# The Voronoi summation formula

Jordan Bell
September 16, 2015

## 1 Mellin transform

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

 $\mathscr{M}(f)(s)=\int_{0}^{\infty}x^{s-1}f(x)dx.$

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

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

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

(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 $\mathscr{S}(\mathbb{R})$ the Fréchet space of Schwartz functions $\mathbb{R}\to\mathbb{C}$.

###### Theorem 1.

Let $a$ be arithmetic function and define

 $L(a,s)=\sum_{n=1}^{\infty}a(n)n^{-s},\qquad\Re(s)>1.$

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

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

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

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

Let $f\in\mathscr{S}(\mathbb{R})$ and define for $x>0$,

 $K(x)=\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}\frac{\gamma(s)}{\gamma(1-s)}x^{% -s}ds,\qquad g(x)=\int_{0}^{\infty}f(y)K(xy)dy.$

Then,

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

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

 $\mathscr{M}(f)(s)=\int_{0}^{\infty}x^{s-1}f(x)dx=f(x)\frac{x^{s}}{s}\bigg{|}_{% 0}^{\infty}-\int_{0}^{\infty}f^{\prime}(x)\frac{x^{s}}{s}dx=-\frac{1}{s}% \mathscr{M}(f^{\prime})(s+1).$

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

 $\displaystyle\sum_{n=1}^{\infty}a(n)f(n)$ $\displaystyle=\sum_{n=1}^{\infty}a(n)\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}% n^{-s}F(s)ds$ $\displaystyle=\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}F(s)\sum_{n=1}^{\infty}% a_{n}n^{-s}ds$ $\displaystyle=\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}F(s)L(a,s)ds.$

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

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

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

 $-\mathscr{M}(f^{\prime})(1)=-\int_{0}^{\infty}f^{\prime}(x)dx=-(f(\infty)-f(0)% )=f(0),$

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 $-\frac{1}{2}\leq\Re(s)\leq\frac{3}{2}$ as $|\Im(s)|\to\infty$,

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

Define

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

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

 $\displaystyle\frac{1}{2\pi i}\int_{\Re(s)=-\frac{1}{2}}F(s)L(a,s)ds$ $\displaystyle=\frac{1}{2\pi i}\int_{\Re(s)=-\frac{1}{2}}F(s)\frac{\gamma(1-s)}% {\gamma(s)}L(a,1-s)ds$ $\displaystyle=\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}F(1-s)\frac{\gamma(s)}{% \gamma(1-s)}L(a,s)ds$ $\displaystyle=\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}G(s)L(a,s)ds.$

Furthermore, define

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

which satisfies

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

We have

 $\displaystyle\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}x^{-s}G(s)ds$ $\displaystyle=\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}x^{-s}F(1-s)\frac{% \gamma(s)}{\gamma(1-s)}ds$ $\displaystyle=\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}x^{-s}\left(-\frac{1}{1% -s}\mathscr{M}(f^{\prime})(2-s)\right)\frac{\gamma(s)}{\gamma(1-s)}ds$ $\displaystyle=\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}x^{-s}\left(-\frac{1}{1% -s}\int_{0}^{\infty}y^{1-s}f^{\prime}(y)dy\right)\frac{\gamma(s)}{\gamma(1-s)}ds$ $\displaystyle=-\frac{1}{x}\int_{0}^{\infty}f^{\prime}(y)\frac{1}{2\pi i}\int_{% \Re(s)=\frac{3}{2}}\frac{1}{1-s}\frac{\gamma(s)}{\gamma(1-s)}(xy)^{1-s}ds$ $\displaystyle=-\frac{1}{x}\int_{0}^{\infty}f^{\prime}(y)J(xy)dy$ $\displaystyle=-\frac{1}{x}f(y)J(xy)\bigg{|}_{0}^{\infty}+\frac{1}{x}\int_{0}^{% \infty}f(y)J^{\prime}(xy)xdy$ $\displaystyle=0+\int_{0}^{\infty}f(y)J^{\prime}(xy)dy$ $\displaystyle=\int_{0}^{\infty}f(y)K(xy)dy$ $\displaystyle=g(x).$

