)^{2}}{t^{4}-1-\cfrac{t^{2}(t^{2}-1)^{2}}{t^{5}-1-\cfrac{t^{2}(t^{3}-1)^{% 2}}{t^{6}-1-\cfrac{t^{3}(t^{3}-1)^{2}}{t^{7}-1-\textrm{etc.}}}}}}}}$

Expressing Lambert series using continued fractions is relevant to the irrationality of the value of the series. See Borwein [3]. See also Zudilin [90].

Möbius [62]

## 12 Jacobi

Jacobi’s Fundamenta nova [44, §40, 66 and p. 185]

Chandrasekharan [20, Chapter X]: using Lambert series to prove the four squares theorem.

Dirichlet [25]

Fischer [29]

## 14 Cauchy

Cauchy [13] and [14] two memoirs in the same volume.

## 15 Burhenne

Burhenne [8] says the following about Lambert series. For

 $F(x)=\sum_{n=1}^{\infty}d(n)x^{n},$

we have

 $d(n)=\frac{F^{(n)}(0)}{n!}.$

Define

 $F_{k}(x)=\frac{x^{k}}{1-x^{k}},$

so that

 $F(x)=\sum_{k=1}^{\infty}F_{k}(x).$

It is apparent that if $k>n$, then

 $F_{k}^{(n)}(0)=0,$

hence

 $F^{(n)}(0)=\sum_{k=1}^{n}F_{k}^{(n)}(0).$

The above suggests finding explicit expressions for $F_{k}^{(n)}(0)$. Burhenne cites Sohncke [74, pp. 32–33]: for even $k$,

 $\displaystyle\frac{d^{n}\left(\frac{x^{p}}{x^{k}-a^{k}}\right)}{dx^{n}}$ $\displaystyle=(-1)^{n}\frac{n!}{ka^{k-p-1}}\left(\frac{1}{(x-a)^{n+1}}-(-1)^{p% }\frac{1}{(x+a)^{n+1}}\right)$ $\displaystyle+(-1)^{n}\frac{n!}{\frac{1}{2}ka^{k-p-1}}\sum_{h=1}^{\frac{1}{2}k% -1}\frac{\cos\left(\frac{2h(p+1)\pi}{k}+(n+1)\phi_{h}\right)}{\sqrt{\left(x^{2% }-2xa\cos\frac{2h\pi}{n}+a^{2}\right)^{n+1}}}$

and for odd $k$,

 $\displaystyle\frac{d^{n}\left(\frac{x^{p}}{x^{k}-a^{k}}\right)}{dx^{n}}$ $\displaystyle=(-1)^{n}\frac{n!}{ka^{k-p-1}}\frac{1}{(x-a)^{n+1}}$ $\displaystyle+(-1)^{n}\frac{n!}{\frac{1}{2}ka^{k-p-1}}\sum_{h=1}^{\frac{k-1}{2% }}\frac{\cos\left(\frac{2h(p+1)\pi}{k}+(n+1)\phi_{h}\right)}{\sqrt{\left(x^{2}% -2xa\cos\frac{2h\pi}{n}+a^{2}\right)^{n+1}}},$

where

 $\cos\phi_{h}=\frac{x-a\cos\frac{2h\pi}{k}}{\sqrt{x^{2}-2xa\cos\frac{2h\pi}{k}+% a^{2}}},\quad\sin\phi_{h}=\frac{a\sin\frac{2h\pi}{k}}{\sqrt{x^{2}-2xa\cos\frac% {2h\pi}{k}+a^{2}}}.$

For $a=1$ and $x=0$,

 $\cos\phi_{h}=-\cos\frac{2h\pi}{k},\qquad\sin\phi_{h}=\sin\frac{2h\pi}{k},$

from which

 $\phi_{h}=\pi-\frac{2h\pi}{k},$

and thus

 $\displaystyle\cos\left(\frac{2h(k+1)\pi}{k}+(n+1)\phi_{h}\right)$ $\displaystyle=\cos\left(\frac{2h(k+1)\pi}{k}+(n+1)\left(\pi-\frac{2h\pi}{k}% \right)\right)$ $\displaystyle=\cos\left(2h\pi+\frac{2h\pi}{k}+\pi-\frac{2h\pi}{k}+n\left(\pi-% \frac{2h\pi}{k}\right)\right)$ $\displaystyle=\cos\left((n+1)\pi-\frac{2nh\pi}{k}\right)$ $\displaystyle=(-1)^{n+1}\cos\frac{2nh\pi}{k}.$

For even $k$, taking $p=k$ we have

 $\frac{d^{n}\left(\frac{x^{k}}{1-x^{k}}\right)}{dx^{n}}=(-1)^{n+1}\frac{n!}{k}% \left(\frac{1}{(-1)^{n+1}}-1\right)+(-1)^{n+1}\frac{n!}{\frac{1}{2}k}\sum_{h=1% }^{\frac{1}{2}k-1}(-1)^{n+1}\cos\frac{2nh\pi}{k},$

i.e.,

 $F_{k}^{(n)}(0)=\frac{n!}{k}(1-(-1)^{n+1})+\frac{2\cdot n!}{k}\sum_{h=1}^{\frac% {1}{2}k-1}\cos\frac{2nh\pi}{k}.$

For odd $k$, taking $p=k$ we have

 $\frac{d^{n}\left(\frac{x^{k}}{1-x^{k}}\right)}{dx^{n}}=(-1)^{n+1}\frac{n!}{k}% \frac{1}{(-1)^{n+1}}+(-1)^{n+1}\frac{n!}{\frac{1}{2}k}\sum_{h=1}^{\frac{k-1}{2% }}(-1)^{n+1}\cos\frac{2nh\pi}{k},$

i.e.,

 $F_{k}^{(n)}(0)=\frac{n!}{k}+\frac{2\cdot n!}{k}\sum_{h=1}^{\frac{k-1}{2}}\cos% \frac{2nh\pi}{k}.$

Using the identity, for $h\not\in 2\pi\mathbb{Z}$,

 $\sum_{h=1}^{M}\cos h\theta=-\frac{1}{2}+\frac{\sin\left(M+\frac{1}{2}\right)% \theta}{2\sin\frac{\theta}{2}}=-\frac{1}{2}+\frac{1}{2}\left(\sin M\theta\cot% \frac{\theta}{2}+\cos M\theta\right),$

we get for even $k$,

 $\displaystyle F_{k}^{(n)}(0)$ \displaystyle=\begin{cases}\hidden@noalign{}\textstyle\frac{n!}{k}\cot\frac{n% \pi}{k}\sin n\pi&k\not|n\\ \hidden@noalign{}\textstyle\frac{n!}{k}(1-(-1)^{n+1})+\frac{2\cdot n!}{k}\left% (\frac{1}{2}k-1\right)&k|n\end{cases} \displaystyle=\begin{cases}\hidden@noalign{}\textstyle 0&k\not|n\\ \hidden@noalign{}\textstyle n!-\frac{n!}{k}(1+(-1)^{n+1})&k|n.\end{cases}

For odd $k$,

 $\displaystyle F_{k}^{(n)}(0)$ \displaystyle=\begin{cases}\hidden@noalign{}\textstyle\frac{n!}{k}\csc\frac{n% \pi}{k}\sin n\pi&k\not|n\\ \hidden@noalign{}\textstyle\frac{n!}{k}+\frac{2\cdot n!}{k}\frac{k-1}{2}&k|n.% \end{cases} \displaystyle=\begin{cases}\hidden@noalign{}\textstyle 0&k\not|n\\ \hidden@noalign{}\textstyle n!&k|n.\end{cases}

Zehfuss [88]

## 17 Bernoulli numbers

The Bernoulli polynomials are defined by

 $\frac{te^{tx}}{e^{t}-1}=\sum_{m=0}^{\infty}B_{m}(x)\frac{t^{m}}{m!}.$

The Bernoulli numbers are defined by $B_{m}=B_{m}(0)$.

We denote by $[x]$ the greatest integer $\leq x$, and we define $\{x\}=x-[x]$, namely, the fractional part of $x$. We define $P_{m}(x)=B_{m}(\{x\})$, the periodic Bernoulli functions.

## 18 Euler-Maclaurin summation formula

Euler E47 and E212, §142, for the summation formula. Euler’s studies the gamma function in E368. In particular, in §12 he gives Stirling’s formula, and in §14 he obtains $\Gamma^{\prime}(1)=-\gamma$. Euler in §142 of E212 states that

 $\gamma=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}B_{2n}}{2n}.$

Bromwich [6, Chapter XII]

The Euler-Maclaurin summation formula [5, p. 280, Ch. VI, Eq. 35] tells us that for $f\in C^{\infty}([0,1])$,

 $f(0)=\int_{0}^{1}f(t)dt+B_{1}(f(1)-f(0))+\sum_{m=1}^{k}\frac{1}{(2m)!}B_{2m}(f% ^{(2m-1)}(1)-f^{(2m-1)}(0))+R_{2k},$

where

 $R_{2k}=-\int_{0}^{1}\frac{P_{2k}(1-\eta)}{(2k)!}f^{(2k)}(\eta)d\eta.$

Poisson and Jacobi on the Euler-Maclaurin summation formula.

## 19 Schlömilch

Schlömilch [69] and [71, p. 238], [70]

For $m\geq 1$,

 $\int_{0}^{\infty}\frac{t^{2m-1}}{e^{2\pi t}-1}dt=(-1)^{m+1}\frac{B_{2m}}{4m}.$ (1)

For $\alpha>0$,

 $\int_{0}^{\infty}\frac{\sin\alpha t}{e^{2\pi t}-1}dt=\frac{1}{4}+\frac{1}{2}% \left(\frac{1}{e^{\alpha}-1}-\frac{1}{\alpha}\right)$ (2)

and

 $\int_{0}^{\infty}\frac{1-\cos\alpha t}{e^{2\pi t}-1}\frac{dt}{t}=\frac{1}{4}% \alpha+\frac{1}{2}\left(\log(1-e^{-\alpha})-\log\alpha\right).$ (3)

For $\xi>0$ and $n\geq 1$, using (2) with $\alpha=\xi,2\xi,3\xi,\ldots,2n\xi$ and also using

 $\sum_{k=1}^{N}\sin k\theta=\frac{1}{2}\cot\frac{\theta}{2}-\frac{\cos(N+\frac{% 1}{2})\theta}{2\sin\frac{\theta}{2}},$

we get

 $\displaystyle\sum_{m=1}^{2n}\left(\frac{1}{e^{m\xi}-1}-\frac{1}{m\xi}\right)$ $\displaystyle=\sum_{m=1}^{2n}\left(-\frac{1}{2}+2\int_{0}^{\infty}\frac{\sin m% \xi t}{e^{2\pi t}-1}dt\right)$ $\displaystyle=-n+\int_{0}^{\infty}\frac{1}{e^{2\pi t}-1}\sum_{m=1}^{2n}2\sin m% \xi tdt$ $\displaystyle=-n+\int_{0}^{\infty}\frac{1}{e^{2\pi t}-1}\left(\cot\frac{\xi t}% {2}-\frac{\cos(2n+\frac{1}{2})\xi t}{\sin\frac{\xi t}{2}}\right)dt.$

