A series of secants

Let $ℌ=\{\tau \in ℂ:\mathrm{Im}\tau >0\}$. Define $C:ℌ\to ℂ$ by

$$C(\tau )=2\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{e^{\pi in\tau }+q^{-\pi in\tau }}=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\sec \pi n\tau ,\tau \in ℌ.$$

We take as granted that $C$ is holomorphic on $ℌ$.

First we calculate the Fourier transform of $x\mapsto \mathrm{sech}\pi x$.

Lemma 1.

For $\xi \mathrm{\in }\mathrm{R}$,

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-2\pi i\xi x}\mathrm{sech}\pi x𝑑x=\mathrm{sech}\pi \xi .$$

Proof.

Let $\xi \in ℝ$ and define

$$f(z)=\frac{e^{-2\pi iz\xi }}{\cosh \pi z}.$$

The poles of $f$ are those $z$ at which $\cosh \pi z=0$, thus $z=ni+\frac{i}{2}$, $n\in \mathbb{Z}$. Taking ${\gamma }_R$ to be the contour going from $-R$ to $R$, from $R$ to $R+2i$, from $R+2i$ to $-R+2i$, and from $-R+2i$ to $-R$, the poles of $f$ inside ${\gamma }_R$ are $\frac{i}{2}$ and $\frac{3i}{2}$. Because ${(\cosh \pi z)}^{\prime }=\pi \sinh \pi z$, we work out

$${\mathrm{Res}}_{z=i/2}f(z)=\frac{e^{-2\pi i\cdot \frac{i}{2}\xi }}{\pi \sinh \pi \frac{i}{2}}=\frac{e^{\pi \xi }}{\pi i\sin \frac{\pi }{2}}=\frac{e^{\pi \xi }}{\pi i}$$

and

$${\mathrm{Res}}_{z=3i/2}f(z)=\frac{e^{-2\pi i\cdot \frac{3i}{2}\xi }}{\pi \sinh \pi \frac{3i}{2}}=\frac{e^{3\pi \xi }}{\pi i\sin \frac{3\pi }{2}}=\frac{e^{3\pi \xi }}{-\pi i}.$$

We bound the integrals on the vertical sides as follows. For $z=-R+iy$,

$$|\cosh \pi z|=\frac{|e^{\pi z}+e^{-\pi z}|}{2}\ge \frac{||e^{\pi z}|-|e^{-\pi z}||}{2}=\frac{|e^{-R\pi }-e^{R\pi }|}{2}=\frac{e^{R\pi }-e^{-R\pi }}{2},$$

and, for $0\le y\le 2$,

$$|e^{-2\pi iz\xi }|=e^{2\pi y\xi }=e^{4\pi \xi }.$$

For $z=R+iy$,

$$|\cosh \pi z|=\frac{|e^{\pi z}+e^{-\pi z}|}{2}\ge \frac{||e^{\pi z}|-|e^{-\pi z}||}{2}=\frac{|e^{R\pi }-e^{-R\pi }|}{2}=\frac{e^{R\pi }-e^{-R\pi }}{2},$$

and, for $0\le y\le 2$,

$$|e^{-2\pi iz\xi }|=e^{2\pi y\xi }=e^{4\pi \xi }.$$

Therefore

$$\left|{\int }_{-R}^{-R+2i}f(z)𝑑z\right|\le {\int }_{-R}^{-R+2i}|f(z)|𝑑z\le 2\cdot e^{4\pi \xi }\cdot \frac{2}{e^{R\pi }-e^{-R\pi }}=\frac{e^{4\pi \xi }}{e^{R\pi }-e^{-R\pi }}$$

and likewise

$$\left|{\int }_R^{R+2i}f(z)𝑑z\right|\le \frac{e^{4\pi \xi }}{e^{R\pi }-e^{-R\pi }}.$$

As $R\to \mathrm{\infty }$, each of these tends to $0$. Therefore,

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(z)𝑑z+{\int }_{\mathrm{\infty }+2i}^{-\mathrm{\infty }+2i}f(z)𝑑z=2\pi i\left(\frac{e^{\pi \xi }}{\pi i}+\frac{e^{3\pi \xi }}{-\pi i}\right)=-2e^{2\pi \xi }(e^{\pi \xi }-e^{-\pi \xi }),$$

i.e.,

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(z)𝑑z={\int }_{-\mathrm{\infty }+2i}^{\mathrm{\infty }+2i}f(z)𝑑z-2e^{2\pi \xi }(e^{\pi \xi }-e^{-\pi \xi }).$$