Therefore,

 $\displaystyle\sum_{n=1}^{\infty}a(n)g(n)$ $\displaystyle=\sum_{n=1}^{\infty}a(n)\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}% n^{-s}G(s)ds$ $\displaystyle=\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}G(s)\sum_{n=1}^{\infty}% a(n)n^{-s}ds$ $\displaystyle=\frac{1}{2\pi i}\int_{\Re(s)=\frac{3}{2}}G(s)L(a,s)ds.$

Thus we have

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

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

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

The Riemann zeta function satisfies the functional equation

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

So with

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

we have

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

Using

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

and

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

we have

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

and so

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

Therefore

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

But, taking as known

 $\int_{0}^{\infty}\cos(2\pi x)x^{s-1}dx=(2\pi)^{-s}\cos\frac{\pi s}{2}\Gamma(s),$

it follows that

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

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

 $\sum_{n=1}^{\infty}f(n)=f(0)\zeta(0)+\mathrm{Res}_{s=1}\mathscr{M}(f)(s)\zeta(% s)+2\sum_{n=1}^{\infty}\int_{0}^{\infty}f(y)\cos(2\pi ny)dy,$

i.e.,

 $\sum_{n=1}^{\infty}f(n)=-\frac{1}{2}f(0)+\int_{0}^{\infty}f(x)dx+2\sum_{n=1}^{% \infty}\int_{0}^{\infty}f(y)\cos(2\pi ny)dy.$

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

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

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

For

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

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

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

whence

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

Taking as given two identities for Bessel functions

 $\int_{0}^{\infty}x^{s-1}K_{0}(4\pi x^{1/2})dx=\frac{1}{2}(2\pi)^{-2s}\Gamma(s)% ^{2}$

and

 $\int_{0}^{\infty}x^{s-1}Y_{0}(4\pi x^{1/2})dx=-\frac{1}{\pi}(2\pi)^{-2s}\cos% \pi s\Gamma(s)^{2},$

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 $f\in\mathscr{S}(\mathbb{R})$,

 $\displaystyle\sum_{n=1}^{\infty}d(n)f(n)$ $\displaystyle=f(0)\zeta^{2}(0)+\mathrm{Res}_{s=1}\mathscr{M}(f)(s)\zeta^{2}(s)$ $\displaystyle+\sum_{n=1}^{\infty}d(n)\int_{0}^{\infty}f(y)\left(4K_{0}(4\pi(ny% )^{1/2})-2\pi Y_{0}(4\pi(ny)^{1/2})\right)dy.$

Using

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

and

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

we have

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

and so

 $\displaystyle\sum_{n=1}^{\infty}d(n)f(n)$ $\displaystyle=\frac{1}{4}f(0)+\int_{0}^{\infty}f(x)(2\gamma+\log x)dx$ $\displaystyle+\sum_{n=1}^{\infty}d(n)\int_{0}^{\infty}f(y)\left(4K_{0}(4\pi(ny% )^{1/2})-2\pi Y_{0}(4\pi(ny)^{1/2})\right)dy.$

## 3 Bernoulli numbers

The Bernoulli polynomials are defined by

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

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 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 $\lambda<\frac{1}{2}\pi$ 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 $\Re z>0$ we have

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

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,

 $\int_{C}F(s)ds=2\pi i\left(\frac{\gamma}{z}-\frac{\log z}{z}+\frac{1}{4}+\sum_% {n=0}^{N-1}-\frac{B_{2n+2}^{2}}{(2n+2)!(2n+2)}z^{2n+1}\right).$ (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 $M\to\infty$. For $s=\sigma+it$ with $-2N\leq\sigma\leq 2$, Stirling’s formula [14, 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 [13, 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 (1) and (2), 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

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

## 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]

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