Evaluating the sum of the series $\sum_{n=1}^\infty \frac{1}{{((2 n - 1) \pi)^2 - m^2}}$ manually.

Question

This sum $\sum_{n=1}^\infty \frac{1}{{((2 n - 1) \pi)^2 - m^2}}$ came up while I was trying to prove that $\displaystyle\int_{0}^{\infty} \frac{e^{mx}-e^{-mx}}{e^{\pi x}-e^{-\pi x}}dx=\frac{1}{2}\tan{\frac{m}{2}}$.

Wolfram Mathematica gives the answer $\displaystyle\sum_{n=1}^\infty \frac{1}{{((2 n - 1) \pi)^2 - m^2}}=\frac{1}{4m}\tan\frac{m}{2}$, $m>0$.

$\mathbf{Question \ 1}$: Is there a simple proof of the equation regarding the definite integral?

My attempt:

I have tried doing $\displaystyle\frac{e^{mx}-e^{-mx}}{e^{\pi x}-e^{-\pi x}}={(1-e^{-2mx})}{(1-e^{-2\pi x})^{-1}e^{(m-\pi)x}}={(1-e^{-2mx})}{(1+e^{-2\pi x}+e^{-4\pi x}+e^{-6\pi x}+...)e^{(m-\pi)x}}$. Then, taking the integral from $0$ to $\infty$ of the product and then the series comes up.

$\mathbf{Question \ 2}$: Is there any way to evaluate the sum manually?

I could not come up with anything for this one.

Any help would be much appreciated. Thank you.

Answer 1

Since you have access to Mathematica, you could try the partial sum $$S_p=\sum_{n=1}^p \frac{1}{{((2 n - 1) \pi)^2 - m^2}}$$

I suppose that you will obtain $$S_p=\frac{\psi \left(p-\frac{m}{2 \pi }+\frac{1}{2}\right)-\psi \left(p+\frac{m}{2 \pi }+\frac{1}{2}\right)-\psi \left(\frac{1}{2}-\frac{m}{2 \pi }\right)+\psi \left(\frac{m}{2 \pi }+\frac{1}{2}\right)}{4 \pi m}$$ which can write $$S_p=\frac{\psi \left(p-\frac{m}{2 \pi }+\frac{1}{2}\right)-\psi \left(p+\frac{m}{2 \pi }+\frac12\right)+\pi \tan \left(\frac{m}{2}\right)}{4 \pi m}$$

Now, for large $p$, $$\psi \left(p-\frac{m}{2 \pi }+\frac{1}{2}\right)-\psi \left(p+\frac{m}{2 \pi }+\frac12\right)=-\frac{m}{\pi p}+\frac{(\pi ^2 -m^2)m}{12 \pi ^3 p^3}+O\left(\frac{1}{p^5}\right)$$ make $$S_p=\frac{1}{4 m}\tan \left(\frac{m}{2}\right)-\frac{1}{4 \pi ^2 p}+\frac{(\pi ^2 -m^2)}{48 \pi ^4 p^3}+O\left(\frac{1}{p^5}\right)$$

For illustartion purposes, using $m=3$, some results $$\left( \begin{array}{ccc} p & \text{approximation} & \text{exact} \\ 1 & 1.149974019 & 1.149948182 \\ 2 & 1.162476429 & 1.162475360 \\ 3 & 1.166681785 & 1.166681633 \\ 4 & 1.168788661 & 1.168788624 \\ 5 & 1.170053758 & 1.170053745 \\ 6 & 1.170897474 & 1.170897469 \\ 7 & 1.171500257 & 1.171500255 \\ 8 & 1.171952405 & 1.171952404 \\ 9 & 1.172304107 & 1.172304106 \\ 10 & 1.172585485 & 1.172585485 \end{array} \right)$$

Answer 2

$$\int_0^\infty \frac{e^{mx}-e^{-mx}}{e^{\pi x}-e^{-\pi x}}dx=\sum_{n=0}^\infty\int_0^\infty\left(e^{-[(2n+1)\pi-m]x}-e^{-[(2n+1)\pi+m]x}\right)dx $$ $$\sum_{n=0}^\infty\left(\frac{1}{(2n+1)\pi-m}-\frac{1}{(2n+1)\pi+m}\right)=2m\sum_{n=0}^\infty\left(\frac{1}{(2n+1)^2\pi^2-m^2}\right)$$

$$m\sum_{n=-\infty}^\infty\left(\frac{1}{(2n+1)^2\pi^2-m^2}\right)=-m\pi \sum Res \left( \frac{cot(\pi z)}{(2z+1)^2\pi^2-m^2}\right)=\frac{tan \left(\frac{m}{2}\right)}{4m}$$

For further information search for Residue Theorem. https://en.wikipedia.org/wiki/Residue_theorem