For the top horizontal side,

${\displaystyle {\int }_{-R+2i}^{R+2i}}f(z)𝑑z$ $={\displaystyle {\int }_{-R}^R}{\displaystyle \frac{e^{-2\pi i(x+2i)\xi }}{\cosh (\pi x+2\pi i)}}𝑑x$

$={\displaystyle {\int }_{-R}^R}{\displaystyle \frac{e^{-2\pi ix\xi }{e}^{4\pi \xi }}{\cosh (\pi x)\cosh (2\pi i)+\sinh (\pi x)\sinh (2\pi i)}}𝑑x$

$=e^{4\pi \xi }{\displaystyle {\int }_{-R}^R}{\displaystyle \frac{e^{-2\pi ix\xi }}{\cosh \pi x}}𝑑x$

$=e^{4\pi \xi }{\displaystyle {\int }_{-R}^R}f(x)𝑑x.$

Writing

$$I={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(z)𝑑z,$$

this gives us

$$I=e^{4\pi \xi }I-2e^{2\pi \xi }(e^{\pi \xi }-e^{-\pi \xi }),$$

and so

$$I=-2e^{2\pi \xi }\frac{e^{\pi \xi }-e^{-\pi \xi }}{1-e^{4\pi \xi }}=2\frac{e^{\pi \xi }-e^{-\pi \xi }}{e^{2\pi \xi }-e^{-2\pi \xi }}=2\frac{e^{\pi \xi }-e^{-\pi \xi }}{(e^{\pi \xi }-e^{-\pi \xi })(e^{\pi \xi }+e^{-\pi \xi })}=\mathrm{sech}\pi \xi ,$$

which is what we wanted to show. ∎

Corollary 2.

For $t\mathrm{>}\mathrm{0}$ and $a\mathrm{\in }\mathrm{R}$,

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-2\pi i\xi x}{e}^{-2\pi iax}\mathrm{sech}\frac{\pi x}{t}𝑑x=t\mathrm{sech}(\pi (\xi +a)t),\xi \in ℝ.$$

Proof.

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi ix\xi }{e}^{-2\pi iax}\mathrm{sech}{\displaystyle \frac{\pi x}{t}}𝑑x$ $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi i(\xi +a)x}\mathrm{sech}{\displaystyle \frac{\pi x}{t}}𝑑x$

$=t{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi i(\xi +a)tx}\mathrm{sech}\pi x𝑑x$

$=t\mathrm{sech}(\pi (\xi +a)t).$

∎

Theorem 3.

For all $\tau \mathrm{\in }\mathrm{H}$,

$$C(\tau )=\frac{i}{\tau }C\left(-\frac{1}{\tau }\right).$$

Proof.

For $f\in L^1(ℝ)$, we define $\widehat{f}:ℝ\to ℂ$ by

$$\widehat{f}(\xi )={\int }_{ℝ}{e}^{-2\pi i\xi x}f(x)𝑑x,\xi \in ℝ.$$

Following Stein and Shakarchi, for $a>0$, define ${𝔉}_a$ to be the set of those functions $f$ defined on some neighborhood of $ℝ$ in $ℂ$ such that $f$ is holomorphic on the set and for which there is some $A>0$ such that

and we set $𝔉={\bigcup }_{a>0}{𝔉}_a$. The Poisson summation formula states that for $f\in 𝔉$,

$$\sum _{n\in \mathbb{Z}}f(n)=\sum _{n\in \mathbb{Z}}\widehat{f}(n).$$

For $z=x+iy$ with ,

$|\mathrm{sech}{\displaystyle \frac{\pi z}{t}}|$ $={\displaystyle \frac{2}{|e^{\pi (x+iy)}-e^{-\pi (x+iy)}|}}$

