A curious integral from @infseriesbot


       I was today looking my twitter when suddenly I came across this curious looking integral from the @infseriesbot:



It immediately caught my attention to try proving it. Coicidently, it applies many techniques from the two previous posts, among them the integral representation of the digamma function


\boxed{\psi(x)=\int_{0}^{\infty} \frac{e^{-t}}{t}-\frac{e^{-x t}}{1-e^{-t}} d t}


       

Lets start!

I=\int_{0}^{\pi / 2} \frac{\tan (x)}{e^{\tan x}-1} d x=-\frac{1}{2}\left(\ln (2 \pi)+\pi+\psi\left(\frac{1}{2 \pi}\right)\right)

First consider the substitution

u=\tan (x) \Longrightarrow d u=\sec ^{2}(x) d x

using the relation

\color{red}\boxed{ {\sec ^{2} (x) =1+\tan ^{2}(x) }}

we get

d x=\frac{d u}{\sec ^{2}(x)}=\frac{d u}{1+u^{2}}

I=\int_{0}^{\pi / 2} \frac{\tan (x)}{e^{\tan x}-1} d x=\int_{0}^{\infty}\frac{u}{e^u-1}\cdot\frac{1}{1+u^2}du

now, from a previous post we already know that

\color{red}{\frac{u}{1+u^2}=\int_{0}^{\infty}e^{-t}\sin(ut)dt}

and the integral can be rewritten as

I=\int_{0}^{\infty}\frac{1}{e^u-1}\cdot \color{red}{\int_{0}^{\infty}e^{-t}\sin(ut)dt} \,\,\color{black}du

I=\int_{0}^{\infty}e^{-t}\cdot \color{red}{\int_{0}^{\infty}\frac{\sin(ut)}{e^u-1}du} \,\,\color{black}dt

I=\int_{0}^{\infty}e^{-t}\cdot \color{red}{\int_{0}^{\infty}\frac{e^{-u}\sin(ut)}{1-e^{-u}}du} \,\,\color{black}dt(1)

Lets concetrate in the inner integral in red

\color{red}{\int_{0}^{\infty}\frac{e^{-u}\sin(ut)}{1-e^{-u}}du}=\int_{0}^{\infty}\sin(ut) e^{-u}\sum_{k=0}^{\infty}e^{-ku}du

\color{red}=\int_{0}^{\infty}\sin(ut) \sum_{k=0}^{\infty}e^{-(k+1)u}du

\color{red}=\int_{0}^{\infty}\sin(ut) \sum_{k=1}^{\infty}e^{-ku}du

\color{red}=\sum_{k=1}^{\infty}\int_{0}^{\infty}e^{-ku}\sin(ut) du

Again from the previous post, we know the result for the integral

\color{red}\boxed{{\int_{0}^{\infty}\frac{e^{-u}\sin(ut)}{1-e^{-u}}du}=\sum_{k=1}^{\infty}\frac{t}{k^2+t^2}}

The sum on the right hand side was evaluated in another post, equation (5), therefore we can write

\color{red}\sum_{k=1}^{\infty}\frac{t}{k^2+t^2}=\frac{\pi \coth(\pi t)}{2}-\frac{1}{2t}

Plugging back this result in (1)

I=\int_{0}^{\infty}e^{-t}\cdot \color{red}{\Big[\frac{\pi \coth(\pi t)}{2}-\frac{1}{2t} \Big]} \,\,\color{black}dt

I= - \frac{1}{2}\int_{0}^{\infty}e^{-t}\Big( \frac{1}{t}-\pi \coth( \pi t)\Big)dt

Now, make the substitution \pi t = w

I= - \frac{1}{2}\int_{0}^{\infty}e^{-\frac{w}{\pi}}\Big( \frac{\pi}{w}-\pi \coth( w)\Big)\frac{dw}{\pi}

I= - \frac{1}{2}\int_{0}^{\infty}e^{-\frac{w}{\pi}}\Big( \frac{1}{w}- \coth( w)\Big){dw}(2)

Now, from the definition of \coth(w)