Using $\cos(a+b)=\cos a\cos b-\sin a\sin b$, this becomes

 $\displaystyle\sum_{m=1}^{2n}\left(\frac{1}{e^{m\xi}-1}-\frac{1}{m\xi}\right)$ $\displaystyle=-n+\int_{0}^{\infty}\frac{1}{e^{2\pi t}-1}(1-\cos 2n\xi t)\cot% \frac{\xi t}{2}dt$ (4) $\displaystyle+\int_{0}^{\infty}\frac{1}{e^{2\pi t}-1}\sin 2n\xi tdt.$

For $\alpha=2n\xi$, (3) tells us

 $\int_{0}^{\infty}\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}\frac{dt}{t}=\frac{1}{4}% \cdot 2n\xi+\frac{1}{2}\left(\log(1-e^{-2n\xi})-\log 2n\xi\right).$

Rearranging,

 $\frac{\log 2n}{\xi}=n+\frac{\log(1-e^{-2n\xi})-\log\xi}{\xi}-\frac{2}{\xi}\int% _{0}^{\infty}\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}\frac{dt}{t}$ (5)

 $\begin{split}&\sum_{m=1}^{2n}\frac{1}{e^{m\xi}-1}-\frac{1}{\xi}\left(-\log 2n+% \sum_{m=1}^{2n}\frac{1}{m}\right)\\ =&\frac{\log(1-e^{-2n\xi})-\log\xi}{\xi}-\int_{0}^{\infty}\left(\frac{2}{\xi t% }-\cot\frac{\xi t}{2}\right)\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}dt\\ &+\int_{0}^{\infty}\frac{1}{e^{2\pi t}-1}\sin 2n\xi tdt.\end{split}$

Writing

 $C_{n}=-\log n+\sum_{m=1}^{n}\frac{1}{m}$

and using (2) this becomes

 $\begin{split}&\sum_{m=1}^{2n}\frac{1}{e^{m\xi}-1}-\frac{1}{\xi}C_{2n}\\ =&\frac{\log(1-e^{-2n\xi})-\log\xi}{\xi}-2\int_{0}^{\infty}\left(\frac{1}{\xi t% }-\frac{1}{2}\cot\frac{\xi t}{2}\right)\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}dt\\ &+\frac{1}{4}+\frac{1}{2}\left(\frac{1}{e^{2n\xi}-1}-\frac{1}{2n\xi}\right).% \end{split}$

We write

 $I_{2n}(\xi)=2\int_{0}^{\infty}\left(\frac{1}{\xi t}-\frac{1}{2}\cot\frac{\xi t% }{2}\right)\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}dt,$

and we shall obtain an asymptotic formula for $I_{2n}(\xi)$.

We apply the Euler-Maclaurin summation formula. Let $h>0$, and for $f(t)=\cos ht$ we have $f^{\prime}(t)=-h\sin ht$, and for $m\geq 1$ we have $f^{(2m)}(t)=(-1)^{m}h^{2m}\cos ht$ and $f^{(2m-1)}(t)=(-1)^{m}h^{2m-1}\sin ht$. Thus the Euler-Maclaurin formula yields

 $1=\int_{0}^{1}\cos htdt-\frac{1}{2}(\cos h-1)+\sum_{m=1}^{k}\frac{1}{(2m)!}B_{% 2m}(-1)^{m}h^{2m-1}\sin h+R_{2k}.$

Using the identity $\cot\frac{\theta}{2}=\frac{1+\cos\theta}{\sin\theta}$ and dividing by $\sin h$, this becomes

 $\frac{1}{2}\cot\frac{h}{2}=\frac{1}{h}+\sum_{m=1}^{k}\frac{1}{(2m)!}B_{2m}(-1)% ^{m}h^{2m-1}+\frac{1}{\sin h}R_{2k}.$ (6)

Because $P_{m}(1-\eta)=P_{m}(\eta)$ for even $m$,

 $\displaystyle R_{2k}$ $\displaystyle=-\int_{0}^{1}\frac{P_{2k}(\eta)}{(2k)!}(-1)^{k}h^{2k}\cos h\eta d\eta$ $\displaystyle=-B_{2k}\int_{0}^{1}\frac{1}{(2k)!}(-1)^{k}h^{2k}\cos h\eta d\eta% -\int_{0}^{1}\frac{(P_{2k}(\eta)-B_{2k})}{(2k)!}(-1)^{k}h^{2k}\cos h\eta d\eta$ $\displaystyle=(-1)^{k+1}\frac{B_{2k}h^{2k}}{(2k)!}\frac{\sin h}{h}+(-1)^{k+1}% \frac{h^{2k}}{(2k)!}\int_{0}^{1}(P_{2k}(\eta)-B_{2k})\cos h\eta d\eta.$

Since $P_{2k}(\eta)-B_{2k}$ does not change sign on $(0,1)$, by the mean-value theorem for integration there is some $\theta=\theta(h,k)$, $0<\theta<1$, such that (using $\int_{0}^{1}P_{2k}(\eta)d\eta=0$)

 $\int_{0}^{1}(P_{2k}(\eta)-B_{2k})\cos h\eta d\eta=\cos h\theta\int_{0}^{1}(P_{% 2k}(\eta)-B_{2k})d\eta=-B_{2k}\cos h\theta.$

Therefore (6) becomes

 $\displaystyle\frac{1}{2}\cot\frac{h}{2}-\frac{1}{h}$ $\displaystyle=\sum_{m=1}^{k}\frac{1}{(2m)!}B_{2m}(-1)^{m}h^{2m-1}$ $\displaystyle+(-1)^{k+1}\frac{B_{2k}h^{2k-1}}{(2k)!}+(-1)^{k+2}\frac{h^{2k}}{(% 2k)!\sin h}B_{2k}\cos h\theta,$

i.e.,

 $\frac{1}{2}\cot\frac{h}{2}-\frac{1}{h}=\sum_{m=1}^{k-1}\frac{1}{(2m)!}B_{2m}(-% 1)^{m}h^{2m-1}+(-1)^{k}\frac{h^{2k}}{(2k)!\sin h}B_{2k}\cos h\theta.$

Write

 $E_{k}(h)=(-1)^{k+1}\frac{h^{2k}}{(2k)!\sin h}B_{2k}\cos h\theta.$

We apply the above to $I_{2n}(\xi)$, and get, for any $k\geq 1$,

 $\displaystyle I_{2n}(\xi)$ $\displaystyle=2\int_{0}^{\infty}\left(E_{k}(\xi t)-\sum_{m=1}^{k-1}\frac{1}{(2% m)!}B_{2m}(-1)^{m}(\xi t)^{2m-1}\right)\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}dt$ $\displaystyle=-2\sum_{m=1}^{k-1}\frac{1}{(2m)!}B_{2m}(-1)^{m}\xi^{2m-1}\int_{0% }^{\infty}t^{2m-1}\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}dt$ $\displaystyle+2\int_{0}^{\infty}E_{k}(\xi t)\frac{1-\cos 2n\xi t}{e^{2\pi t}-1% }dt.$

Using (1),

 $\displaystyle\int_{0}^{\infty}t^{2m-1}\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}dt$ $\displaystyle=\int_{0}^{\infty}\frac{t^{2m-1}}{e^{2\pi t}-1}dt-\int_{0}^{% \infty}\frac{t^{2m-1}\cos 2n\xi t}{e^{2\pi t}-1}dt$ $\displaystyle=(-1)^{m+1}\frac{B_{2m}}{4m}-\int_{0}^{\infty}\frac{t^{2m-1}\cos 2% n\xi t}{e^{2\pi t}-1}dt.$

Let

 $f(x)=\frac{1}{e^{x}-1}-\frac{1}{x}.$

By (2),

 $f(x)+\frac{1}{2}=2\int_{0}^{\infty}\frac{\sin xt}{e^{2\pi t}-1}dt.$

For $m\geq 1$,

 $f^{(2m-1)}(x)=2\int_{0}^{\infty}\frac{(-1)^{m-1}t^{2m-1}\cos xt}{e^{2\pi t}-1}dt,$

which for $x=2n\xi$ becomes

 $\frac{(-1)^{m-1}}{2}f^{(2m-1)}(2n\xi)=\int_{0}^{\infty}\frac{t^{2m-1}\cos 2n% \xi t}{e^{2\pi t}-1}dt.$

Therefore

 $2\int_{0}^{\infty}t^{2m-1}\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}dt=(-1)^{m+1}% \frac{B_{2m}}{2m}+(-1)^{m}f^{(2m-1)}(2n\xi).$

Thus $I_{2n}(\xi)$ is

 $\displaystyle I_{2n}(\xi)$ $\displaystyle=-\sum_{m=1}^{k-1}\frac{1}{(2m)!}B_{2m}(-1)^{m}\xi^{2m-1}\left((-% 1)^{m+1}\frac{B_{2m}}{2m}+(-1)^{m}f^{(2m-1)}(2n\xi)\right)$ $\displaystyle+2\int_{0}^{\infty}E_{k}(\xi t)\frac{1-\cos 2n\xi t}{e^{2\pi t}-1% }dt$ $\displaystyle=\sum_{m=1}^{k-1}\frac{B_{2m}^{2}}{(2m)!2m}\xi^{2m-1}-\sum_{m=1}^% {k-1}\frac{B_{2m}}{(2m)!}\xi^{2m-1}f^{(2m-1)}(2n\xi)$ $\displaystyle+2\int_{0}^{\infty}E_{k}(\xi t)\frac{1-\cos 2n\xi t}{e^{2\pi t}-1% }dt.$

But

 $\displaystyle\left|\int_{0}^{\infty}E_{k}(\xi t)\frac{1-\cos 2n\xi t}{e^{2\pi t% }-1}dt\right|$ $\displaystyle=\left|\int_{0}^{\infty}(-1)^{k+1}\frac{(\xi t)^{2k}}{(2k)!\sin% \xi t}B_{2k}\cos\xi t\theta\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}dt\right|$ $\displaystyle\leq\frac{|B_{2k}|}{(2k)!}\int_{0}^{\infty}\frac{(\xi t)^{2k}}{|% \sin\xi t|}\frac{1-\cos 2n\xi t}{e^{2\pi t}-1}dt.$

It is a fact that for all $u\in\mathbb{R}$,

 $\frac{1-\cos 2nu}{|\sin u|}\leq\frac{\pi}{2}\frac{1-\cos 2nu}{u},$

we obtain

 $\begin{split}&\left|\int_{0}^{\infty}E_{k}(\xi t)\frac{1-\cos 2n\xi t}{e^{2\pi t% }-1}dt\right|\\ \leq&\frac{\pi}{2}\frac{|B_{2k}|}{(2k)!}\int_{0}^{\infty}(\xi t)^{2k-1}\frac{1% -\cos 2n\xi t}{e^{2\pi t}-1}dt\\ =&\frac{\pi}{2}\frac{|B_{2k}|}{(2k)!}\xi^{2k-1}\cdot\frac{1}{2}\left((-1)^{k+1% }\frac{B_{2k}}{2k}+(-1)^{k}f^{(2k-1)}(2n\xi)\right).\end{split}$

Hence

 $\displaystyle I_{2n}(\xi)$ $\displaystyle=\sum_{m=1}^{k-1}\frac{B_{2m}^{2}}{(2m)!2m}\xi^{2m-1}-\sum_{m=1}^% {k-1}\frac{B_{2m}}{(2m)!}\xi^{2m-1}f^{(2m-1)}(2n\xi)$ $\displaystyle+O\left(\frac{B_{2k}^{2}}{(2k)!2k}\xi^{2k-1}\right)+O\left(\frac{% |B_{2k}|}{(2k)!}\xi^{2k-1}f^{(2k-1)}(2n\xi)\right).$

