The Voronoi summation formula

1 Mellin transform

The Mellin transform of $f:(0,\mathrm{\infty })\to ℂ$ is defined by

$$ℳ(f)(s)={\int }_0^{\mathrm{\infty }}{x}^{s-1}f(x)𝑑x.$$

For example, $s\mapsto \mathrm{\Gamma }(s)$ is the Mellin transform of $x\mapsto e^{-x}$.

Suppose that $f$ continuous on $(0,\mathrm{\infty })$, that there is some $\alpha \in ℝ$ such that $f(x)=O(x^{-\alpha })$ as $x\to 0$, and that for any $n\ge 1$, $\frac{f(x)}{x^n}\to 0$ as $x\to \mathrm{\infty }$. Then [4, p. 107, Proposition 9.7.7] $ℳ(f)(s)$ is holomorphic on $\mathrm{\Re }(s)>\alpha$, and for $\sigma >\alpha$ and $x>0$,

$$f(x)=\frac{1}{2\pi i}{\int }_{\mathrm{\Re }(s)=\sigma }{x}^{-s}ℳ(f)(s)𝑑s.$$

(The Mellin inversion formula.)

2 Generalized Poisson summation formula

Cohen [4, pp. 177–182, §10.2.5] presents a “generalized Poisson summation formula” which yields both the Poisson summation formula and the Voronoi summation formula.

We denote by $𝒮(ℝ)$ the Fréchet space of Schwartz functions $ℝ\to ℂ$.

Theorem 1.

Let $a$ be arithmetic function and define

$$L(a,s)=\sum _{n=1}^{\mathrm{\infty }}a(n)n^{-s},\mathrm{\Re }(s)>1.$$

Suppose that $L(a,s)$ has an analytic continuation to $ℂ$ whose only possible pole is at $s=1$. Suppose also that there are $A,a_1,\mathrm{\dots },a_g>0$ such that for

$$\gamma (s)=A^s\prod _{j=1}^g\mathrm{\Gamma }(a_{j}s),$$

$L(a,s)$ satisfies the functional equation

$$\gamma (s)L(a,s)=\gamma (1-s)L(a,1-s).$$

Let $f\in 𝒮(ℝ)$ and define for $x>0$,

$$K(x)=\frac{1}{2\pi i}{\int }_{\mathrm{\Re }(s)=\frac{3}{2}}\frac{\gamma (s)}{\gamma (1-s)}{x}^{-s}𝑑s,g(x)={\int }_0^{\mathrm{\infty }}f(y)K(xy)𝑑y.$$

Then,

$$\sum _{n=1}^{\mathrm{\infty }}a(n)f(n)=f(0)L(a,0)+{\mathrm{Res}}_{s=1}ℳ(f)(s)L(a,s)+\sum _{n=1}^{\mathrm{\infty }}a(n)g(n).$$

Proof.

Since $f$ is a Schwartz function, $ℳ(f)$ is holomorphic on $\mathrm{\Re }(s)>0$. Futhermore, for $\mathrm{\Re }(s)>0$, integrating by parts,

$$ℳ(f)(s)={\int }_0^{\mathrm{\infty }}{x}^{s-1}f(x)𝑑x={f(x)\frac{x^s}{s}|}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}{f}^{\prime }(x)\frac{x^s}{s}𝑑x=-\frac{1}{s}ℳ(f^{\prime })(s+1).$$

It follows that $ℳ(f)$ has an analytic continuation to $ℂ$ possibly with poles at $0,-1,-2,-3,\mathrm{\dots }$. Write $F=ℳ(f)$. By the Mellin inversion formula we get

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}a(n)f(n)$ $={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}a(n){\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}{n}^{-s}F(s)𝑑s$

$={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}F(s){\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{a}_n{n}^{-s}ds$

