Show that, for $0<\alpha<1$: $$\int_{-\infty}^{\infty} \frac{e^{\alpha x}}{{e^x+1}}\text{d}x=\frac{\pi}{\sin(\pi\alpha)}.$$ Hint: Use the recangular path $S_r=\left[-r, r, r+2\pi, -r+2\pi,-r \right]$ (see fig. 1).
My attempt: Denoting $f(z)=\frac{e^{\alpha x}}{{e^x+1}}$, we notice that $f$ has only simple poles at $z=\pi i+ 2\pi i\mathbb{Z}$. Using the suggested path, we have
$$ \oint_{S_r} f(z) \ \text{d}z=\int_{-r}^{r} \frac{e^{\alpha x}}{{e^x+1}}\text{d}x+\int_{[r+2\pi i,-r+2\pi i]}f(z) \ \text{d} z+\int_{[r,r+2\pi i]}f(z) \ \text{d} z+\int_{[-r+2\pi i, -r]}f(z) \ \text{d}z. $$
The integrals on the right and left paths, denoted $\gamma_1$ and $\gamma_2$ in fig. 1, become insignificant as $r\to\infty$: $$ \left|\int_{\gamma_1} f(z) \ \text{d}z \right|\le \max_{z\in[r,r+2\pi i]} \left| \frac{e^{\alpha r}e^{i 2\pi\alpha t}}{e^r e^{i2\pi t}+1} \right|\ell(\gamma_1)\le \frac{e^{\alpha r}}{e^r+1}\cdot 2\pi\xrightarrow[r\to\infty]{}0. $$
A similar argument applies for $\gamma_2$. But what do I do with the path $\Gamma_1$?
Just parametrize it. This is the path $\Gamma_1:[-r,r]\to\mathbb{C}$ defined by $\Gamma_1(t)=-t+2\pi i$. The derivative if $\Gamma_1'(t)=-1$. So:
$\int_{\Gamma_1} f=\int_{-r}^r \frac{e^{\alpha(-t+2\pi i)}}{e^{-t+2\pi i}+1} (-dt)=\{u=-t\}=-\int_{-r}^r \frac{e^{\alpha u}e^{\alpha 2\pi i}}{e^ue^{2\pi i}+1}du=-e^{2\pi \alpha i}\int_{-r}^r \frac{e^{\alpha u}}{e^u+1}du\to$
$\to -e^{2\pi\alpha i}\int_{-\infty}^\infty f(u)du$