Evaluate the integral $\int_{\mathbb{R}}\frac{e^{\gamma x}}{(1+e^x)}dx$.

Question

I'd like to compute the integral

$$\int_{-\infty}^{+\infty}\frac{e^{\gamma x}}{1+e^x}dx$$ for some $0<\gamma <1.$

First thing that I tried with is the Residue Theorem, but the denominator $1+e^z$ has infinite poles...

Thank you for any hint.

Answer 1

With substitution $\dfrac{1}{1+e^x}=u$ we have $$\int_{-\infty}^{+\infty}e^{\gamma x}(1+e^x)^{-1}dx=\int_0^1(1-u)^{\gamma-1}u^{-\gamma}du=B(1-\gamma,\gamma)=\Gamma(\gamma)\Gamma(1-\gamma)=\dfrac{\pi}{\sin\pi\gamma}$$ for $0<\gamma <1$. See here for details about beta function.