Here is an integral that I'm having trouble with. I have got to this integral by trying to prove Jacobi's sum of two squares by complex analysis. And here is it: $$I:= \int_{-\infty}^{\infty} \frac{\operatorname{csch}(x) \sin (2x)}{\cos (2x) - \cosh (\pi)}\mathrm{d}x=\pi \coth\left(\frac{\pi}{2}\right) - \frac{1}{2}B \left(\frac{1}{4}, \frac{1}{4}\right)$$ My step so far: It is well-known that: $$\sum_{n=1}^\infty 2 \pi (-1)^{n-1} \mathrm{e}^{-\omega \pi n} \sin(\pi \kappa n) = \frac{2 \pi e^{\pi \omega } \sin (\pi \kappa )}{2 e^{\pi \omega } \cos (\pi \kappa )+e^{2 \pi \omega }+1}$$$$=\frac{\pi \sin (\pi \kappa )}{\cos (\pi \kappa )+\cosh\left( \pi \omega \right)}$$ By substitution $k = t+1$ cause we know that $\cos\pi (t+1) = - \cos \pi t$: $$2\cdot\sum_{n=1}^\infty \mathrm{e}^{-\omega \pi n} \sin(\pi t n) = \frac{ \sin (\pi t )}{\cosh\left( \pi \omega \right)-\cos (\pi t)}$$ Now let $\omega = 1$ and $ t = \frac{2x}{\pi}$. We obtain: $$2\cdot\sum_{n=1}^\infty \mathrm{e}^{-\pi n} \sin\left( 2nx\right) = \frac{ \sin (2x )}{\cos (2x)-\cosh\left( \pi \right)}$$ Plug this infinite representation in our integral: $$ I = -2\cdot\sum_{n=1}^\infty \mathrm{e}^{-\pi n} \int_{-\infty}^{\infty} \frac{\sin (2nx)}{\sinh(x)} \mathrm{d}x$$ Each integral inside is fairly easy to compute. Indeed, one can use Laplace transform: $$\int_{-\infty}^{\infty} \frac{\sin (2nx)}{\sinh(x)} \mathrm{d}x = 2 \int_{0}^{\infty} \frac{\sin (2nx)}{\sinh(x)} \mathrm{d}x=4 \int_{0}^{\infty} \frac{e^{-x}\sin (2nx)}{1- e^{-2x}} \mathrm{d}x$$$$=4\sum_{i=0}^{\infty} \int_{0}^{\infty} \sin(2nx) e^{-(2n+1)x} \mathrm{d}x= \pi \tanh\left(n\pi \right) $$ The last inequality is obtained by the fact that: $\tanh \left(\frac{\pi x}{2}\right) = \frac{4x}{\pi} \sum_{k\geq 1} \frac{1}{(2k-1)^2 + x^2}$. Therefore: $$I: = 2\pi \sum_{n=1}^{\infty} e^{-\pi n} \tanh(\pi n)$$ And I'm stuck here because I don't know how to connect this sum with the result above. Indeed, I can derive a further result by using $\tanh(x)$ generating summation which is: $$\tanh (x) = 1+ 2\sum_{n=1}^{\infty}\frac{(-1)^n}{e^{2nx}}$$ However, I can't still be able to get the Beta Function to appear. Hope anyone can help me to derive the result above. Thank you so much.

