Calculating $\text{PV}\int_{-\infty}^{+\infty}\frac{e^{\alpha x}}{e^{2x}-1}\mathrm d x$

I am trying to show that for $0<\alpha<2,$ $$\text{PV}\left (\int_{-\infty}^{+\infty} \frac{e^{\alpha x}}{e^{2x}-1}\mathrm d x\right )=-\frac{\pi}{2}\cot\left (\frac{\alpha\pi}{2} \right )\tag{\star}$$ to gain some familiarity with the concept of Principal Value.

My attempts

• First of all I started by expanding the integral. Let $R>0$ be a positive real number. Then \begin{align} \int_{-R}^R\frac{e^{\alpha x}}{e^{2x}-1}\mathrm dx&= \int_{-R}^R e^{\alpha x}\left (\sum_{n\ge 1}e^{-2nx}\right )\mathrm dx=\sum_{n\ge 1}\int_{-R}^R e^{(\alpha-2n)x}\mathrm d x \\ &=\sum_{n\ge 1}\left .\frac{e^{(\alpha-2n)x}}{\alpha-2n} \right |^{x=R}_{x=-R}=\sum_{n\ge 1}\frac{1}{\alpha-2n}\left ( e^{(\alpha-2n)R} -e^{-(\alpha-2n)R}\right) \end{align} but I don't know how to continue from here. I tried evaluating the series through complex analytic methods but I was not successful.

• I tried substituting $e^{2x}=u$. The integral becomes $$\int_{-R}^R\frac{e^{2x\frac{\alpha}{2}}}{e^{2x}-1}\mathrm dx=\frac{1}{2}\int_{e^{-2R}}^{e^{2R}}\frac{u^{\frac{\alpha}{2}-1}}{u-1}\mathrm du$$ but this doesn't look very promising. I was unable to manipulate this expression to evaluate the integral.

Question: How can I evaluate the principal value $(\star)$?

• have you proved that your integral does converge? – Dr. Sonnhard Graubner Jun 29 '17 at 14:27
• Do you know contour integration/residue calculus? – icurays1 Jun 29 '17 at 14:30
• @icurays Yes, I do. – Lonidard Jun 29 '17 at 14:31
• @Dr. Sonnhard Graubner I have not, but Mathematica computed its value and I took existence for granted. – Lonidard Jun 29 '17 at 14:32

$$\begin{eqnarray*}\text{PV}\int_{-\infty}^{+\infty}\frac{e^{\alpha x}}{e^{2x}-1}\,dx &=& \int_{0}^{+\infty}\left(\frac{e^{\alpha x}}{e^{2x}-1}+\frac{e^{-\alpha x}}{e^{-2x}-1}\right)\,dx\\&=&\int_{0}^{+\infty}\frac{\sinh((\alpha-1)x)}{\sinh x}\,dx\\(\text{De Moivre})&=&\sum_{n\geq 0}\left(\frac{1}{2-\alpha+2n}-\frac{1}{\alpha+2n}\right)\\(\text{Herglotz})&=&-\frac{\pi}{2}\cot\left(\frac{\pi \alpha}{2}\right)\end{eqnarray*}$$ since the meromorphic functions $\sum_{n\geq 0}\left(\frac{1}{2-\alpha+2n}-\frac{1}{\alpha+2n}\right)$ and $-\frac{\pi}{2}\cot\left(\frac{\pi \alpha}{2}\right)$ have simple poles at the same points with the same residues. Herglotz' trick is summarized here.
Note that \eqalign{\int_{|x|>\epsilon}\frac{e^{ax}}{e^{2x}-1}dx&= \int_{\epsilon}^\infty\frac{e^{ax}}{e^{2x}-1 }dx+\int_{-\infty}^{-\epsilon}\frac{e^{ax}}{e^{2x}-1}dx\cr &=\int_{\epsilon}^\infty\frac{e^{ax}}{e^{2x}-1 }dx+\int_\epsilon^{\infty}\frac{e^{-ax}}{e^{-2x}-1}dx\cr &=\int_{\epsilon}^\infty\frac{e^{ax}-e^{(2-a)x}}{e^{2x}-1 }dx } The integrand on the right can be extended by continuity at $x=0$ so $$V.P \int_{-\infty}^{\infty}\frac{e^{ax}}{e^{2x}-1}dx=\lim_{\epsilon\to0}\int_{|x|>\epsilon}\frac{e^{ax}}{e^{2x}-1}dx=\int_{0}^\infty\frac{e^{ax}-e^{(2-a)x}}{e^{2x}-1 }dx$$ Now, the last integral can be evaluated as follows: \eqalign{\int_{0}^\infty\frac{e^{ax}-e^{(2-a)x}}{e^{2x}-1 }dx&=\int_{0}^\infty(e^{ax}-e^{(2-a)x})\left(\sum_{n=1}^\infty e^{-2n x}\right)dx\cr &=\sum_{n=1}^\infty\left(\frac{1}{2n-a}-\frac{1}{2n+a-2}\right)\cr &=-\frac1a-\sum_{n=1}^\infty\frac{a}{a^2-4n^2}=-\frac{\pi}{2}\cot\frac{\pi a}{2} }
$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{P.V.}\int_{-\infty}^{\infty}{\expo{\alpha x} \over \expo{2x} - 1}\,\dd x & = \int_{0}^{\infty}\pars{% {\expo{\alpha x} \over \expo{2x} - 1} + {\expo{-\alpha x} \over \expo{-2x} - 1}} \,\dd x = \int_{0}^{\infty} {\expo{-\pars{1 - \alpha/2}2x} - \expo{-\pars{\alpha/2}2x} \over 1 - \expo{-2x}}\,\dd x \\[5mm] & \stackrel{x\ =\ -\ln\pars{t}/2}{=}\,\,\, \int_{1}^{0}{t^{1 - \alpha/2} - t^{\alpha/2} \over 1 - t} \,\pars{-\,{1 \over 2t}}\dd t \\[5mm] & = -\,{1 \over 2}\pars{\int_{0}^{1}{1 - t^{-\alpha/2} \over 1 - t}\,\dd t - \int_{0}^{1}{1 - t^{-1 + \alpha/2} \over 1 - t}\,\dd t} \\[5mm] & = -\,{1 \over 2}\pars{H_{-\alpha/2} - H_{\alpha/2 - 1}}\qquad \pars{~H_{z}:\ Harmonic Number~} \\[5mm] & = -\,{1 \over 2}\bracks{\pi\cot\pars{\pi\,{\alpha \over 2}}}\qquad \pars{~Euler\ Reflection\ Formula~} \\[5mm] & = \bbx{-\,{\pi \over 2}\,\cot\pars{\alpha\pi \over 2}} \end{align}
• @Lonidard Thanks. They are very useful. Many exponential like integrals can be reduced to an integration over $\left(0,1\right)$. We don't need to "reinvent the wheel". – Felix Marin Jun 29 '17 at 22:32