$={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}F(s)L(a,s)𝑑s.$

The only possible pole of $L(a,s)$ is at $s=1$. From

$$ℳ(f)(s)=-\frac{1}{s}ℳ(f^{\prime })(s+1),$$

the only possible pole of $F(s)$ in the half-plane $\mathrm{\Re }(s)>-1$ is at $s=0$, and the residue of $F(s)L(a,s)$ at $s=0$ is

$$-ℳ(f^{\prime })(1)=-{\int }_0^{\mathrm{\infty }}{f}^{\prime }(x)𝑑x=-(f(\mathrm{\infty })-f(0))=f(0),$$

so the residue of $F(s)L(a,s)$ at $s=0$ is

$$f(0)L(a,0).$$

Therefore, by the residue theorem, taking as given that $F(s)L(a,s)\to 0$ uniformly in $-\frac{1}{2}\le \mathrm{\Re }(s)\le \frac{3}{2}$ as $|\mathrm{\Im }(s)|\to \mathrm{\infty }$,

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}a(n)f(n)$ $=f(0)L(a,0)+{\mathrm{Res}}_{s=1}F(s)L(a,s)+{\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=-\frac{1}{2}}}F(s)L(a,s)𝑑s.$

Define

$$G(s)=F(1-s)\frac{\gamma (s)}{\gamma (1-s)}.$$

Using the functional equation for $L(a,s)$,

${\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=-\frac{1}{2}}}F(s)L(a,s)𝑑s$ $={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=-\frac{1}{2}}}F(s){\displaystyle \frac{\gamma (1-s)}{\gamma (s)}}L(a,1-s)𝑑s$

$={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}F(1-s){\displaystyle \frac{\gamma (s)}{\gamma (1-s)}}L(a,s)𝑑s$

$={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}G(s)L(a,s)𝑑s.$

Furthermore, define

$$J(x)=\frac{1}{2\pi i}{\int }_{\mathrm{\Re }(s)=\frac{3}{2}}\frac{1}{1-s}\frac{\gamma (s)}{\gamma (1-s)}{x}^{1-s}𝑑s,$$

which satisfies

$$J^{\prime }(x)=K(x).$$

We have

${\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}{x}^{-s}G(s)𝑑s$ $={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}{x}^{-s}F(1-s){\displaystyle \frac{\gamma (s)}{\gamma (1-s)}}𝑑s$

$={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}{x}^{-s}\left(-{\displaystyle \frac{1}{1-s}}ℳ(f^{\prime })(2-s)\right){\displaystyle \frac{\gamma (s)}{\gamma (1-s)}}𝑑s$

$={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}{x}^{-s}\left(-{\displaystyle \frac{1}{1-s}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{y}^{1-s}{f}^{\prime }(y)𝑑y\right){\displaystyle \frac{\gamma (s)}{\gamma (1-s)}}𝑑s$

$=-{\displaystyle \frac{1}{x}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{f}^{\prime }(y){\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}{\displaystyle \frac{1}{1-s}}{\displaystyle \frac{\gamma (s)}{\gamma (1-s)}}{(xy)}^{1-s}𝑑s$

$=-{\displaystyle \frac{1}{x}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{f}^{\prime }(y)J(xy)𝑑y$

$=-{{\displaystyle \frac{1}{x}}f(y)J(xy)|}_0^{\mathrm{\infty }}+{\displaystyle \frac{1}{x}}{\displaystyle {\int }_0^{\mathrm{\infty }}}f(y)J^{\prime }(xy)x𝑑y$

$=0+{\displaystyle {\int }_0^{\mathrm{\infty }}}f(y)J^{\prime }(xy)𝑑y$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}f(y)K(xy)𝑑y$

$=g(x).$

Therefore,

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}a(n)g(n)$ $={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}a(n){\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}{n}^{-s}G(s)𝑑s$

$={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}G(s){\displaystyle \sum _{n=1}^{\mathrm{\infty }}}a(n)n^{-s}ds$

$={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{\mathrm{\Re }(s)=\frac{3}{2}}}G(s)L(a,s)𝑑s.$

Thus we have

$$\sum _{n=1}^{\mathrm{\infty }}a(n)f(n)=f(0)L(a,0)+{\mathrm{Res}}_{s=1}F(s)L(a,s)+\sum _{n=1}^{\mathrm{\infty }}a(n)g(n)$$

∎

Take $a(n)=1$ for all $n$. Then,

$$L(a,s)=\sum _{n=1}^{\mathrm{\infty }}{n}^{-s}=\zeta (s).$$

The Riemann zeta function satisfies the functional equation

$${\pi }^{-s/2}\mathrm{\Gamma }\left(\frac{s}{2}\right)\zeta (s)={\pi }^{-(1-s)/2}\mathrm{\Gamma }\left(\frac{1-s}{2}\right)\zeta (1-s).$$