UPDATE: So I am trying to prove the relation without using Jacobi's two square theorem or elliptical function: $$\left(\sum_{n=-\infty}^{\infty}e^{-\pi n^2}\right)^2 = \sum_{n = - \infty}^{\infty} \frac{1}{\cosh \left(\pi n\right)}$$ Actually, I intend proving this by constructing an complex integral and make use of Residue Theorem of the function $\displaystyle f(z) = \frac{1}{2i} \cot (\pi z) \mathrm{sech} (\pi z) $ over the contour which has a square shape from $-N - \frac{1}{4}$ to $N + \frac{1}{4}$ on the real axis and likewise on the imaginary axis. Since $N \to \infty$, $f(z) \to 0$, by calculating the residue, I obtained: $$\sum_{n=-\infty}^{\infty} \mathrm{sech}(\pi n) = \sum_{n=-\infty}^{\infty} (-1)^n \coth \left(\frac{\pi}{2}(2n+1)\right)$$ Then, in order to evaluate the LHS of $(1)$, I used Abel–Plana formula for the divergent series on the RHS which is: $$\sum_{n=0}^{\infty} (-1)^n f(n) = \frac{f(0)}{2} + i\int_0^{\infty} \frac{f(it) - f(-it)}{2\mathrm{sinh}(\pi t)} \mathrm{d}t$$ Rewrite the RHS of $(1)$ and then apply this formula since $\coth (z)$ is holomorphic in the region $\Re{(z)} \geq 0$. Finally, I use the result of: $$\sum_{n=-\infty}^{\infty}e^{-\pi n^2}=\frac{\sqrt[4]{\pi}}{\Gamma\left(\frac{3}{4}\right)}$$ By Euler's reflection formula, I got the integral above. Therefore, if one can prove the result above indenpendently, we can have another way proving Jacobi's two square theorem via complex analysis.

  • 1
    $\begingroup$ If $q=e^{-\pi} $ then the integral is $$2\pi\sum_{n\geq 1}\frac{q^{n}(1-q^{2n}) }{1+q^{2n}}$$ The sum can be split using numerator and $\sum q^n/(1+q^{2n})$ has a closed form. $\endgroup$
    – Paramanand Singh
    Commented Nov 9, 2020 at 11:37
  • $\begingroup$ Can one solve it without the use of elliptical function? $\endgroup$ Commented Nov 9, 2020 at 12:28
  • $\begingroup$ Well explicit evaluations do require elliptic function theory. I am writing a somewhat lengthy comment as an answer. Do wait and give feedback. $\endgroup$
    – Paramanand Singh
    Commented Nov 9, 2020 at 12:29
  • $\begingroup$ Perhaps you will need some different approach for that integral. Converting it into a sum leads back to the same series of $1/\cosh (n\pi) $. $\endgroup$
    – Paramanand Singh
    Commented Nov 10, 2020 at 2:33

1 Answer 1


This is a long comment (exposition of my comment to the question). Please do respond by updating your question.

The Jacobi two square theorem is equivalent to the following identity $$\left(\sum_{n\in\mathbb {Z}} q^{n^2}\right)^2=1+4\sum_{n\geq 1}\frac{q^n}{1+q^{2n}}\tag{1}$$ The final sum expression for your integral $I$ can be written as $$I=-2\pi\sum_{n\geq 1}\frac{q^n(1-q^{2n})}{1+q^{2n}}=2\pi\sum_{n\geq 1}\left(q^n-\frac{2q^n}{1+q^{2n}}\right)=\frac{2\pi q} {1-q}-4\pi\sum_{n\geq 1}\frac{q^n}{1+q^{2n}}$$ where $q=e^{-\pi} $ (you have a minor sign typo in your question).

Now an explicit evaluation of the above sum is itself based on the equation $(1)$ and the integral becomes $$I=\pi(1-\vartheta_{3}^2(q))+\frac{2\pi q} {1-q}$$ where $\vartheta_3(q)=\sum_{n\in\mathbb {Z}} q^{n^2}$ is a theta function defined by Jacobi.

Explicit evaluation of $\vartheta_{3}(q)$ is possible if $q=\exp(-\pi\sqrt{r}), r\in\mathbb {Q} ^{+} $ via a complicated theorem of Selberg and Chowla. The evaluation is easy and famous for $q=e^{-\pi} $.

However I don't see an explicit evaluation without using the Jacobi two square theorem $(1)$. Can you give some more details about your version of Jacobi two square theorem and some indication of how your arrive at your integral while trying to prove it. A direct algebraic proof of $(1)$ is available on this website.

  • $\begingroup$ I edited my post. Thank you for your contribution so much sir. $\endgroup$ Commented Nov 9, 2020 at 15:07

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