Show that $\int_{-\infty}^\infty \frac{e^x}{e^{2x}+e^{2a}}\frac{1}{x^2+\pi^2}dx = \frac{2\pi e^{-a}}{4a^2+\pi^2}-\frac{1}{1+e^{2a}}$

Question

Show that\begin{align*} \int_{-\infty}^\infty \frac{e^x}{e^{2x}+e^{2a}}\frac{1}{x^2+\pi^2}dx = \frac{2\pi e^{-a}}{4a^2+\pi^2}-\frac{1}{1+e^{2a}} \end{align*}where $a\in \mathbb{R}$.

My small try

Let $\displaystyle f(z) = \frac{e^z}{(e^{2z}+e^{2a})z}$. Let $\Gamma$ be the positively oriented rectangle in the complex plane with vertices $R-i\pi$, $R+i\pi$, $-R+i\pi$ and $-R-i\pi$ where $R>|a|$. There are three first order poles of $f(z)$ lying inside $\Gamma$ at $z= 0, \frac{i\pi}{2}+a$ and $-\frac{i\pi}{2}+a$. Then, by the residue theorem, we have \begin{align*} \int_\Gamma f(z)\; dz &= 2\pi i \left( \mathop{\text{Res}}_{z=0} \; f(z) + \mathop{\text{Res}}_{z=\frac{i\pi}{2}+a} \; f(z) + \mathop{\text{Res}}_{z=-\frac{i\pi}{2}+a} \; f(z)\right) \\ &= 2\pi i \left( \frac{1}{1+e^{2a}} - \frac{i e^{-a}}{2a+i\pi} + \frac{i e^{-a}}{2a-i\pi}\right) \\ &= 2\pi i \left( \frac{1}{1+e^{2a}} - \frac{2\pi e^{-a}}{4a^2 +\pi^2}\right) \tag{1} \end{align*}

Answer 1

Suppose $a\ge 0$. Let $\displaystyle f(z) = \frac{e^z}{(e^{2z}+e^{2a})(z^2+\pi^2)}$. Then $f(z)$ will have poles $$ z=\pi i, z_k=(k+\frac12)\pi i+a,k=0,1,2,\cdots.$$ in the upper half plane.

Set $R_n=\sqrt{(k+1)^2\pi^2+a^2}$. Let $\Gamma_n$ be the union of the upper semi-circle $\gamma_n$ with radius $R_n$ centered at $0$ and the segment from $-R_n$ to $R_n$. Clearly $f(z)$ has poles $$ z=\pi i, z_k=(k+\frac12)\pi i+a,k=0,1,2,\cdots, n $$ and analytic in $\gamma_n$. Then \begin{align*} \int_{-R_n}^{R_n}f(x)dx+\int_{\gamma_n} f(z)\; dz &= 2\pi i \left( \mathop{\text{Res}}_{z=\pi i} \; f(z) + \sum_{k=0}^n\mathop{\text{Res}}_{z=z_k} \; f(z)\right) \\ &= 2\pi i \left( \frac{i}{2\pi(1+e^{2a})}+\sum_{k=0}^n\frac{i (-1)^{k+1}e^{-a}}{2} \frac{1}{z_k^2+\pi^2}\right) \\ &=-\frac{1}{1+e^{2a}}+\pi e^{-a}\sum_{k=0}^n \frac{(-1)^{k}}{z_k^2+\pi^2}. \end{align*} Now letting $n\to\infty$ gives \begin{align*} \int_\infty^\infty f(x)\; dx &=-\frac{1}{1+e^{2a}}+\pi e^{-a}\sum_{k=0}^\infty \frac{(-1)^{k}}{z_k^2+\pi^2}. \end{align*} Note $$ \begin{eqnarray} &&\sum_{k=0}^\infty \frac{(-1)^{k}}{z_k^2+\pi^2}=\sum_{k=0}^\infty \frac{(-1)^{k}}{((k+\frac12)\pi i+a)^2+\pi^2}\\ &=&\frac{1}{\pi^2}\sum_{k=0}^\infty (-1)^{k-1} \frac{1}{(k+(\frac12-\frac{ai}{\pi}))^2-1}\\ &=&\frac{1}{2\pi^2}\sum_{k=0}^\infty (-1)^{k-1}\bigg(\frac{1}{k-\frac12-\frac{ai}{\pi}}-\frac{1}{k+\frac{3}2-\frac{ai}{\pi}}\bigg)\\ &=&-\frac{1}{2\pi^2}\bigg(\frac{1}{-\frac12-\frac{ai}{\pi}}-\frac{1}{\frac12-\frac{ai}{\pi}}\bigg)\\ &=&\frac{2}{4a^2+\pi^2}. \end{eqnarray}$$

Answer 2

Substitute $t = e^x$, along with the shorthand $b= e^a$ \begin{align*} I= &\int_{-\infty}^\infty \frac{e^x}{e^{2x}+e^{2a}}\frac{1}{x^2+\pi^2}dx =\int_0^\infty \frac1{(t^2+b^2)(\ln^2t+\pi^2)}dt\\ =& \int_0^\infty \frac{(1+t)-\frac{2b^2}{1+b^2}(t-\frac1t) }{(t^2+b^2)(\ln^2t+\pi^2) } - \frac{\frac1{1+b^2}}{t( \ln^2t+\pi^2) }\ dt \end{align*} Then, utilize the integral expressions $$\frac{1+t}{\ln^2t+\pi^2}= \frac1\pi\int_0^1 \sin(\pi y)t^{1-y}dy$$ $$\frac{t-\frac1t}{\ln^2t+\pi^2}=\frac2\pi\int_0^1 \sin(2\pi y)t^{1-2y}dy$$ and $ \int_0^\infty \frac1{t(\ln^2t+\pi^2)}dt=1 $ to evaluate \begin{align*} I= & \int_0^\infty \frac1{t^2+b^2} \left(\frac1\pi\int_0^1 \sin\pi y \ {t^{1-y}} -\frac{2b^2\sin2\pi y\ t^{1-2y}}{1+b^2}\ dy\right)dt\\ &\>\>\>\>\> -\frac1{1+b^2}\int_0^\infty \frac1{t(\ln^2t+\pi^2)}dt\\ = & \int_0^1 \int_0^\infty \frac{\sin\pi y \ {t^{1-y}}}{\pi(t^2+b^2)} -\frac{2b^2\sin2\pi y\ t^{1-2y}}{\pi(1+b^2)(t^2+b^2)}\ dt\ dy-\frac1{1+b^2}\\ =& \int_0^1 \cos\frac{\pi y}2 b^{-y}dy - \frac{2b^2}{1+b^2} \int_0^1 \cos\pi y\ b^{-2y}dy - \frac1{1+b^2}\\ =&\ \frac{2\pi+4b\ln b}{b(4\ln^2 b+\pi^2)} -\frac{4\ln b}{4\ln^2 b+\pi^2}- \frac1{1+b^2}\\ =&\ \frac{2\pi}{b(4\ln^2 b+\pi^2)}- \frac1{1+b^2} = \frac{2\pi e^{-a}}{4a^2+\pi^2}-\frac{1}{1+e^{2a}} \end{align*}