$\textbf{Problem statement}$. Inspired by the computations at this nospoon we introduce the following integral:

$$\int_{0}^{\infty }\frac{~\theta _{4}^{2}\left( \exp \left( -\pi \,y\,\beta \right) \right) }{1+y^{2}}dy\; \tag{1}\label{1}$$ which is as far as I know only calculated for $\beta =1$. $$\int_{0}^{\infty }\frac{~\theta _{4}^{2}\left( \exp \left( -\pi \,y\right) \right) }{1+y^{2}}dx=1 \tag{2}\label{2} $$ My goal is to calculate the integral of \eqref{1} for any $\beta $.

$\textbf{Ansatz}$.With the aid of the well-known representation of the square power of $ \theta_{4}$ Dieckmann $$\theta _{4}^{2}\left( \exp \left( -\pi \,y\right) \right) =1+2\sum_{k=1}^{\infty }(-1)^{k}~s{ech}(\pi k\,y) \tag{3}\label{3}$$ and the transformation $y=\frac{x}{\beta }$, for \eqref{1} follows: $$\frac{\pi }{2}+2\sum_{k=1}^{\infty }(-1)^{k}\int_{0}^{\infty }\frac{~\beta ~s% {ech}(\pi k\,x)}{\beta ^{2}+x^{2}}\,dx \tag{4}\label{4}$$ The integral in the sum: $$\int_{0}^{\infty }\frac{\beta ~s{ech}(\pi k\,x)}{\beta ^{2}+x^{2}}% \,dx=\int_{0}^{\infty }\frac{\beta ~}{\beta ^{2}+x^{2}}\frac{1\,}{\cosh \left( \pi k\,x\right) }dx \tag{5}\label{5}$$ is done in Sangchul Lee and returns the solution of \eqref{1}: $$\mathcal{I}\left( \beta \right) =\frac{\pi }{2}+\sum_{k=1}^{\infty }(-1)^{k}\left( \psi \left( \frac{k\,\beta \ }{2}+\frac{3}{4}\right) -\psi \left( \frac{k\,\beta \ }{2}+\frac{1}{4}\right) \right) \; \tag{6}\label{6}$$

Using nospoon returns $\mathcal {I} \left(1\right) = 1 $. A proof is upon request.

For the readability, here some of the steps performed in Sangchul Lee. Transformation of \eqref{5} with $y=\frac{x}{% \beta }$ leads to: $$\int_{0}^{\infty }\frac{\beta ~}{\beta ^{2}+x^{2}}\frac{1\,}{\cosh \left( \pi k\,x\right) }dx=\int_{0}^{\infty }\frac{1~}{1+y^{2}}\frac{1\,}{\cosh \left( a\,y\right) }dy \tag{7}\label{7}$$ with $a=k~\beta \,\pi $. Transformation with $z=$ $\frac{a\,y}{\pi }$ leads to: $$\int_{0}^{\infty }\frac{1~}{1+y^{2}}\frac{1\,}{\cosh \left( a\,y\right) }dy=% \frac{\pi a}{2}\int_{-\infty }^{\infty }\frac{dz}{\left( a^{2}+\pi ^{2}z^{2}\right) \cosh \left( \pi \,z\right) } \tag{8}\label{8}$$

In the following, we need the Fourier transform of $f$ $$\widehat{f}\left( \xi \right) =\mathcal{F}\left[ f\left( z\right) \right] =\int_{\mathcal{R}}f\left( z\right) \exp \left( -2\pi i\xi z\right) ~dz \tag{9}\label{9}$$ Let $$f\left( z\right) =s{ech}(\pi \,z),\;g\left( z\right) =\frac{1}{a^{2}+\pi ^{2}z^{2}} \tag{10}\label{10}$$ and $$\widehat{f}\left( \xi \right) =s{ech}(\pi \,\xi ),\;\widehat{g}\left( \xi \right) =\frac{1}{a}\exp \left( -2a\left\vert \xi \right\vert \right) \tag{11}\label{11}$$

Also, if both $f$ and $g$ are in $L^{2}$, then $$\int_{\mathcal{R}}\widehat{f}~g=\int_{\mathcal{R}}f~\widehat{g} \tag{12}\label{12}$$ results to $$\frac{\pi }{2}+\pi a\sum_{k=1}^{\infty }(-1)^{k}\int_{-\infty }^{\infty }% \frac{dz}{\left( a^{2}+\pi ^{2}z^{2}\right) \cosh \left( \pi \,z\right) }=% \frac{\pi }{2}+2\pi \sum_{k=1}^{\infty }(-1)^{k}\int_{0}^{\infty }\frac{\exp \left( -2\pi k\beta ~z\right) }{\cosh \left( \pi \,z\right) }dz \tag{13}\label{13}$$

Further transformations then leads finally to the solution \eqref{6}.