Therefore we have

 $\begin{split}&\sum_{m=1}^{2n}\frac{1}{e^{m\xi}-1}-\frac{1}{\xi}C_{2n}\\ =&\frac{\log(1-e^{-2n\xi})-\log\xi}{\xi}+\frac{1}{4}+\frac{1}{2}\left(\frac{1}% {e^{2n\xi}-1}-\frac{1}{2n\xi}\right)-I_{2n}(\xi)\\ =&\frac{\log(1-e^{-2n\xi})-\log\xi}{\xi}+\frac{1}{4}+\frac{1}{2}\left(\frac{1}% {e^{2n\xi}-1}-\frac{1}{2n\xi}\right)\\ &-\sum_{m=1}^{k-1}\frac{B_{2m}^{2}}{(2m)!2m}\xi^{2m-1}+\sum_{m=1}^{k-1}\frac{B% _{2m}}{(2m)!}\xi^{2m-1}f^{(2m-1)}(2n\xi)\\ &+O\left(\frac{B_{2k}^{2}}{(2k)!2k}\xi^{2k-1}\right)+O\left(\frac{|B_{2k}|}{(2% k)!}\xi^{2k-1}f^{(2k-1)}(2n\xi)\right).\end{split}$

Taking $n\to\infty$,

 $\sum_{m=1}^{\infty}\frac{1}{e^{m\xi}-1}-\frac{\gamma}{\xi}=-\frac{\log\xi}{\xi% }+\frac{1}{4}-\sum_{m=1}^{k-1}\frac{B_{2m}^{2}}{(2m)!2m}\xi^{2m-1}+O\left(% \frac{B_{2k}^{2}}{(2k)!2k}\xi^{2k-1}\right).$

## 20 Voronoi summation formula

The Voronoi summation formula [22, p. 182] states that if $f:\mathbb{R}\to\mathbb{C}$ is a Schwartz function, then

 $\displaystyle\sum_{n=1}^{\infty}d(n)f(n)$ $\displaystyle=\int_{0}^{\infty}f(t)(\log t+2\gamma)dt+\frac{f(0)}{4}$ $\displaystyle+\sum_{n=1}^{\infty}d(n)\int_{0}^{\infty}f(t)(4K_{0}(4\pi(nt)^{1/% 2})-2\pi Y_{0}(4\pi(nt)^{1/2}))dt,$

where $K_{0}$ and $Y_{0}$ are Bessel functions.

Let $0. For $f(t)=e^{-tx}$, we compute

 $\begin{split}&\int_{0}^{\infty}f(t)(4K_{0}(4\pi(nt)^{1/2})-2\pi Y_{0}(4\pi(nt)% ^{1/2}))dt\\ =&-\frac{2}{x}\exp\left(\frac{4\pi^{2}n}{x}\right)\mathrm{Ei}\left(-\frac{4\pi% ^{2}n}{x}\right)-\frac{2}{x}\exp\left(-\frac{4\pi^{2}n}{x}\right)\mathrm{Ei}% \left(\frac{4\pi^{2}n}{x}\right),\end{split}$

where

 $\mathrm{Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}{t}dt,\qquad x\neq 0,$

the exponential integral. Then the Voronoi summation formula yields

 $\begin{split}&\sum_{n=1}^{\infty}d(n)e^{-nx}\\ =&\frac{\gamma}{x}-\frac{\log x}{x}+\frac{1}{4}\\ &+\sum_{n=1}^{\infty}d(n)\left(-\frac{2}{x}\exp\left(\frac{4\pi^{2}n}{x}\right% )\mathrm{Ei}\left(-\frac{4\pi^{2}n}{x}\right)-\frac{2}{x}\exp\left(-\frac{4\pi% ^{2}n}{x}\right)\mathrm{Ei}\left(\frac{4\pi^{2}n}{x}\right)\right).\end{split}$

Egger and Steiner [26] give a proof of the Voronoi summation formula involving Lambert series.

Kluyver [47] and [48]

Guinand [36]

Curtze [23]

Laguerre [52]

## 23 V. A. Lebesgue

V. A. Lebesgue [56]:

Bouniakowsky [4]

Chebyshev [80]

## 26 Catalan

Catalan [9]

Catalan [10, p. 89]

Catalan [11, p. 119, §CXXIV] and [12, pp. 38–39, §CCXXVI]

Pincherle [63]

## 28 Glaisher

Glaisher [34, p. 163]

## 29 Günther

Günther [37, p. 83] and [38, p. 178]

Stieltjes [78]

cf. Zhang [89]

## 31 Rogel

Rogel [65] and [66]

## 32 Cesàro

Cesàro [15]

Cesàro [16]

Cesàro [17] and [18, pp. 181–184]

Bromwich [6, p. 201, Chapter VIII, Example B, 35]

## 33 de la Vallée-Poussin

de la Vallée-Poussin [24]

Torelli [83]

Landau [54]

Knopp [49]

## 37 Generating functions

Hardy and Wright [40, p. 258, Theorem 307]:

###### Theorem 1.

For $f(s)=\sum_{n=1}^{\infty}a_{n}n^{-s}$ and $g(s)=\sum_{n=1}^{\infty}b_{n}n^{-s}$,

 $\sum_{n=1}^{\infty}a_{n}\frac{x^{n}}{1-x^{n}}=\sum_{n=1}^{\infty}b_{n}x^{n},% \qquad|x|<1,$

if and only if there is some $\sigma$ such that

 $\zeta(s)f(s)=g(s),\qquad\mathrm{Re}\,(s)>\sigma.$

For $f(s)=\sum_{n=1}^{\infty}\mu(n)n^{-s}$ and $g(s)=1$, using [40, p. 250, Theorem 287]

 $\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\mu(n)n^{-s},\qquad\mathrm{Re}\,(s)>1,$

we get

 $\sum_{n=1}^{\infty}\frac{\mu(n)x^{n}}{1-x^{n}}=x.$ (7)

For $f(s)=\sum_{n=1}^{\infty}\phi(n)n^{-s}$ and

 $g(s)=\zeta(s-1)=\sum_{n=1}^{\infty}n^{-s+1}=\sum_{n=1}^{\infty}nn^{-s},$

using [40, p. 250, Theorem 288]

 $\frac{\zeta(s-1)}{\zeta(s)}=\sum_{n=1}^{\infty}\phi(n)n^{-s},\qquad\mathrm{Re}% \,(s)>2,$

we get

 $\sum_{n=1}^{\infty}\frac{\phi(n)x^{n}}{1-x^{n}}=\sum_{n=1}^{\infty}nx^{n}=% \frac{x}{(1-x)^{2}}.$

For $n=p_{1}^{a_{1}}\cdots p_{r}^{a_{r}}$, define $\Omega(n)=a_{1}+\cdots+a_{n}$ and

 $\lambda(n)=(-1)^{\Omega(n)}.$

For $f(s)=\sum_{n=1}^{\infty}\lambda(n)n^{-s}$ and

 $g(s)=\zeta(2s)=\sum_{n=1}^{\infty}n^{-2s}=\sum_{n=1}^{\infty}(n^{2})^{-s},$

using [40, p. 255, Theorem 300]

 $\frac{\zeta(2s)}{\zeta(s)}=\sum_{n=1}^{\infty}\lambda(n)n^{-s},\qquad\mathrm{% Re}\,(s)>1,$

we get

 $\sum_{n=1}^{\infty}\frac{\lambda(n)x^{n}}{1-x^{n}}=\sum_{n=1}^{\infty}x^{n^{2}}.$

We define the von Mangoldt function $\Lambda:\mathbb{N}\to\mathbb{R}$ by $\Lambda(n)=\log p$ if $n$ is some positive integer power of a prime $p$, and $\Lambda(n)=0$ otherwise. For example, $\Lambda(1)=0$, $\Lambda(12)=0$, $\Lambda(125)=\log 5$. It is a fact [40, p. 254, Theorem 296] that for any $n$, the von Mangoldt function satisfies

 $\sum_{m|n}\Lambda(m)=\log n.$ (8)

For $f(s)=\sum_{n=1}^{\infty}\Lambda(n)n^{-s}$ and

 $g(s)=-\zeta^{\prime}(s)=\sum_{n=1}^{\infty}\log nn^{-s},$

using [40, p. 253, Theorem 294]

 $-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\sum_{n=1}^{\infty}\Lambda(n)n^{-s},$

we obtain

 $\sum_{n=1}^{\infty}\frac{\Lambda(n)x^{n}}{1-x^{n}}=\sum_{n=1}^{\infty}\log nx^% {n}.$

## 38 Mertens

For $\mathrm{Re}\,s>1$, we define

 $P(s)=\sum_{p}\frac{1}{p^{s}}.$

We also define

 $H=\sum_{m=2}^{\infty}\sum_{p}\frac{1}{mp^{m}}.$

Mertens [61] proves the following.

###### Theorem 2.

As $\varrho\to 0$,

 $P(1+\rho)=\log\left(\frac{1}{\rho}\right)-H+o(1).$
###### Proof.

As $\varrho\to 0$,

 $\zeta(1+\varrho)=\frac{1}{\varrho}+\gamma+O(\varrho)=\frac{1}{\varrho}(1+% \gamma\varrho+O(\varrho^{2})).$

Taking the logarithm,

 $\log\zeta(1+\varrho)=\log\left(\frac{1}{\varrho}\right)+\log(1+\gamma\varrho+O% (\varrho^{2}))=\log\left(\frac{1}{\varrho}\right)+\gamma\varrho+O(\varrho^{2}).$ (9)

On the other hand, for $\varrho>0$,

 $\zeta(1+\varrho)=\prod_{p}\frac{1}{1-\frac{1}{p^{1+\varrho}}},$

and taking the logarithm,

 $\displaystyle\log\zeta(1+\varrho)$ $\displaystyle=-\sum_{p}\log\left(1-\frac{1}{p^{1+\varrho}}\right)$ $\displaystyle=\sum_{p}\sum_{m=1}^{\infty}\frac{1}{mp^{m(1+\varrho)}}$ $\displaystyle=P(1+\rho)+\sum_{m=2}^{\infty}\sum_{p}\frac{1}{mp^{m(1+\varrho)}}.$

Then as $\varrho\to 0$,

 $\log\zeta(1+\varrho)=P(1+\varrho)+H+o(1).$

Combining this with (9) we get that as $\varrho\to 0$,

 $P(1+\rho)=\log\left(\frac{1}{\rho}\right)-H+o(1).$

Mertens [61] also proves that for any $x$ there is some

 $|\delta|<\frac{4}{\log(x+1)}+\frac{2}{x\log x}$

such that

 $\sum_{p\leq x}\frac{1}{p}=\log\log x+\gamma-H+\delta.$

Thus,

 $\sum_{p\leq x}\frac{1}{p}=\log\log x+\gamma-H+O\left(\frac{1}{\log x}\right).$

Mertens shows that

 $H=-\sum_{n=2}^{\infty}\mu(n)\frac{\log\zeta(n)}{n}.$

This can be derived using (7), and we do this now; see [58].

###### Lemma 3.

For $\mathrm{Re}\,s>1$,

 $\frac{1}{s}\log\zeta(s)=\int_{2}^{\infty}\frac{\pi(t)dt}{t(t^{s}-1)}.$
###### Proof.