So with

$$\gamma (s)={\pi }^{-s/2}\mathrm{\Gamma }\left(\frac{s}{2}\right),$$

we have

$$\gamma (s)\zeta (s)=\gamma (1-s)\zeta (1-s).$$

Using

$$\mathrm{\Gamma }(1-z)\mathrm{\Gamma }(z)=\frac{\pi }{\sin \pi z}$$

and

$$\mathrm{\Gamma }(z)\mathrm{\Gamma }\left(z+\frac{1}{2}\right)=2^{1-2z}\sqrt{\pi }\mathrm{\Gamma }(2z),$$

we have

$\mathrm{\Gamma }\left({\displaystyle \frac{1-s}{2}}\right)$ $=\mathrm{\Gamma }\left(1-{\displaystyle \frac{s+1}{2}}\right)$

$={\displaystyle \frac{\pi }{\sin \frac{\pi (s+1)}{2}\mathrm{\Gamma }\left(\frac{s+1}{2}\right)}}$

$={\displaystyle \frac{\pi }{\sin \frac{\pi (s+1)}{2}\mathrm{\Gamma }\left(\frac{s}{2}+\frac{1}{2}\right)}}$

$={\displaystyle \frac{\pi \mathrm{\Gamma }\left(\frac{s}{2}\right)}{\sin \frac{\pi (s+1)}{2}{2}^{1-s}\sqrt{\pi }\mathrm{\Gamma }(s)}},$

and so

${\displaystyle \frac{{\pi }^{-s/2}\mathrm{\Gamma }\left(\frac{s}{2}\right)}{{\pi }^{-(1-s)/2}\mathrm{\Gamma }\left(\frac{1-s}{2}\right)}}$ $={\pi }^{-s+\frac{1}{2}}\mathrm{\Gamma }\left({\displaystyle \frac{s}{2}}\right)\cdot {\displaystyle \frac{\sin \frac{\pi (s+1)}{2}{2}^{1-s}\sqrt{\pi }\mathrm{\Gamma }(s)}{\pi \mathrm{\Gamma }\left(\frac{s}{2}\right)}}$

$=\sin {\displaystyle \frac{\pi (s+1)}{2}}\cdot 2{(2\pi )}^{-s}\mathrm{\Gamma }(s)$

$=\cos {\displaystyle \frac{\pi s}{2}}\cdot 2{(2\pi )}^{-s}\mathrm{\Gamma }(s).$

Therefore

$$K(x)=\frac{1}{2\pi i}{\int }_{\mathrm{\Re }(s)=\frac{3}{2}}\frac{\gamma (s)}{\gamma (1-s)}{x}^{-s}𝑑s=\frac{1}{2\pi i}{\int }_{\mathrm{\Re }(s)=\frac{3}{2}}\cos \frac{\pi s}{2}\cdot 2{(2\pi )}^{-s}\mathrm{\Gamma }(s)x^{-s}𝑑s$$

But, taking as known

$${\int }_0^{\mathrm{\infty }}\cos (2\pi x)x^{s-1}𝑑x={(2\pi )}^{-s}\cos \frac{\pi s}{2}\mathrm{\Gamma }(s),$$

it follows that

$$K(x)=2\cos 2\pi x.$$

Thus Theorem 1 tells us that for $f\in 𝒮(ℝ)$,

$$\sum _{n=1}^{\mathrm{\infty }}f(n)=f(0)\zeta (0)+{\mathrm{Res}}_{s=1}ℳ(f)(s)\zeta (s)+2\sum _{n=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}f(y)\cos (2\pi ny)𝑑y,$$

i.e.,

$$\sum _{n=1}^{\mathrm{\infty }}f(n)=-\frac{1}{2}f(0)+{\int }_0^{\mathrm{\infty }}f(x)𝑑x+2\sum _{n=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}f(y)\cos (2\pi ny)𝑑y.$$

If $f:ℝ\to ℂ$ is even, this is the Poisson summation formula.