$\le {\displaystyle \frac{2}{||e^{\pi (x+iy)}|-|e^{-\pi (x+iy)}||}}$

$={\displaystyle \frac{2}{|e^{\pi x}-e^{-\pi x}|}}$

$=\mathrm{sech}\pi |x|.$

Let $t>0$. Because the zeros of $\cosh \pi z$ are $ni+\frac{i}{2}$, $n\in \mathbb{Z}$, the function $f(z)=\mathrm{sech}\frac{\pi z}{t}$ belongs to ${𝔉}_{\frac{t}{2}}$. Corollary 2 with $a=0$ gives us

$$\widehat{f}(\xi )=t\mathrm{sech}\pi \xi t,$$

so applying the Poisson summation formula we get

$$\sum _{n\in \mathbb{Z}}\mathrm{sech}\frac{\pi n}{t}=t\sum _{n\in \mathbb{Z}}\mathrm{sech}\pi nt,$$

or,

$$\sum _{n\in \mathbb{Z}}\sec \frac{\pi in}{t}=t\sum _{n\in \mathbb{Z}}\sec \pi int,$$

i.e.,

$$C\left(\frac{i}{t}\right)=tC(it).$$

For $\tau =it$ this reads

$$C(\tau )=\frac{i}{\tau }C\left(-\frac{1}{\tau }\right).$$

But $\tau \mapsto C(\tau )$ and $\tau \mapsto \frac{i}{\tau }C\left(-\frac{1}{\tau }\right)$ are holomorphic on $ℌ$, so by analytic continuation this identity is true for all $\tau \in ℌ$. ∎

Theorem 4.

$$C\left(1-\frac{1}{\tau }\right)\sim \frac{4\tau }{i}{e}^{\frac{\pi i\tau }{2}},\mathrm{Im}\tau \to +\mathrm{\infty }.$$

Proof.

Let $t>0$ and define $f(z)=e^{-\pi iz}\mathrm{sech}\frac{\pi z}{t}$, which we check belongs to ${𝔉}_{\frac{t}{2}}$. Corollary 2 with $a=\frac{1}{2}$ tells us that for $t>0$,

$$\widehat{f}(\xi )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-2\pi i\xi x}{e}^{-\pi ix}\mathrm{sech}\frac{\pi x}{t}𝑑x=t\mathrm{sech}\left(\pi \left(\xi +\frac{1}{2}\right)t\right),\xi \in ℝ.$$

Thus the Poisson summation formula gives, as ${(-1)}^n=e^{-i\pi n}$,

$$\sum _{n\in \mathbb{Z}}{(-1)}^n\mathrm{sech}\frac{\pi n}{t}=t\sum _{n\in \mathbb{Z}}\mathrm{sech}\left(\pi \left(n+\frac{1}{2}\right)t\right),$$

or

$$\sum _{n\in \mathbb{Z}}{(-1)}^n\sec \frac{\pi in}{t}=t\sum _{n\in \mathbb{Z}}\sec \left(\pi i\left(n+\frac{1}{2}\right)t\right).$$

For $\tau =it$ this reads

$$\sum _{n\in \mathbb{Z}}{(-1)}^n\sec \frac{\pi n}{\tau }=\frac{\tau }{i}\sum _{n\in \mathbb{Z}}\sec \left(\pi \left(n+\frac{1}{2}\right)\tau \right).$$

Now,

$$\sec \left(\pi n\left(1-\frac{1}{\tau }\right)\right)=\frac{1}{\cos \pi n\cos \frac{-\pi n}{\tau }-\sin \pi n\sin \frac{-\pi n}{\tau }}={(-1)}^n\sec \frac{\pi n}{\tau },$$

so the above states that for $\tau =it$, $t>0$,

$$C\left(1-\frac{1}{\tau }\right)=\frac{\tau }{i}\sum _{n\in \mathbb{Z}}\sec \left(\pi \left(n+\frac{1}{2}\right)\tau \right).$$ (1)

We assert that both sides of (1) are holomorphic on $ℌ$, and thus by analytic continuation that (1) is true for all $\tau \in ℌ$.

Write $\tau =\sigma +it$. For $\nu >0$,

$$\sec \pi \nu \tau =\frac{2}{e^{i\pi \nu \tau }+e^{-i\pi \nu \tau }}=\frac{2}{e^{-i\pi \nu \tau }(e^{2\pi i\nu \tau }+1)}=2e^{i\pi \nu \tau }(1+O(|e^{2\pi i\nu \tau }|)),$$

or,

$$\sec \pi \nu \tau =2e^{i\pi \nu \tau }+O(|e^{3\pi i\nu \tau }|).$$

Now,

$$|e^{\frac{3\pi i\tau }{2}}|=e^{\frac{-3\pi t}{2}},$$

so,

$$\sec \pi \nu \tau =2e^{i\pi \nu \tau }+O(e^{\frac{-3\pi t}{2}}).$$

For ,

$$\sec \pi \nu \tau =\sec (-\pi \nu \tau )=2e^{-i\pi \nu \tau }+O(e^{\frac{-3\pi t}{2}}).$$