For $p$ prime and $\mathrm{Re}\,s>0$,

 $\displaystyle\int_{p}^{\infty}\frac{dt}{t(t^{s}-1)}$ $\displaystyle=\int_{p}^{\infty}t^{-s-1}\frac{1}{1-t^{-s}}dt$ $\displaystyle=\int_{p}^{\infty}t^{-s-1}\sum_{n=0}^{\infty}(t^{-s})^{n}dt$ $\displaystyle=\sum_{n=0}^{\infty}\int_{p}^{\infty}t^{-ns-s-1}dt$ $\displaystyle=\sum_{n=0}^{\infty}\frac{t^{-ns-s}}{-ns-s}\bigg{|}_{p}^{\infty}$ $\displaystyle=\frac{1}{s}\sum_{n=1}^{\infty}\frac{p^{-ns}}{n}$ $\displaystyle=-\frac{1}{s}\log(1-p^{-s}),$

hence

 $\log\left(\frac{1}{1-p^{-s}}\right)=s\int_{p}^{\infty}\frac{dt}{t(t^{s}-1)}.$

On the one hand,

 $\sum_{p}\int_{p}^{\infty}\frac{dt}{t(t^{s}-1)}=\int_{2}^{\infty}\frac{\pi(t)dt% }{t(t^{s}-1)}.$

On the other hand, for $\mathrm{Re}\,s>1$ we have

 $\sum_{p}\log\left(\frac{1}{1-p^{-s}}\right)=\log\prod_{p}\left(\frac{1}{1-p^{-% s}}\right)=\log\zeta(s).$

Combining these, for $\mathrm{Re}\,s>1$,

 $\frac{1}{s}\log\zeta(s)=\int_{2}^{\infty}\frac{\pi(t)dt}{t(t^{s}-1)}.$

###### Theorem 4.
 $H=-\sum_{n=2}^{\infty}\mu(n)\frac{\log\zeta(n)}{n}.$
###### Proof.

For any prime $p$ and for $m\geq 1$,

 $\int_{p}^{\infty}t^{-m-1}dt=\frac{t^{-m}}{-m}\bigg{|}_{p}^{\infty}=\frac{1}{mp% ^{m}},$

and using this we have

 $\displaystyle H$ $\displaystyle=\sum_{m=2}^{\infty}\sum_{p}\frac{1}{mp^{m}}$ $\displaystyle=\sum_{m=2}^{\infty}\sum_{p}\int_{p}^{\infty}t^{-m-1}dt$ $\displaystyle=\sum_{m=2}^{\infty}\int_{2}^{\infty}\pi(t)t^{-m-1}dt$ $\displaystyle=\int_{2}^{\infty}\pi(t)\left(\sum_{m=2}^{\infty}t^{-m-1}\right)dt$ $\displaystyle=\int_{2}^{\infty}\pi(t)\frac{1}{t^{2}(t-1)}dt$

Rearranging (7),

 $\frac{x^{2}}{1-x}=-\sum_{n=2}^{\infty}\frac{\mu(n)x^{n}}{1-x^{n}}.$

With $x=t^{-1}$,

 $\frac{1}{t(t-1)}=-\sum_{n=2}^{\infty}\frac{\mu(n)}{t^{n}-1},$

so

 $\frac{1}{t^{2}(t-1)}=-\sum_{n=2}^{\infty}\frac{\mu(n)}{t(t^{n}-1)}.$

Thus we have

 $H=-\int_{2}^{\infty}\pi(t)\left(\sum_{n=2}^{\infty}\frac{\mu(n)}{t(t^{n}-1)}% \right)dt=-\sum_{n=2}^{\infty}\mu(n)\int_{2}^{\infty}\frac{\pi(t)dt}{t(t^{n}-1% )}dt.$

Using Lemma 3 for $s=2,3,4,\ldots$,

 $H=-\sum_{n=2}^{\infty}\mu(n)\cdot\frac{1}{n}\log\zeta(n),$

completing the proof. ∎

## 39 Preliminaries on prime numbers

We define

 $\vartheta(x)=\sum_{p\leq x}\log p=\log\prod_{p\leq x}p$

and

 $\psi(x)=\sum_{p^{m}\leq x}\log p=\sum_{n\leq x}\Lambda(n).$

One sees that

 $\psi(x)=\sum_{p\leq x}[\log_{p}x]\log p=\sum_{p\leq x}\left[\frac{\log x}{\log p% }\right]\log p.$

As well,

 $\psi(x)=\sum_{m=1}^{\infty}\sum_{p\leq x^{1/m}}\log p=\sum_{m=1}^{\infty}% \vartheta(x^{1/m});$ (10)

there are only finitely many terms on the right-hand side, as $\vartheta(x^{1/m})=0$ if $x<2^{m}$.

###### Theorem 5.
 $\psi(x)=\vartheta(x)+O(x^{1/2}(\log x)^{2}).$
###### Proof.

For $x\geq 2$, $\vartheta(x), giving

 $\displaystyle\sum_{2\leq m\leq\frac{\log x}{\log 2}}\vartheta(x^{1/m})$ $\displaystyle<\sum_{2\leq m\leq\frac{\log x}{\log 2}}x^{1/m}\frac{1}{m}\log x$ $\displaystyle\leq x^{1/2}\log x\sum_{2\leq m\leq\frac{\log x}{\log 2}}\frac{1}% {m}$ $\displaystyle=O(x^{1/2}(\log x)^{2}).$

Thus, using (10) we have

 $\psi(x)=\vartheta(x)+\sum_{2\leq m\leq\frac{\log x}{\log 2}}\vartheta(x^{1/m})% =\vartheta(x)+O(x^{1/2}(\log x)^{2}).$

We prove that if $\lim_{x\to\infty}\frac{\vartheta(x)}{x}=1$ then $\frac{\pi(x)}{x/\log x}=1$.

###### Theorem 6.
 $\liminf_{x\to\infty}\frac{\pi(x)}{x/\log x}=\liminf_{x\to\infty}\frac{% \vartheta(x)}{x}$

and

 $\limsup_{x\to\infty}\frac{\pi(x)}{x/\log x}=\limsup_{x\to\infty}\frac{% \vartheta(x)}{x}.$
###### Proof.

From (10), $\vartheta(x)\leq\psi(x)$. And,

 $\psi(x)=\sum_{p\leq x}\left[\frac{\log x}{\log p}\right]\log p\leq\sum_{p\leq x% }\frac{\log x}{\log p}\log p=\log x\sum_{p\leq x}.$

Hence

 $\frac{\vartheta(x)}{x}\leq\frac{\pi(x)\log x}{x},$

whence

 $\liminf_{x\to\infty}\frac{\vartheta(x)}{x}\leq\liminf_{x\to\infty}\frac{\pi(x)% }{x/\log x}$

and

 $\limsup_{x\to\infty}\frac{\vartheta(x)}{x}\leq\limsup_{x\to\infty}\frac{\pi(x)% }{x/\log x}.$

Let $0<\alpha<1$. For $x>1$,

 $\vartheta(x)=\sum_{p\leq x}\log p\geq\sum_{x^{\alpha}\sum_{x^{% \alpha}

As $\pi(x^{\alpha}),

 $\vartheta(x)>\alpha\pi(x)\log x-\alpha x^{\alpha}\log x,$

i.e.,

 $\frac{\vartheta(x)}{x}>\alpha\frac{\pi(x)\log x}{x}-\alpha\frac{\log x}{x^{1-% \alpha}}.$

This yields

 $\liminf_{x\to\infty}\frac{\vartheta(x)}{x}\geq\alpha\liminf_{x\to\infty}\frac{% \pi(x)\log x}{x}-\alpha\liminf_{x\to\infty}\frac{\log x}{x^{1-\alpha}}=\alpha% \liminf_{x\to\infty}\frac{\pi(x)\log x}{x}$

and

 $\limsup_{x\to\infty}\frac{\vartheta(x)}{x}\geq\alpha\limsup_{x\to\infty}\frac{% \pi(x)\log x}{x}-\alpha\limsup_{x\to\infty}\frac{\log x}{x^{1-\alpha}}=\alpha% \limsup_{x\to\infty}\frac{\pi(x)\log x}{x}.$

Since these are true for all $0<\alpha<1$, we obtain respectively

 $\liminf_{x\to\infty}\frac{\vartheta(x)}{x}\geq\liminf_{x\to\infty}\frac{\pi(x)% \log x}{x}$

and

 $\limsup_{x\to\infty}\frac{\vartheta(x)}{x}\geq\limsup_{x\to\infty}\frac{\pi(x)% \log x}{x}.$

## 40 Wiener’s tauberian theorem

Wiener [85, Chapter III].

Wiener-Ikehara [19]

Rudin [67, p. 229, Theorem 9.7]

We say that a function $s:(0,\infty)\to\mathbb{R}$ is slowly decreasing if

 $\liminf(s(\rho v)-s(v))\geq 0,\qquad v\to\infty,\quad\rho\to 1^{+}.$

Widder [84, p. 211, Theorem 10b]: Wiener’s tauberian theorem tells us that if $a\in L^{\infty}(0,\infty)$ and is slowly decreasing and if $g\in L^{1}(0,\infty)$ satisfies

 $\int_{0}^{\infty}t^{ix}g(t)dt\neq 0,\qquad x\in\mathbb{R},$

then

 $\lim_{x\to\infty}\frac{1}{x}\int_{0}^{\infty}g\left(\frac{t}{x}\right)a(t)dt=A% \int_{0}^{\infty}g(t)dt$

implies that

 $\lim_{v\to\infty}a(v)=A.$

It is straightforward to check the following by rearranging summation.

###### Lemma 7.

If $\sum_{n=1}^{\infty}a_{n}z^{n}$ has radius of convergence $\geq 1$, then for $|z|<1$,

 $\sum_{n=1}^{\infty}a_{n}\frac{z^{n}}{1-z^{n}}=\sum_{n=1}^{\infty}\left(\sum_{m% |n}a_{m}\right)z^{n}.$

Using Lemma 7 with $a_{n}=\Lambda(n)$ and $z=e^{-x}$ and applying (8), we get

 $\sum_{n=1}^{\infty}\Lambda(n)\frac{z^{n}}{1-z^{n}}=\sum_{n=1}^{\infty}\log(n)z% ^{n}.$ (11)

From (11), and Lemma 7 with $a_{n}=1$, we have

 $\sum_{n=1}^{\infty}(\Lambda(n)-1)\frac{e^{-nx}}{1-e^{-nx}}=\sum_{n=1}^{\infty}% (\log n-d(n))e^{-nx}.$

We follow Widder [84, p. 231, Theorem 16.6].

###### Theorem 8.

As $x\to 0^{+}$,

 $\sum_{n=1}^{\infty}(\log n-d(n))e^{-nx}=-\frac{2\gamma}{x}+O(x^{-1/2}).$
###### Proof.