Take $a(n)=d(n)$ for all $n$. Then,

$$L(d,s)=\sum _{n=1}^{\mathrm{\infty }}d(n)n^{-s}={\zeta }^2(s).$$

For

$$\gamma (s)={\pi }^{-s}\mathrm{\Gamma }{\left(\frac{s}{2}\right)}^2,$$

it follows from the functional equation for the Riemann zeta function that $L(d,s)$ satisfies the functional equation

$$\gamma (s)L(d,s)=\gamma (1-s)L(d,1-s).$$

We worked out above that

$$\frac{{\pi }^{-s/2}\mathrm{\Gamma }\left(\frac{s}{2}\right)}{{\pi }^{-(1-s)/2}\mathrm{\Gamma }\left(\frac{1-s}{2}\right)}=\cos \frac{\pi s}{2}\cdot 2{(2\pi )}^{-s}\mathrm{\Gamma }(s),$$

whence

${\displaystyle \frac{\gamma (s)}{\gamma (1-s)}}$ $={(2\pi )}^{-2s}4{\cos }^2{\displaystyle \frac{\pi s}{2}}\mathrm{\Gamma }{(s)}^2$

$={(2\pi )}^{-2s}(2+2\cos \pi s)\mathrm{\Gamma }{(s)}^2.$

Taking as given two identities for Bessel functions

$${\int }_0^{\mathrm{\infty }}{x}^{s-1}{K}_0(4\pi x^{1/2})𝑑x=\frac{1}{2}{(2\pi )}^{-2s}\mathrm{\Gamma }{(s)}^2$$

and

$${\int }_0^{\mathrm{\infty }}{x}^{s-1}{Y}_0(4\pi x^{1/2})𝑑x=-\frac{1}{\pi }{(2\pi )}^{-2s}\cos \pi s\mathrm{\Gamma }{(s)}^2,$$

it follows that

$$K(x)=4K_0(4\pi x^{1/2})-2\pi Y_0(4\pi x^{1/2}).$$

Thus Theorem 1 tells us that for $f\in 𝒮(ℝ)$,

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}d(n)f(n)$ $=f(0){\zeta }^2(0)+{\mathrm{Res}}_{s=1}ℳ(f)(s){\zeta }^2(s)$

$+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}d(n){\displaystyle {\int }_0^{\mathrm{\infty }}}f(y)\left(4K_0(4\pi {(ny)}^{1/2})-2\pi Y_0(4\pi {(ny)}^{1/2})\right)𝑑y.$

Using

$${\zeta }^2(s)=\frac{1}{{(s-1)}^2}+\frac{2\gamma }{s-1}+O(1),s\to 1,$$

and

$$x^{s-1}=1+(s-1)\log x+O({|s-1|}^2),$$

we have

$${\mathrm{Res}}_{s=1}ℳ(f)(s){\zeta }^2(s)=2\gamma +\log x,$$

and so

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}d(n)f(n)$ $={\displaystyle \frac{1}{4}}f(0)+{\displaystyle {\int }_0^{\mathrm{\infty }}}f(x)(2\gamma +\log x)𝑑x$

$+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}d(n){\displaystyle {\int }_0^{\mathrm{\infty }}}f(y)\left(4K_0(4\pi {(ny)}^{1/2})-2\pi Y_0(4\pi {(ny)}^{1/2})\right)𝑑y.$

3 Bernoulli numbers

The Bernoulli polynomials are defined by

$$\frac{t{e}^{tx}}{e^t-1}=\sum _{m=0}^{\mathrm{\infty }}{B}_m(x)\frac{t^m}{m!}.$$

The Bernoulli numbers are defined by $B_m=B_m(0)$.

We denote by $[x]$ the greatest integer $\le x$, and we define $\{x\}=x-[x]$, namely, the fractional part of $x$. We define $P_m(x)=B_m(\{x\})$, the Bernoulli functions.

4 Wigert

The following result is proved by Wigert [18]. Our proof follows Titchmarsh [13, p. 163, Theorem 7.15]. Cf. Landau [10].

Theorem 2.

For and $N\ge 1$,

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

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

Proof.

