Prove $\int_{-\infty}^{\infty} \frac{e^{ax}}{e^{x}+1} dx = \frac{\pi}{\sin a\pi}$ ($0 < a < 1$)

Question

The question is from Complex Variables by Levinson and Redheffer and is in Chapter 4, Section 3, Problem Q7.

Notice that $$\frac{\pi}{\sin a\pi} = \frac{2\pi i}{e^{ia\pi}-e^{-ia\pi}} = 2\pi i\frac{e^{-ia\pi}}{1-e^{-2ia\pi}} = 2\pi i \sum_{k=1}^{\infty} e^{-i(2k+1)\pi}\qquad (*)$$

If I let C be the semicircle with radius equal to R with the bottom lying on [$-R, R$], then as $R \rightarrow \infty$, I am tempted to write $e^z + 1$ in the form of $g(x)(x - i\pi)(x - i3\pi)\dots$ where $g(x)$ is a function that does not vanish on any $i(2k + 1)\pi, k \geq 0$. If this idea is correct, then $\Large \int_{C} \frac{e^{az}}{e^z + 1} dz$ will be equal to ($\large *$) but then I do not know how to show that $\Large \int_{C_R} \frac{e^{az}}{e^z + 1} dz$ vanish where $C_R$ is the arc of the semicircle mentioned above.

The hint given in the textbook says integrating $\Large \frac{e^{az}}{e^z + 1}$ over the rectangle with vertices at $-R, R, R+2i\pi, -R+2i\pi$ will help. In this case, I can write $e^z + 1$ in the form of $h(z)(z - i\pi)$ where $h(z)$ is an analytic function that does not vanish at $i\pi$. Then I do not know how to derive the result on the right.

Answer 1

Choosing the upper half-plane counterclockwise oriented as our integration path and $f(z)=\frac{e^{az}}{e^z+1} $, we have that:

$$\oint f(z)dz=\int_{-R}^Rf(z)dz+\int_{\Gamma_R}f(z)dz$$ 1) Applying the ML Inequality: $$\left|\int_{\Gamma_R}f(z)dz\right|≤ \lim_{R\rightarrow \infty} \int_0^{\pi}\left|\frac{e^{aRe^{it}}}{e^{Re^{it}}+1}\right||iRe^{it}dt|≤\pi \lim_{R\rightarrow \infty}\frac{Re^{Ra}}{e^R}\rightarrow0$$

2) Computing the residues: $$\oint f(z)dz=2\pi i \sum_{k=0}^\infty \lim_{z\rightarrow i\pi(2k+1)}\frac{[z-i\pi(2k+1)]e^{az}}{e^z+1}=2\pi i \sum_{k=0}^\infty \frac{e^{ai\pi(2k+1)}}{e^{i\pi(2k+1)}}$$

$$\oint f(z)dz=-2\pi i \sum_{k=0}^\infty {e^{ai\pi(2k+1)}}=-2\pi i \frac{e^{ai\pi}}{1-e^{2ai\pi}}=\frac{\pi}{sin\left(\pi a\right)}$$

Hence: $$\int_{-\infty}^{\infty}\frac{e^{az}}{e^z+1}dz=\frac{\pi}{sin\left(\pi a\right)}$$

You could also prove this using the Beta Function: $$\int_{-\infty}^{\infty}\frac{e^{az}}{e^z+1}dz\overbrace{=}^{e^z=x}\int_{0}^{\infty}\frac{x^{a-1}}{x+1}dx=\mathfrak{B}(a,1-a)=\frac{\Gamma(a)\Gamma(1-a)}{\Gamma(1)}$$ Using the Reflection Formula, you will conclude that: $$\int_{-\infty}^{\infty}\frac{e^{az}}{e^z+1}dz=\frac{\pi}{sin\left(\pi a\right)}$$

Answer 2

Assume that $\Im(a) > 0,\Re(a)\in (0,1)$ so that $$\int_{-\infty}^\infty \frac{e^{ax}}{e^x+1}dx=\lim_{R\to \infty} \int_{|z|=R,\Im(z)\ge 0}\frac{e^{az}}{e^z+1}dz=2i\pi \sum Res(...)$$ Then extend to $a\in (0,1)$ by continuity.