Generally,

 $\displaystyle(1-z)\sum_{n=1}^{\infty}z^{n}\sum_{m=1}^{n}a_{m}$ $\displaystyle=(1-z)\sum_{m=1}^{\infty}a_{m}\sum_{n=m}^{\infty}z^{n}$ $\displaystyle=(1-z)\sum_{m=1}^{\infty}a_{m}\frac{z^{m}}{1-z}$ $\displaystyle=\sum_{m=1}^{\infty}a_{m}z^{m}.$

Using this with $a_{m}=\log m-d(m)$ and $z=e^{-x}$ gives

 $\displaystyle\sum_{n=1}^{\infty}(\log n-d(n))e^{-nx}$ $\displaystyle=(1-e^{-x})\sum_{n=1}^{\infty}e^{-nx}\left(\sum_{m=1}^{n}\log m-% \sum_{m=1}^{n}d(m)\right)$ $\displaystyle=(1-e^{-x})\sum_{n=1}^{\infty}e^{-nx}\left(\log(n!)-\sum_{m=1}^{n% }d(m)\right).$

Using

 $\log(n!)=n\log n-n+O(\log n)$

and

 $\sum_{m=1}^{n}d(m)=n\log n+(2\gamma-1)n+O(n^{1/2}),$

we get

 $\log(n!)-\sum_{m=1}^{n}d(m)=-2\gamma n+O(n^{1/2}).$

Therefore,

 $\sum_{n=1}^{\infty}(\log n-d(n))e^{-nx}=(1-e^{-x})\sum_{n=1}^{\infty}e^{-nx}(-% 2\gamma n+O(n^{1/2})).$

One proves that there is some $K$ such that for all $0\leq y<1$,

 $(1-y)\left(\log\frac{1}{y}\right)^{1/2}\sum_{n=1}^{\infty}n^{1/2}y^{n}\leq K,$

whence, with $y=e^{-x}$,

 $\sum_{n=1}^{\infty}n^{1/2}e^{-nx}\leq K\frac{x^{-1/2}}{1-e^{-x}}.$

Also,

 $\sum_{n=1}^{\infty}ne^{-nx}=\frac{e^{-x}}{(1-e^{-x})^{2}},$

and thus we have

 $\displaystyle\sum_{n=1}^{\infty}(\log n-d(n))e^{-nx}$ $\displaystyle=-2\gamma\frac{e^{-x}}{1-e^{-x}}+O(x^{-1/2})$ $\displaystyle=-2\gamma\frac{1}{e^{x}-1}+O(x^{-1/2}).$

But

 $\frac{1}{e^{x}-1}=\frac{1}{x}-\frac{1}{2}+O(x),$

so

 $\sum_{n=1}^{\infty}(\log n-d(n))e^{-nx}=-\frac{2\gamma}{x}+O(x^{-1/2}).$

Define

 $f(x)=\sum_{n=1}^{\infty}(\Lambda(n)-1)\frac{e^{-nx}}{1-e^{-nx}},$

and

 $h(x)=\sum_{n\leq x}\frac{\Lambda(n)-1}{n},$

and

 $g(t)=\frac{d}{dt}\left(\frac{te^{-t}}{1-e^{-t}}\right).$

First we show that $h$ is slowly decreasing.

###### Lemma 9.

$h(x)$ is slowly decreasing.

###### Proof.

Using

 $\sum_{1\leq n\leq x}\frac{1}{n}=\log x+\gamma+O(n^{-1}),\qquad x\to\infty,$

we have, for $0 and $\rho>1$,

 $\displaystyle h(\rho x)-h(x)$ $\displaystyle=\sum_{x $\displaystyle\geq-\sum_{x $\displaystyle=-\sum_{1\leq n\leq\rho x}\frac{1}{n}+\sum_{1\leq n\leq x}\frac{1% }{n}$ $\displaystyle=-\log(\rho x)+\log x+O((\rho x)^{-1})+O(x^{-1})$ $\displaystyle=-\log\rho+O((\rho x)^{-1})+O(x^{-1}).$

Hence as $x\to\infty$ and $\rho\to 1^{+}$,

 $h(\rho x)-h(x)\to 0,$

which shows that $h$ is slowly decreasing. ∎

The following is from Widder [84, pp. 231–232].

###### Lemma 10.

As $x\to\infty$,

 $\frac{1}{x}\int_{0}^{\infty}g\left(\frac{t}{x}\right)h(t)dt=2\gamma+O(x^{-1/2}).$
###### Proof.

Let $I(t)=0$ for $t<0$ and $I(t)=1$ for $t\geq 0$. Writing

 $h(x)=\sum_{n=1}^{\infty}I(x-n)\frac{\Lambda(n)-1}{n},$

we check that for $x>0$,

 $\displaystyle\int_{0}^{\infty}\frac{te^{-xt}}{1-e^{-xt}}dh(t)$ $\displaystyle=\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{te^{-xt}}{1-e^{-xt}}% \frac{\Lambda(n)-1}{n}d(I(t-n))$ $\displaystyle=\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{te^{-xt}}{1-e^{-xt}}% \frac{\Lambda(n)-1}{n}d\delta_{n}(t)$ $\displaystyle=\sum_{n=1}^{\infty}\frac{ne^{-nx}}{1-e^{-nx}}\frac{\Lambda(n)-1}% {n}$ $\displaystyle=f(x).$

On the other hand, integrating by parts,

 $\displaystyle f(x)$ $\displaystyle=\int_{0}^{\infty}\frac{te^{-xt}}{1-e^{-xt}}dh(t)$ $\displaystyle=\int_{0}^{\infty}\frac{1}{x}\frac{xte^{-xt}}{1-e^{xt}}dh(t)$ $\displaystyle=\int_{0}^{\infty}\frac{1}{x}\frac{xte^{-xt}}{1-e^{-xt}}dh(t)$ $\displaystyle=\int_{0}^{\infty}\frac{1}{x}\frac{te^{-t}}{1-e^{-t}}dh\left(% \frac{t}{x}\right)$ $\displaystyle=\frac{1}{x}\frac{te^{-t}}{1-e^{-t}}h\left(\frac{t}{x}\right)% \bigg{|}_{0}^{\infty}-\int_{0}^{\infty}\frac{1}{x}g(t)h\left(\frac{t}{x}\right% )dt$ $\displaystyle=-\int_{0}^{\infty}\frac{1}{x}g(t)h\left(\frac{t}{x}\right)dt$ $\displaystyle=-\int_{0}^{\infty}g(xt)h(t)dt.$

By Theorem 8, as $x\to 0^{+}$,

 $f(x)=-\frac{2\gamma}{x}+O(x^{-1/2}),$

i.e., as $x\to 0^{+}$,

 $\int_{0}^{\infty}g(xt)h(t)dt=\frac{2\gamma}{x}+O(x^{-1/2}).$

Thus, as $x\to\infty$,

 $\int_{0}^{\infty}g\left(\frac{t}{x}\right)h(t)dt=2\gamma x+O(x^{1/2}).$

The following is from Widder [84, p. 232].

###### Lemma 11.
 $\int_{0}^{\infty}t^{-ix}g(t)dt=\begin{cases}-1&x=0\\ ix\zeta(1-ix)\Gamma(1-ix)&x\neq 0.\end{cases}$
###### Proof.
 $\displaystyle\int_{0}^{\infty}t^{-ix}g(t)dt$ $\displaystyle=\int_{0}^{\infty}t^{-ix}\frac{d}{dt}\left(\frac{te^{-t}}{1-e^{-t% }}\right)dt$ $\displaystyle=\lim_{\delta\to 0}\int_{0}^{\infty}t^{-ix+\delta}\frac{d}{dt}% \left(\frac{te^{-t}}{1-e^{-t}}\right)dt$ $\displaystyle=\lim_{\delta\to 0}\left(t^{-ix+\delta}\frac{te^{-t}}{1-e^{-t}}% \bigg{|}_{0}^{\infty}+(ix-\delta)\int_{0}^{\infty}t^{-ix+\delta-1}\frac{te^{-t% }}{1-e^{-t}}dt\right)$ $\displaystyle=\lim_{\delta\to 0}(ix-\delta)\int_{0}^{\infty}t^{-ix+\delta-1}% \frac{te^{-t}}{1-e^{-t}}dt$ $\displaystyle=\lim_{\delta\to 0}(ix-\delta)\int_{0}^{\infty}\frac{t^{(-ix+% \delta+1)-1}e^{-t}}{1-e^{-t}}dt.$

Using

 $\int_{0}^{\infty}\frac{t^{s-1}}{e^{t}-1}dt=\zeta(s)\Gamma(s),\qquad\mathrm{Re}% \,(s)>1,$

this becomes

 $\int_{0}^{\infty}t^{-ix}g(t)dt=\lim_{\delta\to 0^{+}}(ix-\delta)\zeta(1+\delta% -ix)\Gamma(1+\delta-ix).$

If $x=0$, then using

 $\zeta(s)=\frac{1}{s-1}+\gamma+O(|s-1|),\qquad s\to 1,$

we get

 $\lim_{\delta\to 0^{+}}(-\delta)\zeta(1+\delta)\Gamma(1+\delta)=-1.$

If $x>0$, then

 $\lim_{\delta\to 0^{+}}(ix-\delta)\zeta(1+\delta-ix)\Gamma(1+\delta-ix)=ix\zeta% (1-ix)\Gamma(1-ix).$

By Wiener’s tauberian theorem, it follows that

 $\sum_{n=1}^{\infty}\frac{\Lambda(n)-1}{n}=-2\gamma.$
###### Lemma 12.
 $h(x)=\int_{\frac{1}{2}}^{x}\frac{d(\psi(t)-[t])}{t}.$
###### Proof.

Let $I(t)=0$ for $t<0$ and $I(t)=1$ for $t\geq 0$. Writing

 $\psi(x)=\sum_{n=1}^{\infty}I(x-n)\Lambda(n),\qquad[x]=\sum_{n=1}^{\infty}I(x-n),$

we have

 $\displaystyle\int_{\frac{1}{2}}^{x}\frac{d(\psi(t)-[t])}{t}$ $\displaystyle=\int_{\frac{1}{2}}^{x}\frac{1}{t}d\left(\sum_{n=1}^{\infty}I(t-n% )(\Lambda(n)-1)\right)$ $\displaystyle=\int_{\frac{1}{2}}^{x}\frac{1}{t}\sum_{n=1}^{\infty}(\Lambda(n)-% 1)d\delta_{n}(t)$ $\displaystyle=\sum_{1\leq n\leq x}\frac{\Lambda(n)-1}{n}$ $\displaystyle=h(x).$

Thus, we have established that

 $\int_{\frac{1}{2}}^{\infty}\frac{d(\psi(t)-[t])}{t}=-2\gamma.$

Hermite [42]

Hermite [43]

## 42 Gerhardt

Gerhardt [33, p. 196] refers to Lambert’s Architectonic.

Levi-Civita [57]

## 44 Franel

Franel [32] and [31]

The next theorem shows that the set of points on the unit circle that are singularities of $\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}$ is dense in the unit circle. Titchmarsh [82, pp. 160–161, §4.71].

###### Theorem 13.

For $|z|<1$, define

 $f(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}.$

Suppose that $p>0,q>1$ are relatively prime integers. As $r\to 1^{-}$,

 $(1-r)f(re^{2\pi i/q})\to\infty.$
###### Proof.

