# A series of secants

Jordan Bell
November 3, 2014

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

 $C(\tau)=2\sum_{n=-\infty}^{\infty}\frac{1}{e^{\pi in\tau}+q^{-\pi in\tau}}=% \sum_{n=-\infty}^{\infty}\sec\pi n\tau,\qquad\tau\in\mathfrak{H}.$

We take as granted that $C$ is holomorphic on $\mathfrak{H}$.

First we calculate the Fourier transform of $x\mapsto\mathrm{sech}\,\pi x$.11 1 Elias M. Stein and Rami Shakarchi, Complex Analysis, p. 81, Example 3.

###### Lemma 1.

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

 $\int_{-\infty}^{\infty}e^{-2\pi i\xi x}\mathrm{sech}\,\pi xdx=\mathrm{sech}\,% \pi\xi.$
###### Proof.

Let $\xi\in\mathbb{R}$ 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 $z=-R+iy$,

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

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

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

For $z=R+iy$,

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

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

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

Therefore

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

and likewise

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

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

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

i.e.,

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

For the top horizontal side,

 $\displaystyle\int_{-R+2i}^{R+2i}f(z)dz$ $\displaystyle=\int_{-R}^{R}\frac{e^{-2\pi i(x+2i)\xi}}{\cosh(\pi x+2\pi i)}dx$ $\displaystyle=\int_{-R}^{R}\frac{e^{-2\pi ix\xi}e^{4\pi\xi}}{\cosh(\pi x)\cosh% (2\pi i)+\sinh(\pi x)\sinh(2\pi i)}dx$ $\displaystyle=e^{4\pi\xi}\int_{-R}^{R}\frac{e^{-2\pi ix\xi}}{\cosh\pi x}dx$ $\displaystyle=e^{4\pi\xi}\int_{-R}^{R}f(x)dx.$

Writing

 $I=\int_{-\infty}^{\infty}f(z)dz,$

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

 $\int_{-\infty}^{\infty}e^{-2\pi i\xi x}e^{-2\pi iax}\mathrm{sech}\,\frac{\pi x% }{t}dx=t\mathrm{sech}\,(\pi(\xi+a)t),\qquad\xi\in\mathbb{R}.$
###### Proof.
 $\displaystyle\int_{-\infty}^{\infty}e^{-2\pi ix\xi}e^{-2\pi iax}\mathrm{sech}% \,\frac{\pi x}{t}dx$ $\displaystyle=\int_{-\infty}^{\infty}e^{-2\pi i(\xi+a)x}\mathrm{sech}\,\frac{% \pi x}{t}dx$ $\displaystyle=t\int_{-\infty}^{\infty}e^{-2\pi i(\xi+a)tx}\mathrm{sech}\,\pi xdx$ $\displaystyle=t\mathrm{sech}\,(\pi(\xi+a)t).$

###### Theorem 3.

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

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

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

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

Following Stein and Shakarchi, for $a>0$, define $\mathfrak{F}_{a}$ to be the set of those functions $f$ defined on some neighborhood of $\mathbb{R}$ in $\mathbb{C}$ such that $f$ is holomorphic on the set $\{z\in\mathbb{C}:|\mathrm{Im}\,z| and for which there is some $A>0$ such that

 $|f(x+iy)|\leq\frac{A}{1+x^{2}},\qquad x\in\mathbb{R},\quad|y|

and we set $\mathfrak{F}=\bigcup_{a>0}\mathfrak{F}_{a}$. The Poisson summation formula22 2 Elias M. Stein and Rami Shakarchi, Complex Analysis, p. 118, Theorem 2.4. states that for $f\in\mathfrak{F}$,

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

For $z=x+iy$ with $|y|<\frac{1}{2}$,

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

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 $\mathfrak{F}_{\frac{t}{2}}$. 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 $\mathfrak{H}$, so by analytic continuation this identity is true for all $\tau\in\mathfrak{H}$. ∎

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

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

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

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 $\mathfrak{H}$, and thus by analytic continuation that (1) is true for all $\tau\in\mathfrak{H}$.

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 $\nu<0$,

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

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

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

proving the claim. ∎

Define $\theta:\mathfrak{H}\to\mathbb{C}$ by

 $\theta(\tau)=\sum_{n\in\mathbb{Z}}e^{\pi in^{2}\tau},\qquad\tau\in\mathfrak{H}.$

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 $C=\theta^{2}$.33 3 Elias M. Stein and Rami Shakarchi, Complex Analysis, p. 304. One reason that $\theta$ is significant is that, for $q=e^{i\pi\tau}$,

 $\displaystyle\theta(\tau)^{2}$ $\displaystyle=\left(\sum_{n_{1}\in\mathbb{Z}}q^{n_{1}^{2}}\right)\left(\sum_{n% _{2}\in\mathbb{Z}}q^{n_{2}^{2}}\right)$ $\displaystyle=\sum_{(n_{1},n_{2})\in\mathbb{Z}\times\mathbb{Z}}q^{n_{1}^{2}+n_% {2}^{2}}$ $\displaystyle=\sum_{n=0}^{\infty}r_{2}(n)q^{n},$

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

 $\displaystyle C(\tau)$ $\displaystyle=2\sum_{n=-\infty}^{\infty}\frac{1}{q^{n}+q^{-n}}$ $\displaystyle=2\sum_{n=-\infty}^{\infty}\frac{q^{n}}{1+q^{2n}}$ $\displaystyle=1+4\sum_{n=1}^{\infty}\frac{q^{n}}{1+q^{2n}}$ $\displaystyle=1+4\sum_{n=1}^{\infty}q^{n}\frac{1-q^{2n}}{1-q^{4n}}$ $\displaystyle=1+4\sum_{n=1}^{\infty}\left(\frac{q^{n}}{1-q^{4n}}-\frac{q^{3n}}% {1-q^{4n}}\right).$

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

 $\sum_{n=0}^{\infty}r_{2}(n)q^{n}=1+4\sum_{n=1}^{\infty}\left(\frac{q^{n}}{1-q^% {4n}}-\frac{q^{3n}}{1-q^{4n}}\right).$

We write

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

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

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

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

 $r_{2}(n)=4(a(n)-b(n)).$