Now, for an equivalent integral representation of \eqref{1}, we first perform the sum in \eqref{13}: $$\mathcal{I}\left( \beta \right) =\frac{\pi }{2}-2\pi \int_{0}^{\infty }\frac{% s{ech}(\pi \,z)}{1+\exp \left( 2\pi \beta ~z\right) }dz \tag{14}\label{14}$$

For $\beta =1$, the known value $\mathcal{I}\left( 1\right) =1$ results. With the aid of Mathematica, further analytical expressions for some fixed $\beta$-values can be calculated. With \eqref{4} the following identity results:

$$\sum_{k=1}^{\infty }(-1)^{k}\int_{0}^{\infty }\frac{s{ech}(k~\pi \,z)}{\beta ^{2}+z^{2}}dz=-\frac{\pi }{\beta }\int_{0}^{\infty }\frac{s{ech}(\pi \,z)}{% 1+\exp \left( 2\pi \beta ~z\right) }dz \tag{15}\label{15}$$

The integral form \eqref{14} can be transformed into other interesting expressions. Using the known identity Kim: $$\,_{2}F_{1}\left( a,a;a+1;\frac{1}{2}\right) =2^{a-1}a~\left( \psi \left( \frac{a}{2}+\frac{1}{2}\right) -\psi \left( \frac{a}{2}\right) \right) \tag{16}\label{16}$$ and \eqref{6}, these expressions can be reproduced and further identities can be derived. For the readability, I omit lots of results, I've found so far. In case of interest, these results can be requested.

$\textbf{1st Question}$

$\textit{Does anybody know how to approach this sum \eqref{6}?} $ $\textit{ Where can I find out more about dealing with the sum?}$ $\textit{Is it possible to derive a simpler expression?}$

$\textbf{2st Question}$

$\textit{Can we find a closed form expression for $\mathcal{I}\left( \beta\right)$, at least for $\beta $ $\in \mathbb{N}$?, distinguishing even/odd $\beta $ ?}$

$\textbf{Bonus Q}$

$\textit{How can I proof the identity \eqref{15} with the help of the Poisson Summation Formula?}$

$\textit{Are there any further results can be obtained by doing so?}$


1 Answer 1


To evaluate the series \begin{equation} S=\sum_{k=1}^{\infty }(-1)^{k}\left( \psi \left( \frac{k\,\beta \ }{2}+\frac{3}{4}\right) -\psi \left( \frac{k\,\beta \ }{2}+\frac{1}{4}\right) \right) \end{equation} we use the Mellin transform representation (Ederlyi 7.3.49 p.353) \begin{equation} \Psi(s+a)-\Psi(s+b)=\int_0^\infty x^{s-1}g(x)\,dx \end{equation} where \begin{equation} g(x)=\begin{cases} \frac{x^b-x^a}{1-x} &\mbox{if } 0<x<1 \\ 0 & \mbox{if } x> 1 \end{cases} \end{equation} and $\Re s>-\Re a,-\Re b$. Here, with $s=k\beta/2$, we can express \begin{align} S&=\sum_{k=1}^{\infty }(-1)^{k}\int_0^1 \frac{x^{1/4}-x^{3/4}}{1-x}x^{\frac{k\beta}{2}-1}\,dx\\ &=\sum_{k=1}^{\infty }(-1)^{k}\int_0^1 \frac{x^{\frac{k\beta}{2}-\frac{3}{4}}}{1+x^{1/2}}\,dx\\ &=4\sum_{k=1}^{\infty }(-1)^{k}\int_0^1 \frac{u^{2k\beta}}{1+u^2}\,du \end{align} By changing the order of summation with the DCT, we find \begin{align} S&=4\int_0^1\frac{du}{1+u^2}\sum_{k=1}^{\infty }(-1)^{k}u^{2k\beta}\\ &=-4\int_0^1\frac{u^{2\beta}}{(1+u^2)(1+u^{2\beta})}\,du\\ &=-4\int_1^\infty\frac{du}{(1+u^2)(1+u^{2\beta})} \end{align} and thus \begin{equation} I(\beta)=\frac{\pi}{2}-4\int_1^\infty\frac{du}{(1+u^2)(1+u^{2\beta})} \end{equation} Which is equivalent to the representation (14) in the OP. One may check, for example, that, using the Laplace transform of $\operatorname{sech}^2$ (Ederlyi 4.9.8) \begin{align} I(1)&=\frac{\pi}{2}-4\int_1^\infty\frac{du}{(1+u^2)^2}\\ &=\frac{\pi}{2}-\int_0^\infty \frac{e^{-s}}{\cosh^2s}\,ds\\ &=\frac{\pi}{2}-\left[\frac{1}{2}\left( \Psi\left(\frac{3}{4} \right)-\Psi\left( \frac{1}{4} \right)\right)-1\right]\\ &=1 \end{align}

Unfortunately, this integral has probably no general closed form solution (see, for example,this answer for a similar integral).