Set $z=re^{2\pi ip/q}$ and write

 $\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}=\sum_{n\equiv 0\pmod{q}}\frac{z^{n}}{% 1-z^{n}}+\sum_{n\not\equiv 0\pmod{q}}\frac{z^{n}}{1-z^{n}}.$

On the one hand,

 $\displaystyle(1-r)\sum_{n\equiv 0\pmod{q}}\frac{z^{n}}{1-z^{n}}$ $\displaystyle=(1-r)\sum_{m=1}^{\infty}\frac{z^{mq}}{1-z^{mq}}$ $\displaystyle=(1-r)\sum_{m=1}^{\infty}\frac{(re^{2\pi ip/q})^{mq}}{1-(re^{2\pi ip% /q})^{mq}}$ $\displaystyle=(1-r)\sum_{m=1}^{\infty}\frac{r^{mq}}{1-r^{mq}}$ $\displaystyle=\frac{1-r}{1-r^{q}}\sum_{m=1}^{\infty}\frac{r^{mq}}{1+r^{q}+% \cdots+r^{(m-1)q}}$ $\displaystyle=\frac{1}{1+r+\cdots+r^{q-1}}\sum_{m=1}^{\infty}\frac{r^{mq}}{1+r% ^{q}+\cdots+r^{(m-1)q}}$ $\displaystyle\geq\frac{1}{q}\sum_{m=1}^{\infty}\frac{r^{mq}}{m}$ $\displaystyle=-\frac{1}{q}\log(1-r^{q})$ $\displaystyle\to\infty$

as $r\to 1$.

On the other hand, for $n\not\equiv 0\pmod{q}$ we have

 $\displaystyle|1-z^{n}|^{2}$ $\displaystyle=|1-r^{n}e^{2\pi ipn/q}|^{2}$ $\displaystyle=(1-r^{n}e^{2\pi ipn/q})(1-r^{n}e^{-2\pi ipn/q})$ $\displaystyle=1-r^{n}(e^{2\pi ipn/q}+e^{-2\pi ipn/q})+r^{2n}$ $\displaystyle=1-2r^{n}\cos 2\pi pn/q+r^{2n}$ $\displaystyle=1-2r^{n}+4r^{n}\sin^{2}\frac{\pi pn}{q}+r^{2n}$ $\displaystyle=(1-r^{n})^{2}+4r^{n}\sin^{2}\frac{\pi pn}{q}.$

So far we have not used the hypothesis that $n\equiv 0\pmod{q}$. We use it to obtain

 $\sin\frac{\pi pn}{q}\geq\sin\frac{\pi}{q}.$

With this we have

 $|1-z^{n}|^{2}\geq 4r^{n}\sin^{2}\frac{\pi}{q},$

and therefore, as $r<1$,

 $\displaystyle(1-r)\left|\sum_{n\not\equiv 0\pmod{q}}\frac{z^{n}}{1-z^{n}}\right|$ $\displaystyle\leq(1-r)\sum_{n\not\equiv 0\pmod{q}}\frac{|z|^{n}}{|1-z^{n}|}$ $\displaystyle\leq(1-r)\sum_{n\not\equiv 0\pmod{q}}\frac{r^{n}}{2r^{n/2}\sin% \frac{\pi}{q}}$ $\displaystyle\leq\frac{1-r}{2\sin\frac{\pi}{q}}\sum_{n=0}^{\infty}r^{n/2}$ $\displaystyle=\frac{1-r}{2\sin\frac{\pi}{q}}\cdot\frac{1}{1-\sqrt{r}}$ $\displaystyle=\frac{1+\sqrt{r}}{2\sin\frac{\pi}{q}}$ $\displaystyle<\frac{1}{\sin\frac{\pi}{q}}.$

## 45 Wigert

The following result is proved by Wigert [86]. Our proof follows Titchmarsh [81, p. 163, Theorem 7.15]. Cf. Landau [55].

###### Theorem 14.

For $\lambda<\frac{1}{2}\pi$ and $N\geq 1$,

 $\sum_{n=1}^{\infty}d(n)e^{-nz}=\frac{\gamma}{z}-\frac{\log z}{z}+\frac{1}{4}-% \sum_{n=0}^{N-1}\frac{B_{2n+2}^{2}}{(2n+2)!(2n+2)}z^{2n+1}+O(|z|^{2N})$

as $z\to 0$ in any angle $|\arg z|\leq\lambda$.

###### Proof.

For $\sigma>1$, $s=\sigma+it$,

 $\zeta^{2}(s)=\sum_{n=1}^{\infty}\frac{d(n)}{n^{s}}.$

Using this, for $\mathrm{Re}\,z>0$ we have

 $\displaystyle\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}\Gamma(s)\zeta^{2}(s)% z^{-s}ds$ $\displaystyle=\sum_{n=1}^{\infty}d(n)\frac{1}{2\pi i}\int_{2-i\infty}^{2+i% \infty}\Gamma(s)(nz)^{-s}ds$ $\displaystyle=\sum_{n=1}^{\infty}d(n)e^{-nz}.$ (12)

Define $F(s)=\Gamma(s)\zeta^{s}(s)z^{-s}$. $F$ has poles at $1,0$, and the negative odd integers. (At each negative even integer, $\Gamma$ has a first order pole but $\zeta^{2}$ has a second order zero.) First we determine the residue of $F$ at $1$. We use the asymptotic formula

 $\zeta(s)=\frac{1}{s-1}+\gamma+O(|s-1|),\qquad s\to 1,$

the asymptotic formula

 $\Gamma(s)=1-\gamma(s-1)+O(|s-1|^{2}),\qquad s\to 1,$

and the asymptotic formula

 $z^{-s}=\frac{1}{z}-\frac{\log z}{z}(s-1)+O(|s-1|^{2}),\qquad s\to 1,$

to obtain

 $\displaystyle\Gamma(s)\zeta^{s}(s)z^{-s}$ $\displaystyle=(1-\gamma(s-1)+O(|s-1|^{2}))\cdot\left(\frac{1}{(s-1)^{2}}+\frac% {2\gamma}{s-1}+O(|s-1|^{2})\right)$ $\displaystyle\cdot\left(\frac{1}{z}-\frac{\log z}{z}(s-1)+O(|s-1|^{2})\right)$ $\displaystyle=\frac{1}{z(s-1)^{2}}-\frac{\gamma}{z(s-1)}+\frac{2\gamma}{z(s-1)% }-\frac{\log z}{z(s-1)}+O(1)$ $\displaystyle=\frac{1}{z(s-1)^{2}}+\frac{\gamma}{z(s-1)}-\frac{\log z}{z(s-1)}% +O(1).$

Hence the residue of $F$ at $1$ is

 $\frac{\gamma}{z}-\frac{\log z}{z}.$

Now we determine the residue of $F$ at $0$. The residue of $\Gamma$ at $0$ is $1$, and hence the residue of $F$ at $0$ is

 $1\cdot\zeta^{2}(0)\cdot z^{0}=\zeta^{2}(0)=\left(-\frac{1}{2}\right)^{2}=\frac% {1}{4}.$

Finally, for $n\geq 0$ we determine the residue of $F$ at $-(2n+1)$. The residue of $\Gamma$ at $-(2n+1)$ is $\frac{(-1)^{2n+1}}{(2n+1)!}$, hence the residue of $F$ at $-(2n+1)$ is

 $\frac{(-1)^{2n+1}}{(2n+1)!}\cdot\zeta^{2}(2n+1)\cdot z^{2n+1}=-\frac{B_{2n+2}^% {2}}{(2n+2)!(2n+2)}z^{2n+1}$

using

 $\zeta(-m)=-\frac{B_{m+1}}{m+1},\qquad m\geq 1.$

Let $M>0$, and let $C$ be the rectangular path starting at $2-iM$, then going to $2+iM$, then going to $-2N+iM$, then going to $-2N-iM$, and then ending at $2-iM$. By the residue theorem,

 $\int_{C}F(s)ds=2\pi i\left(\frac{\gamma}{z}-\frac{\log z}{z}+\frac{1}{4}+\sum_% {n=0}^{N-1}-\frac{B_{2n+2}^{2}}{(2n+2)!(2n+2)}z^{2n+1}\right).$ (13)

Denote the right-hand sideof (13) by $2\pi iR$. We have

 $\int_{C}F(s)ds=\int_{2-iM}^{2+iM}F(s)ds+\int_{2+iM}^{-2N+iM}F(s)ds+\int_{-2N+% iM}^{-2N-iM}F(s)ds+\int_{-2N-iM}^{2-iM}F(s)ds.$

We shall show that the second and fourth integrals tend to $0$ as $M\to\infty$. For $s=\sigma+it$ with $-2N\leq\sigma\leq 2$, Stirling’s formula [82, p. 151] tells us that

 $|\Gamma(s)|\sim\sqrt{2\pi}e^{-\frac{\pi}{2}|t|}|t|^{\sigma-\frac{1}{2}},\qquad% |t|\to\infty.$

As well [81, p. 95], there is some $K>0$ such that in the half-plane $\sigma\geq-2N$,

 $\zeta(s)=O(|t|^{K}).$

Also,

 $\displaystyle z^{-s}$ $\displaystyle=e^{-s\log z}$ $\displaystyle=e^{-(\sigma+it)(\log|z|+i\arg z)}$ $\displaystyle=e^{-\sigma\log|z|+t\arg z-i(\sigma\arg z+t\log|z|)},$

and so for $|\arg z|\leq\lambda$,

 $|z^{-s}|=e^{-\sigma\log|z|+t\arg z}\leq e^{-\sigma\log|z|+\lambda|t|}=|z|^{-% \sigma}e^{\lambda|t|}.$

Therefore

 $\left|\int_{2+iM}^{-2N+iM}F(s)ds\right|\leq(2+2N)\sup_{-2N\leq\sigma\leq 2}|F(% \sigma+iM)|=O(e^{-\frac{\pi}{2}M}M^{\sigma-\frac{1}{2}}M^{2K}|z|^{-\sigma}e^{% \lambda M}),$

and because $\lambda<\frac{\pi}{2}$ this tends to $0$ as $M\to\infty$. Likewise,

 $\left|\int_{-2N-iM}^{2-iM}F(s)ds\right|\to 0$

as $M\to\infty$. It follows that

 $\int_{2-i\infty}^{2+i\infty}F(s)ds+\int_{-2N+i\infty}^{-2N-i\infty}F(s)ds=2\pi iR.$

Hence,

 $\int_{2-i\infty}^{2+i\infty}F(s)ds=2\pi iR+\int_{-2N-i\infty}^{-2N+i\infty}F(s% )ds.$

We bound the integral on the right-hand side. We have

 $\int_{-2N-i\infty}^{-2N+i\infty}F(s)ds=\int_{\sigma=-2N,|t|\leq 1}F(s)ds+\int_% {\sigma=-2N,|t|>1}F(s)ds.$

The first integral satisfies

 $\left|\int_{\sigma=-2N,|t|\leq 1}F(s)ds\right|\leq\int_{\sigma=-2N,|t|\leq 1}|% \Gamma(s)\zeta^{2}(s)||z|^{-\sigma}e^{\lambda|t|}ds=|z|^{2N}\cdot O(1)=O(|z|^{% 2N}),$

because $\Gamma(s)\zeta^{2}(s)$ is continuous on the path of integration. The second integral satisfies