For $\nu =\frac{1}{2}$,

$$\sec \pi \nu \tau =2e^{\frac{i\pi \tau }{2}}+O(e^{\frac{-3\pi t}{2}}),$$

and for $\nu =-\frac{1}{2}$,

$$\sec \pi \nu \tau =2e^{\frac{i\pi \tau }{2}}+O(e^{\frac{-3\pi t}{2}}).$$

It follows that

$$\sum _{n\in \mathbb{Z}}\sec \left(\pi \left(n+\frac{1}{2}\right)\tau \right)=2e^{\frac{i\pi \tau }{2}}+2e^{\frac{i\pi \tau }{2}}+O(e^{\frac{-3\pi t}{2}})=4e^{\frac{i\pi \tau }{2}}+O(e^{\frac{-3\pi t}{2}}).$$

Using this with (1) yields

$$C\left(1-\frac{1}{\tau }\right)=\frac{4\tau }{i}{e}^{\frac{i\pi \tau }{2}}+O(|\tau |e^{\frac{-3\pi t}{2}}),\tau =\sigma +it,$$

proving the claim. ∎

Define $\theta :ℌ\to ℂ$ by

$$\theta (\tau )=\sum _{n\in \mathbb{Z}}{e}^{\pi i{n}^2\tau },\tau \in ℌ.$$

By proving that $\frac{C}{{\theta }^2}$ is a modular form of weight $0$, it follows that it is constant, and one thus finds that $C={\theta }^2$. One reason that $\theta$ is significant is that, for $q=e^{i\pi \tau }$,

$\theta {(\tau )}^2$ $=\left({\displaystyle \sum _{n_1\in \mathbb{Z}}}{q}^{n_1^2}\right)\left({\displaystyle \sum _{n_2\in \mathbb{Z}}}{q}^{n_2^2}\right)$

$={\displaystyle \sum _{(n_1,n_2)\in \mathbb{Z}\times \mathbb{Z}}}{q}^{n_1^2+n_2^2}$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{r}_2(n)q^n,$

where $r_2(n)$ denotes the number of ways that $n$ can be expressed as a sum of two squares. We can write $C(\tau )$ as

$C(\tau )$ $=2{\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}{\displaystyle \frac{1}{q^n+q^{-n}}}$

$=2{\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}{\displaystyle \frac{q^n}{1+q^{2n}}}$

$=1+4{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle \frac{q^n}{1+q^{2n}}}$

$=1+4{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{q}^n{\displaystyle \frac{1-q^{2n}}{1-q^{4n}}}$

$=1+4{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\left({\displaystyle \frac{q^n}{1-q^{4n}}}-{\displaystyle \frac{q^{3n}}{1-q^{4n}}}\right).$

Therefore the identity $\theta {(\tau )}^2=C(\tau )$ can be written as

$$\sum _{n=0}^{\mathrm{\infty }}{r}_2(n)q^n=1+4\sum _{n=1}^{\mathrm{\infty }}\left(\frac{q^n}{1-q^{4n}}-\frac{q^{3n}}{1-q^{4n}}\right).$$

We write

$$\sum _{n=1}^{\mathrm{\infty }}\frac{q^n}{1-q^{4n}}=\sum _{n=1}^{\mathrm{\infty }}{q}^n\sum _{m=0}^{\mathrm{\infty }}{(q^{4n})}^m=\sum _{n=1}^{\mathrm{\infty }}\sum _{m=0}^{\mathrm{\infty }}{q}^{n(4m+1)}=\sum _{k=1}^{\mathrm{\infty }}a(k)q^k,$$

where $a(k)$ denotes the number of divisors of $k$ of the form $4m+1$, and

$$\sum _{n=1}^{\mathrm{\infty }}\frac{q^{3n}}{1-q^{4n}}=\sum _{n=1}^{\mathrm{\infty }}{q}^{3n}\sum _{m=0}^{\mathrm{\infty }}{(q^{4n})}^m=\sum _{n=1}^{\mathrm{\infty }}\sum _{m=0}^{\mathrm{\infty }}{q}^{n(4m+3)}=\sum _{k=1}^{\mathrm{\infty }}b(k)q^k,$$

where $b(k)$ denotes the number of divisors of $k$ of the form $4m+3$. Thus for $n\ge 1$,

$$r_2(n)=4(a(n)-b(n)).$$