Compute the integral $$\int_{-\infty}^\infty \frac{e^x}{1 + e^{4x}}dx$$
My attempt: I tried applying residue integration. There are four simple poles in the upper half plane: $z \in \{\pi/4, 3\pi/4, 5\pi/4, 7\pi/4\}$. However evaluating all of these singularities and summing them gives the value $0$ due to symmetry, so I get that the integral evaluates to $0$ which is incorrect. Not sure what I'm missing in my approach.
Any help is appreciated.
As @2 is even prime proceeds, that method is a bit complicated, imo. Here is a similar alternative.
Without separating the integral, we substitute $e^x=t$.
$$I = \int\limits_0^\infty \frac{\mathrm dt}{1+t^4}$$
Now we substitute $t\mapsto \frac1t $.
$$I = \int\limits_0^\infty \frac{t^2}{1+t^4}\,\mathrm dt$$
Adding the two equations,
$$\begin{align} I &= \frac12 \int\limits_0^\infty\frac{1+t^2}{1+t^4}\,\mathrm dt \\ &= \frac12 \int\limits_0^\infty \frac{1+\frac1{t^2}}{(t-\frac1t)^2+2}\,\mathrm dt \\ &\overset{t-1/t =y}{=} \frac12 \int\limits_{-\infty}^\infty \frac{\mathrm dy}{y^2+2} \\ &= \left. \frac1{2\sqrt 2}\arctan \Big(\frac y{\sqrt 2}\Big) \right|_{-\infty}^\infty \\ I &= \frac\pi{2\sqrt2} \end{align}$$
$$\int_{-\infty}^{\infty}\frac {e^x}{1 + e^{4x}}dx$$ Substitute: $ e^x = \sqrt{\tan\theta} \implies e^xdx = \frac {\sec^2{\theta}d\theta}{2\sqrt{\tan\theta}}$
$$\begin{align*} \frac12\int_0^{\frac\pi2} \sin^{-\frac12}\theta\cos^{\frac12}\theta d\theta & = \frac12\frac12\beta(\frac14, \frac34)\\ & = \frac14\frac\pi{\sin(\frac\pi4)} \end{align*}$$ Which is the Beta function
$$\int_{-\infty}^\infty \frac{e^x}{1 + e^{4x}}dx$$ Split the integral at $0$ and bring them between same bounds i.e $[0,\infty]$. $$\int_{0}^{\infty}\frac{e^{3x}+e^{x}}{e^{4x}+1} dx$$ Substitute $u=e^{x}\implies dx=\frac{du}{u}$. $$\int_{1}^{\infty}\frac{u^2+1}{u^{4}+1} du$$ Substitute $k=u^2\implies du=\frac{dk}{2\sqrt{k}}$. $$\frac{1}{2}\int_{1}^{\infty}\frac{\sqrt{k}+\frac{1}{\sqrt{k}}}{k^2+1} dk$$ Put, $ k=\frac{1}{t} \implies dk=\frac{-dt}{t^2} $ $$\int_{0}^{1}\frac{\sqrt{t}+\frac{1}{\sqrt{t}}}{t^{2}+1} dt$$ This step allows us to use formula for infinite geometric sums but the problem is at the boundary condition. To solve that take a limit to $1$ from negative as follows, $$\lim_{a\to 1^{-}} \int_{0}^{a}\frac{\sqrt{t}+\frac{1}{\sqrt{t}}}{t^{2}+1} dt$$ Write, $\frac{1}{t^2+1}=\sum_{n\ge 0}t^{2n}$. $$\lim_{a\to 1^{-}} \sum_{n\ge 0}\int_{0}^{a} t^{2n-\frac{1}{2}}+t^{2n+\frac{1}{2}} dt$$ I think you can proceed from here.
We start with the substitution $e^x=t$. This yields
$$I = \int\limits_0^\infty\frac{\mathrm dt}{1+t^4} $$
Now using $t^4=y$,
$$\begin{align}I &= \frac14\int\limits_0^\infty \frac{ y^{1/4-1}}{1+y}\,\mathrm dy \\ &= \frac14\mathrm B(1/4,3/4) \\ &= \frac{ \Gamma(1/4)\Gamma(3/4)}{\Gamma(1)}\\ &= \frac{\pi}{4\sin\frac{\pi}4} \\ I &= \frac{\pi}{2\sqrt2}\end{align}$$
$$ \begin{aligned} I &=\int_{-\infty}^{\infty} \frac{d y}{1+y^{4}}, \quad \text{where } y=e^{x} \\ &=\int_{-\infty}^{\infty} \frac{\frac{1}{y^{2}}}{y^{2}+\frac{1}{y^{2}}} d y \\ &=\frac{1}{2} \int_{-\infty}^{\infty} \frac{\left(1+\frac{1}{y^{2}}\right)-\left(1+\frac{1}{y^{2}}\right)}{y^{2}+\frac{1}{y^{2}}} d y \\ &=\frac{1}{2} \int_{-\infty}^{\infty} \frac{d\left(y-\frac{1}{y}\right)}{\left(y-\frac{1}{y}\right)^{2}+2}-\frac{1}{2} \int_{-\infty}^{\infty} \frac{d\left(y+\frac{1}{y}\right)}{\left(y+\frac{1}{y}\right)^{2}-2} \\ &=\frac{1}{2 \sqrt{2}}\left[\tan ^{-1}\left(\frac{y-\frac{1}{y}}{\sqrt{2}}\right)\right]_{-\infty}^{\infty}-\frac{1}{2 \sqrt{2}}\left[\ln \left| \frac{y+\frac{1}{y}-\sqrt{2}}{y+\frac{1}{y}+\sqrt{2}}\right|\right]_{-\infty}^{\infty}\\ &=\frac{\pi}{2 \sqrt{2}} \end{aligned} $$