 $\displaystyle\left|\int_{\sigma=-2N,|t|>1}F(s)ds\right|$ $\displaystyle\leq\int_{\sigma=-2N,|t|>1}e^{-\frac{\pi}{2}|t|}|t|^{\sigma-\frac% {1}{2}}|t|^{K}|z|^{-\sigma}e^{\lambda|t|}ds$ $\displaystyle=|z|^{2N}\int_{\sigma=-2N,|t|>1}e^{-\frac{\pi}{2}|t|}|t|^{-2N-% \frac{1}{2}}|t|^{K}e^{\lambda|t|}dt$ $\displaystyle=|z|^{2N}\cdot O(1)$ $\displaystyle=O(|z|^{2N}),$

because $\lambda<\frac{\pi}{2}$. This establishes

 $\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}F(s)ds=R+O(|z|^{2N}).$

Using (12) and (13), this becomes

 $\sum_{n=1}^{\infty}d(n)e^{-nz}=\frac{\gamma}{z}-\frac{\log z}{z}+\frac{1}{4}-% \sum_{n=0}^{N-1}\frac{B_{2n+2}^{2}}{(2n+2)!(2n+2)}z^{2n+1}+O(|z|^{-2N}),$

completing the proof. ∎

For example, as $B_{2}=\frac{1}{6},B_{4}=-\frac{1}{30},B_{6}=\frac{1}{42}$, the above theorem tells us that

 $\sum_{n=1}^{\infty}d(n)e^{-nz}=\frac{\gamma}{z}-\frac{\log z}{z}+\frac{1}{4}-% \frac{z}{144}-\frac{z^{3}}{86400}-\frac{z^{5}}{7620480}+O(|z|^{6}).$

Steffensen [75]

Szegő [79]

## 48 Pólya and Szegő

Pólya and Szegő [64]

## 49 Partition function

Let

 $F(x)=\sum_{n=0}^{\infty}p(n)x^{n}=\prod_{n=1}^{\infty}\frac{1}{1-x^{n}}.$

Taking the logarithm,

 $\log F(x)=\sum_{n=1}^{\infty}\log\frac{1}{1-x^{n}}=-\sum_{n=1}^{\infty}\log(1-% x^{n})=-\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}-\frac{(x^{n})^{m}}{m},$

and switching the order of summation gives

 $\log F(x)=\sum_{m=1}^{\infty}\frac{1}{m}\sum_{n=1}^{\infty}(x^{m})^{n}=\sum_{m% =1}^{\infty}\frac{1}{m}\frac{x^{m}}{1-x^{m}}.$

On the one hand, for $0 we have $mx^{m-1}(1-x)<1-x^{m}$ and using this,

 $\sum_{m=1}^{\infty}\frac{1}{m}\frac{x^{m}}{1-x^{m}}<\sum_{m=1}^{\infty}\frac{1% }{m}\frac{x^{m}}{mx^{m-1}(1-x)}=\frac{x}{1-x}\sum_{m=1}^{\infty}\frac{1}{m^{2}% }=\frac{\pi^{2}}{6}\frac{x}{1-x}.$

On the other hand, for $-1 we have $1-x^{m}, and using this, for $0 we have

 $\sum_{m=1}^{\infty}\frac{1}{m}\frac{x^{m}}{1-x^{m}}>\sum_{m=1}^{\infty}\frac{1% }{m}\frac{x^{m}}{m(1-x)}=\frac{1}{1-x}\sum_{m=1}^{\infty}\frac{x^{m}}{m^{2}}.$

Thus, for $0,

 $\sum_{m=1}^{\infty}\frac{x^{m}}{m^{2}}<(1-x)\log F(x)<\frac{\pi^{2}}{6}x.$

Taking $x\to 1^{-}$ gives

 $\frac{\pi^{2}}{6}\leq\lim_{x\to 1^{-}}(1-x)\log F(x)\leq\frac{\pi^{2}}{6},$

i.e.,

 $\log F(x)\sim\frac{\pi^{2}}{6}\frac{1}{1-x},\qquad x\to 1^{-}.$

See Stein and Shakarchi [76, p. 311].

Hansen [39]

Kiseljak [45]

## 52 Unsorted

In 1892, in volume VII, no. 23, p. 296 of the weekly Naturwissenschaftliche Rundschau, it is stated that for the year 1893, one of the six prize questions for the Belgian Academy of Sciences in Brussels is to determine the sum of the Lambert series

 $\frac{x}{1-x}+\frac{x^{2}}{1-x^{2}}+\frac{x^{3}}{1-x^{3}}+\cdots,$

or if one cannot do this, to find a differential equation that determines the function.

Gram [35] on distribution of prime numbers.

Hardy [41]

Bohr and Cramer [1, p. 820]

Flajolet, Gourdon and Dumas [30]