However, explicit expressions can be found when $\beta$ is a positive rational number. With $\beta=p/(2q)$, with $p,q$ integers and $q\ge1$, changing $u=v^q$ one obtains \begin{align} J\left(\frac{p}{2q}\right)&=\int_1^\infty\frac{du}{(1+u^2)(1+u^{p/q})}\\ &=q\int_1^\infty\frac{v^{q-1}dv}{(1+v^{2q})(1+v^{p})} \end{align} It can be classically evaluated using the residue theorem by considering the function \begin{equation} f(z)=\frac{z^{q-1}}{(1+z^{2q})(1+z^{p})}\log (z-1) \end{equation} defined in the complex plane with a branch cut on the interval $[1,\infty)$ and by integrating it along the keyhole contour. One can express \begin{equation} I\left(\frac{p}{2q}\right)=\frac{\pi}{2}+4\sum_{z_j}\Re\left[\operatorname{Res}\left( f(z),z_j \right)\right] \end{equation} As an illustration, the case $\beta=2 n$ is calculated below. Here, \begin{equation} f(z)=\frac{\log (z-1)}{(1+z^{2})(1+z^{4n})} \end{equation} The poles $z_{\pm}=\pm i$ and $z_k=\exp(\frac{\pi}{4n}(2k+1)),k=0,1,\ldots,4p-1$ are simple. We have, using $z_\pm^2=-1)$, \begin{align} \operatorname{Res}\left( f(z),z_\pm \right)&=\frac{\log(z_\pm-1)}{2z_\pm(1+z^{4n}_\pm)}\\ &=\frac{z_\pm\log(z_\pm-1)}{2z_\pm^2(1+z^{4n}_\pm)}\\ &=\frac{\mp i\log(\pm i-1)}{4} \end{align} and thus \begin{align} \Re\operatorname{Res}\left( f(z),i \right)&=-\Re\left[\frac{i}{4}\left( \log\sqrt{2}+\frac{3i\pi}{4} \right)\right]\\ &=\frac{3\pi}{16}\\ \Re\operatorname{Res}\left( f(z),-i \right)&=\Re\left[\frac{i}{4}\left( \log\sqrt{2}+\frac{5i\pi}{4} \right)\right]\\ &=-\frac{5\pi}{16} \end{align} Similarly, \begin{align} \operatorname{Res}\left( f(z),z_k \right)&=\frac{\log(z_k-1)}{4nz_k^{4n-1}(1+z^{2}_k)}\\ &=\frac{z_k\log(z_k-1)}{4nz_k^{4n}(1+z^{2}_k)}\\ &=-\frac{\log(z_k-1)}{4n\left( z_k+\frac{1}{z_k} \right)} \end{align} and thus \begin{align} \Re\operatorname{Res}\left( f(z),z_k \right)&=-\frac{1}{16n}\frac{\log\left( 2\left( 1-\cos\left(\frac{\pi}{4n}(2k+1) \right) \right) \right)}{\cos\left(\frac{\pi}{4n}(2k+1) \right)}\\ &=-\frac{1}{8n}\frac{\log\left( 2\sin\left(\frac{\pi}{8n}(2k+1) \right) \right) }{\cos\left(\frac{\pi}{4n}(2k+1) \right)} \end{align} Finally, \begin{equation} I(2n)=-\frac{1}{2n}\sum_{k=0}^{4n-1}\frac{\log\left( 2\sin\left(\frac{\pi}{8n}(2k+1) \right) \right) }{\cos\left(\frac{\pi}{4n}(2k+1) \right)} \end{equation} The case $\beta=2n+1$ is slightly more complicated as the poles $\pm i$ are double.


  • $\begingroup$ Great! Paul, thank you very much. This was the solution I was looking for for a longer time. My approach was to understand a similar problem [stocha] (math.stackexchange.com/questions/3483723/…) and find an equivalent expression. I suppose, since you answer my other question [stocha] (math.stackexchange.com/questions/3348684/…) you already have a litle insight in the background of these questions. $\endgroup$
    – stocha
    Sep 9, 2020 at 15:32
  • $\begingroup$ You're welcome. This ensemble of integrals, summations and series is amazing and very rich. $\endgroup$
    – Paul Enta
    Sep 9, 2020 at 21:22

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