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

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

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

${\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{2-i\mathrm{\infty }}^{2+i\mathrm{\infty }}}\mathrm{\Gamma }(s){\zeta }^2(s)z^{-s}𝑑s$ $={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}d(n){\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{2-i\mathrm{\infty }}^{2+i\mathrm{\infty }}}\mathrm{\Gamma }(s){(nz)}^{-s}𝑑s$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}d(n)e^{-nz}.$ (1)

Define $F(s)=\mathrm{\Gamma }(s){\zeta }^s(s)z^{-s}$. $F$ has poles at $1,0$, and the negative odd integers. (At each negative even integer, $\mathrm{\Gamma }$ 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

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

the asymptotic formula

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

and the asymptotic formula

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

to obtain

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

$\cdot \left({\displaystyle \frac{1}{z}}-{\displaystyle \frac{\log z}{z}}(s-1)+O({|s-1|}^2)\right)$

$={\displaystyle \frac{1}{z{(s-1)}^2}}-{\displaystyle \frac{\gamma }{z(s-1)}}+{\displaystyle \frac{2\gamma }{z(s-1)}}-{\displaystyle \frac{\log z}{z(s-1)}}+O(1)$

$={\displaystyle \frac{1}{z{(s-1)}^2}}+{\displaystyle \frac{\gamma }{z(s-1)}}-{\displaystyle \frac{\log z}{z(s-1)}}+O(1).$

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 $\mathrm{\Gamma }$ 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 $n\ge 0$ we determine the residue of $F$ at $-(2n+1)$. The residue of $\mathrm{\Gamma }$ 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

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

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,

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

Denote the right-hand sideof (2) 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 $M\to \mathrm{\infty }$. For $s=\sigma +it$ with $-2N\le \sigma \le 2$, Stirling’s formula [14, p. 151] tells us that

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

As well [13, p. 95], there is some $K>0$ such that in the half-plane $\sigma \ge -2N$,

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

Also,

$z^{-s}$ $=e^{-s\log z}$

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

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

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

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

Therefore

$$\left|{\int }_{2+iM}^{-2N+iM}F(s)𝑑s\right|\le (2+2N)\underset{-2N\le \sigma \le 2}{sup}|F(\sigma +iM)|=O(e^{-\frac{\pi }{2}M}{M}^{\sigma -\frac{1}{2}}{M}^{2K}{|z|}^{-\sigma }{e}^{\lambda M}),$$

and because this tends to $0$ as $M\to \mathrm{\infty }$. Likewise,

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

as $M\to \mathrm{\infty }$. It follows that

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

Hence,

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

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

$${\int }_{-2N-i\mathrm{\infty }}^{-2N+i\mathrm{\infty }}F(s)𝑑s={\int }_{\sigma =-2N,|t|\le 1}F(s)𝑑s+{\int }_{\sigma =-2N,|t|>1}F(s)𝑑s.$$

The first integral satisfies

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

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

$\left|{\displaystyle {\int }_{\sigma =-2N,|t|>1}}F(s)𝑑s\right|$ $\le {\displaystyle {\int }_{\sigma =-2N,|t|>1}}{e}^{-\frac{\pi }{2}|t|}{|t|}^{\sigma -\frac{1}{2}}{|t|}^K{|z|}^{-\sigma }{e}^{\lambda |t|}𝑑s$

$={|z|}^{2N}{\displaystyle {\int }_{\sigma =-2N,|t|>1}}{e}^{-\frac{\pi }{2}|t|}{|t|}^{-2N-\frac{1}{2}}{|t|}^K{e}^{\lambda |t|}𝑑t$

$={|z|}^{2N}\cdot O(1)$

$=O({|z|}^{2N}),$

because . This establishes

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

Using (1) and (2), this becomes

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

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

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

5 Other works on the Voronoi summation formula

Voronoi’s papers on the Voronoi summation formula are [15] and [17] and [16].

Iwaniec and Kowalski [9, Chaper 4]

Stein and Shakarchi [12, p. 392, Theorem 8.11].

Ivic [8, pp. 83ff., Chapter 3] and [7]

Miller and Schmid [11]

Hejhal [6]

Flajolet, Gourdon and Dumas [5]

Bettin and Conrey [1]

Chandrasekharan and Narasimhan [2]

Chandrasekharan [3, Chapter VIII]