## References

• [1] H. Bohr and H. Cramér (1923–1927) Die neure Entwicklung der analytischen Zahlentheorie. In Encyklopädie der Mathematischen Wissenschaften mit Einschluss ihrer Anwendungen, Band II, 3. Teil, 2. Hälfte, H. Burkhardt, W. Wirtinger, R. Fricke, and E. Hilb (Eds.), pp. 722–849. Cited by: §52.
• [2] K. Bopp (1916) Johann Heinrich Lamberts Monatsbuch mit den zugehörigen Kommentaren, sowie mit einem Vorwort über den Stand der Lambertforschung. Abhandlungen der Königlich Bayerischen Akademie der Wissenschaften. Mathematisch-physikalische Klasse 27, pp. 1–84. Note: 6. Abhandlung Cited by: §3.
• [3] P. B. Borwein (1992) On the irrationality of certain series. Math. Proc. Camb. Phil. Soc. 112, pp. 141–146. Cited by: §10.
• [4] V. Bouniakowsky (1861) Recherches sur quelques fonctions numériques. Mémoires de l’Académie impériale des sciences de St.-Pétersbourg, VIIe série 4 (2), pp. 1–35. Cited by: §24.
• [5] N. Bourbaki (2004) Elements of mathematics. Functions of a real variable: Elementary theory. Springer. Note: Translated from the French by Philip Spain Cited by: §18.
• [6] T. J. I’A. Bromwich (1959) An introduction to the theory of infinite series. second edition, Macmillan, London. Cited by: §18, §32.
• [7] M. Bullynck (2010) Factor tables 1657–1817, with notes on the birth of number theory. Revue d’histoire des mathématiques 16 (2), pp. 133–216. Cited by: §3, §3, §3.
• [8] H. Burhenne (1852) Ueber das Gesetz der Primzahlen. Archiv der Mathematik und Physik 19, pp. 442–449. Cited by: §15.
• [9] E. Catalan (1842) Sur la sommation de quelques séries. Journal de Mathématiques Pures et Appliquées 7, pp. 1–12. Cited by: §26.
• [10] E. Catalan (1873) Recherches sur quelques produits indéfinis. Mémoires de l’Académie Royale des Sciences, des Lettres et des Beaux-Arts de Belgique 40, pp. 1–127. Cited by: §26.
• [11] E. Catalan (1886) Mélanges mathématiques. Mémoires de la Société Royale des Sciences de Liège, deuxième série 13, pp. 1–404. Cited by: §26.
• [12] E. Catalan (1888) Mélanges mathématiques. Mémoires de la Société Royale des Sciences de Liège, deuxième série 14, pp. 1–275. Cited by: §26.
• [13] A. Cauchy (1843) Mémoire sur l’application du calcul des résidus au développement des produits composés d’un nombre infini de facteurs. Comptes rendus hebdomadaires des séances de l’Académie des sciences 17, pp. 572–581. Note: Oeuvres complètes, série 1, tome 8, pp. 55–64 Cited by: §14.
• [14] A. Cauchy (1843) Sur la réduction des rapports de factorielles réciproques aux fonctions elliptiques. Comptes rendus hebdomadaires des séances de l’Académie des sciences 17, pp. 825–837. Note: Oeuvres complètes, série 1, tome 8, pp. 97–110 Cited by: §14.
• [15] E. Cesáro (1886) Sur les nombres de Bernoulli et d’Euler. Nouvelles annales de mathématiques, troisième série 5, pp. 305–327. Cited by: §32.
• [16] E. Cesáro (1888) Sur les transformations de la série de Lambert. Nouvelles annales de mathématiques, troisième série 7, pp. 374–382. Cited by: §32.
• [17] E. Cesáro (1893) La serie di Lambert in aritmetica assintotica. Rendiconto delle adunanze e de’ lavori dell’ Accademia Napolitana delle Scienze 7, pp. 197–204. Cited by: §32.
• [18] E. Cesáro (1894) Corso di analisi algebrica con introduzione al calcolo infinitesimale. Fratelli Bocca Editori, Turin. Cited by: §32.
• [19] K. Chandrasekharan (1968) Introduction to analytic number theory. Die Grundlehren der mathematischen Wissenschaften, Vol. 148, Springer. Cited by: §40.
• [20] K. Chandrasekharan (1985) Elliptic functions. Die Grundlehren der mathematischen Wissenschaften, Vol. 281, Springer. Cited by: §12.
• [21] Th. Clausen (1828) Beitrag zur Theorie der Reihen. J. Reine Angew. Math. 3, pp. 92–95. Cited by: §9.
• [22] H. Cohen (2007) Number theory, volume II: analytic and modern tools. Graduate Texts in Mathematics, Vol. 240, Springer. Cited by: §20.
• [23] M. Curtze (1867-1868) Notes diverses sur la série de Lambert et la loi des nombres premiers. Annali di Matematica Pura ed Applicata 1 (1), pp. 285–292. Cited by: §21.
• [24] C. de la Vallée Poussin (1896) Sur la série de Lambert. Annales de la Société Scientifique de Bruxelles 20, pp. 56–62. Cited by: §33.
• [25] G. L. Dirichlet (1838) Über die Bestimmung asymptotischer Gesetze in der Zahlentheorie. Bericht über die Verhandlangen der Königlich Preussischen Akademie der Wissenschaften, pp. 13–15. Note: Werke, Band I, pp. 351–356 Cited by: §13.
• [26] S. Egger né Endres and F. Steiner (2011) A new proof of the Voronoï summation formula. J. Phys. A 44 (22), pp. 225302. Cited by: §20.
• [27] G. Eisenstein (1844) Transformations remarquables de quelques séries. J. Reine Angew. Math. 27, pp. 193–197. Note: Mathematische Werke, Band I, pp. 35–44 Cited by: §10.
• [28] L. Euler (1738) Methodus generalis summandi progressiones. Commentarii Academiae scientiarum Imperialis Petropolitanae 6, pp. 68–97. Note: E25, Opera omnia I.14, pp. 42–72 Cited by: §2.
• [29] H. Fischer (1994) Dirichlet’s contributions to mathematical probability theory. Historia Math. 21, pp. 39–63. Cited by: §13.
• [30] P. Flajolet, X. Gourdon, and P. Dumas (1995) Mellin transforms and asymptotics: harmonic sums. Theoret. Comput. Sci. 144, pp. 3–58. Cited by: §52.
• [31] J. Franel (1899) Sur la théorie des séries. Math. Ann. 52, pp. 529–549. Cited by: §44.
• [32] J. Franel (1899) Sur une formule utile dans la détermination de certaines valeurs asymptotiques. Math. Ann. 51, pp. 369–387. Cited by: §44.
• [33] C. I. Gerhardt (1877) Geschichte der Wissenschaften in Deutschland. Neuere Zeit. Siebenzehnter Band. Geschichte der Mathematik. R. Oldenbourg, München. Cited by: §42.
• [34] J. W. L. Glaisher (1885) On the square of the series in which the coefficients are the sums of the divisors of the exponents. Messenger of Mathematics 14, pp. 156–163. Cited by: §28.
• [35] J. P. Gram (1881–1886) Undersøgelser angaaende Mængden af Primtal under en given Grænse. Det Kongelige Danske Videnskabernes Selskabs Skrifter, 6. Række, Naturvidenskabelig og Mathematisk Afdeling 2, pp. 183–308. Cited by: §52.
• [36] A. P. Guinand (1944) Functional equations and self-reciprocal functions connected with Lambert series. Q. J. Math. 15, pp. 11–23. Cited by: §20.
• [37] S. Günther (1876) Ziele und Resultate der neueren mathematisch-historischen Forschung. Eduard Besold, Erlangen. Cited by: §29.
• [38] S. Günther (1881) Die Lehre von den gewöhnlichen und verallgemeinerten Hyperbelfunktionen. Louis Nebert, Halle a. S.. Cited by: §29.
• [39] C. Hansen (1901) Note sur la sommation de la série de Lambert. Mathematische Annalen 54, pp. 604–607. Cited by: §50.
• [40] G. H. Hardy and E. M. Wright (1979) An introduction to the theory of numbers. Fifth edition, Oxford University Press. Cited by: §37, §37, §37, §37, §37, §37.
• [41] G. H. Hardy (1991) Divergent series. second edition, AMS Chelsea Publishing, Providence, RI. Cited by: §52.
• [42] C. Hermite (1884) Sur quelques conséquences arithmétiques des formules de la théorie des fonctions elliptiques. Bulletin de l’Académie impériale des sciences de St.-Pétersbourg 29, pp. 325–352. Note: Œuvres, tome IV, pp. 138–168 Cited by: §41.
• [43] C. Hermite (1886) Sur les valeurs asymptotiques de quelques fonctions numériques. J. Reine Angew. Math. 99, pp. 324–328. Note: Œuvres, tome IV, pp. 209–214 Cited by: §41.
• [44] C. G. J. Jacobi (1829) Fundamenta nova theoriae functionum ellipticarum. Sumtibus Fratrum Borntraeger, Königsberg. Cited by: §12.
• [45] M. Kiseljak (1917) Über Anzahlen und Summen von Teilern. Monatshefte für Mathematik und Physik 28 (1), pp. 133–166. Cited by: §51.
• [46] G. S. Klügel, C. B. Mollweide, and J. A. Grunert (Eds.) (1831) Mathematisches Wörterbuch oder Erklärung der Begriffe, Lehrsätze, Aufgaben und Methoden der Mathematik mit den nöthigen Beweisen und literarischen Nachrichten begleitet in alphabetischer Ordnung. Erste Abtheilung. Fünfter Theil. Erster Band. I und II. E. B. Schwickert, Leipzig. Cited by: §7.
• [47] J. C. Kluyver (1919) On Lambert’s series. Koninklijke Akademie van Wetenschappen te Amsterdam, Proceedings 22 (4), pp. 323–330. Cited by: §20.
• [48] J. C. Kluyver (1922) On analytic functions defined by certain Lambert series. Koninklijke Akademie van Wetenschappen te Amsterdam, Proceedings 23 (8), pp. 1226–1233. Cited by: §20.
• [49] K. Knopp (1913) Über Lambertsche Reihen. J. Reine Angew. Math. 142, pp. 283–315. Cited by: §36.
• [50] W. L. Krafft (1794) Essai sur les nombres premiers. Nova Acta Academiae Scientiarum Imperialis Petropolitanae 12, pp. 217–245. Cited by: §4.
• [51] S. F. Lacroix (1819) Traité du calcul différentiel et du calcul intégral, tome troisième. second edition, Veuve Courcier, Paris. Cited by: §6.
• [52] E. Laguerre (1872–1873) Sur quelques théorèmes d’arithmétique. Bulletin de la Société Mathématique de France 1, pp. 77–81. Cited by: §22.
• [53] J. H. Lambert (1771) Anlage zur Architectonic, oder Theorie des Einfachen und des Ersten in der philosophischen und mathematischen Erkenntniß, 2. Band. Johann Friedrich Hartknoch, Riga. Cited by: §3.
• [54] E. Landau (1899) Sur la série des inverses des nombres de Fibonacci. Bulletin de la Société Mathématique de France 27, pp. 298–300. Cited by: §35.
• [55] E. Landau (1918) Über die Wigertsche asymptotische Funktionalgleichung für die Lambertsche Reihe. Archiv der Mathematik und Physik, 3. Reihe 27, pp. 144–146. Note: Collected Works, volume 7, pp. 135–137 Cited by: §45.
• [56] V. Lebesgue (1853) Démonstration d’une formule d’Euler, sur les diviseurs d’un nombre. Nouvelles annales de mathématiques 12, pp. 232–235. Cited by: §23.
• [57] T. Levi-Civita (1895) Di una espressione analitica atta a rappresentare il numero dei numeri primi compresi in un determinato intervallo. Atti della Reale Accademia dei Lincei, serie quinta. Rendiconti: Classe di scienze fisiche, matematiche e naturali 4, pp. 303–309. Note: Opere matematiche, volume primo, pp. 153–158 Cited by: §43.
• [58] P. Lindqvist and J. Peetre (1997) On the remainder in a series of Mertens. Expo. Math. 15 (5), pp. 467–478. Cited by: §38.
• [59] W. Lorey (1928) Johann Heinrich Lambert. Sitzungsberichte der Berliner Mathematischen Gesellschaft 18, pp. 2–27. Cited by: §3.
• [60] F. Löwenhaupt (Ed.) (1943) Johann Heinrich Lambert: Leistung und Leben. Braun & Co., Mühlhausen. Cited by: §3.
• [61] F. Mertens (1874) Ein Beitrag zur analytischen Zahlentheorie. J. Reine Angew. Math. 78 (1), pp. 46–62. Cited by: §38, §38.
• [62] A. F. Möbius (1832) Über eine besondere Art von Umkehrung der Reihen. J. Reine Angew. Math. 9 (2), pp. 105–123. Cited by: §11.
• [63] S. Pincherle (1882) Sopra alcuni sviluppi in serie per funzioni analitiche. Memorie della Accademia delle Scienze dell’Istituto di Bologna, serie quarta 3, pp. 149–180. Note: Opere scelte, vol. 1, pp. 64–91 Cited by: §27.
• [64] G. Pólya and G. Szegő (1976) Problems and theorems in analysis, volume II. Die Grundlehren der mathematischen Wissenschaften, Vol. 216, Springer. Note: Translated from the German by C. E. Billigheimer Cited by: §48.
• [65] F. Rogel (1890) Darstellung der harmonischen Reihen durch Factorenfolgen. Archiv der Mathematik und Physik (2) 9, pp. 297–319. Cited by: §31.
• [66] F. Rogel (1891) Darstellungen zalentheoretischer Functionen durch trigonometrische Reihen. Archiv der Mathematik und Physik (2) 10, pp. 62–83. Cited by: §31.
• [67] W. Rudin (1991) Functional analysis. second edition, International Series in Pure and Applied Mathematics, McGraw-Hill. Cited by: §40.
• [68] H. F. Scherk (1832) Bemerkungen über die Lambertsche Reihe $\frac{x}{1-x}+\frac{x^{2}}{1-x^{2}}+\frac{x^{3}}{1-x^{3}}+\frac{x^{4}}{1-x^{4}% }+\mathrm{etc}$.. J. Reine Angew. Math. 9, pp. 162–168. Cited by: §9.
• [69] O. Schlömilch (1861) Ueber die Lambert’sche Reihe. Zeitschrift für Mathematik und Physik 6, pp. 407–415. Cited by: §19.
• [70] O. Schlömilch (1863) Extrait d’une Lettre adressée à M. Liouville par M. Schlömilch. Journal de Mathématiques Pures et Appliquées (2) 8, pp. 99–101. Cited by: §19.
• [71] O. Schlömilch (1874) Compendium der höheren Analysis, zweiter Band. second edition, Friedrich Vieweg und Sohn, Braunschweig. Cited by: §19.
• [72] F. Servois (1814) Essai sur un nouveau mode d’exposition des principes du calcul différentiel. P. Blachier-Belle, Nismes. Cited by: §5.
• [73] F. Servois (1814-1815) Réflexions sur les divers systèmes d’exposition des principes du calcul différentiel, et, en particulier, sur la doctrine des infiniment petits. Annales de Mathématiques pures et appliquées 5, pp. 141–170. Cited by: §5.
• [74] L. A. Sohncke (1850) Sammlung von Aufgaben aus der Differential- und Integralrechnung. H. W. Schmidt, Halle. Cited by: §15.
• [75] J. F. Steffensen (1914) Über Potenzreihen, im besonderen solche, deren Koeffizienten zahlentheoretische Funktionen sind. Rendiconti del Circolo Matematico di Palermo 38 (1), pp. 376–386. Cited by: §46.
• [76] E. M. Stein and R. Shakarchi (2003) Complex analysis. Princeton Lectures in Analysis, Vol. II, Princeton University Press. Cited by: §49.
• [77] M. Stern (1840) Beiträge zur Combinationslehre und deren Anwendung auf die Theorie der Zahlen. J. Reine Angew. Math. 21, pp. 177–192. Cited by: §8.
• [78] T.-J. Stieltjes (1886) Recherches sur quelques séries semi-convergentes. Annales scientifiques de l’École Normale Supérieure, Sér. 3 3, pp. 201–258. Note: Œuvres complètes, tome II, pp. 2–58 Cited by: §30.
• [79] G. Szegő (1920) Über Potenzreihen, deren Koeffizienten zahlentheoretische Funktionen sind. Mathematische Zeitschrift 8, pp. 36–51. Cited by: §47.
• [80] P. L. Tchebychef (1851) Note sur différentes séries. Journal de Mathématiques Pures et Appliquées 16, pp. 337–346. Note: Œuvres, tome I, pp. 99–108 Cited by: §25.
• [81] E. C. Titchmarsh (1986) The theory of the Riemann zeta-function. second edition, Clarendon Press, Oxford. Cited by: §45, §45.
• [82] E. C. Titchmarsh (2002) The theory of functions. second edition, Oxford University Press. Cited by: §44, §45.
• [83] G. Torelli (1901) Sulla totalità dei numeri primi fino ad un limite assegnato. Atti dell’Accademia delle scienze fisiche e matematiche. Sezione della Società Reale di Napoli (2) 11 (1), pp. 1–222. Cited by: §34.
• [84] D. V. Widder (1946) The Laplace transform. Princeton Mathematical Series, Vol. 6, Princeton University Press. Cited by: §40, §40, §40, §40.
• [85] N. Wiener (1933) The Fourier integral and certain of its applications. Cambridge University Press. Cited by: §40.
• [86] S. Wigert (1916) Sur la série de Lambert et son application à la théorie des nombres. Acta Math. 41, pp. 197–218. Cited by: §45.
• [87] A. P. Youschkevitch (1979) Lambert et Léonard Euler. In Colloque international et interdisciplinaire Jean-Henri Lambert, Mulhouse, 26–30 septembre 1977, R. Oberle, A. Thill, and P. Levassort (Eds.), pp. 211–224. Cited by: §3.
• [88] G. Zehfuss (1858) Mathematische Miscellen. Zeitschrift für Mathematik und Physik 3, pp. 247–249. Cited by: §16.
• [89] C. Zhang (2011) On the modular behaviour of the infinite product $(1-x)(1-xq)(1-xq^{2})(1-xq^{3})\dots$. C. R. Math. Acad. Sci. Paris 349 (13-14), pp. 725–730. Cited by: §30.
• [90] W. Zudilin (2002) Remarks on irrationality of $q$-harmonic series. Manuscripta Math. 107 (4), pp. 463–477. Cited by: §10.