\color{red}\coth(w)=\frac{e^{w}+e^{-w}}{e^{w}-e^{-w}}

\color{red}\coth(w)=\frac{e^{-w}}{e^{-w}} \cdot\frac{e^{w}+e^{-w}}{e^{w}-e^{-w}}

\color{red}\coth(w)=\frac{1+e^{-2w}}{1-e^{-2w}}

Plugging this last result in (2)

I= - \frac{1}{2}\int_{0}^{\infty}e^{-\frac{w}{\pi}}\Big( \frac{1}{w}-\color{red}\frac{1+e^{-2w}}{1-e^{-2w}} \color{black}\Big){dw}

Lets do another substitution, w=\frac{x}{2}

I= - \frac{1}{2}\int_{0}^{\infty}e^{-\frac{x}{2 \pi}}\Big( \frac{2}{x}-\frac{1+e^{-x}}{1-e^{-x}} \Big)\frac{dx}{2}

I= - \frac{1}{2}\int_{0}^{\infty}e^{-\frac{x}{2 \pi}}\Big( \frac{1}{x}-\frac{1}{2} \cdot \frac{1+e^{-x}}{(1-e^{-x})} \Big){dx}

I=- \frac{1}{2}\int_{0}^{\infty}\left\{\frac{e^{-\frac{x}{2 \pi}}}{x}-\frac{e^{-\frac{x}{2 \pi}}+e^{-\frac{x}{2 \pi}}e^{-x}}{2(1-e^{-x})}\right\} {dx}

I=- \frac{1}{2}\int_{0}^{\infty}\left\{\frac{e^{-\frac{x}{2 \pi}}\color{red}{+e^{-x}-e^{-x}}}{x}-\frac{e^{-\frac{x}{2 \pi}}+\color{red}{e^{-\frac{x}{2 \pi}}-e^{-\frac{x}{2 \pi}}}\color{black}+e^{-\frac{x}{2 \pi}}e^{-x}}{2(1-e^{-x})}\right\} {dx}

We now rewrite this integral as the sum of three integrals:

I=\left\{- \frac{1}{2}\int_{0}^{\infty}\frac{e^{-x}}{x}-\frac{2e^{-\frac{x}{2 \pi}}}{2(1-e^{-x})}dx+ \int_{0}^{\infty}\frac{e^{-\frac{x}{2 \pi}}-e^{-\frac{x}{2 \pi}}e^{-x}}{2(1-e^{-x})}dx+\int_{0}^{\infty}\frac{e^{-\frac{x}{2 \pi}}-e^{-x}}{x}dx\right\}

I=\left\{- \frac{1}{2}\int_{0}^{\infty}\frac{e^{-x}}{x}-\frac{e^{-\frac{x}{2 \pi}}}{(1-e^{-x})}dx+ \frac{1}{2}\int_{0}^{\infty}{e^{-\frac{x}{2 \pi}}}dx-\int_{0}^{\infty}\frac{e^{-x}-e^{-\frac{x}{2 \pi}}}{x}dx\right\}

It was already proved in this post the first and the third integrals. The first integral is an integral representation of    \psi(\frac{1}{2 \pi}) , the Digamma function, and the third is Frullani´s integral representation of     \log(\frac{1}{2 \pi}).

The second integral is straightforward:

\frac{1}{2}\int_{0}^{\infty}{e^{-\frac{x}{2 \pi}}}dx=\frac{1}{2}\Big[-2 \pi e^{-\frac{x}{2 \pi}} \Big|_{0}^{\infty}\Big]= \pi

Putting all together we finally get that

\boxed{I=- \frac{1}{2}\int_{0}^{\infty}e^{-\frac{w}{\pi}}\Big( \frac{1}{w}- \coth( w)\Big){dw}=-\frac{1}{2}\left(\ln (2 \pi)+\pi+\psi\left(\frac{1}{2 \pi}\right)\right)}

and

\boxed{I=\int_{0}^{\pi / 2} \frac{\tan (x)}{e^{\tan x}-1} d x=-\frac{1}{2}\left(\ln (2 \pi)+\pi+\psi\left(\frac{1}{2 \pi}\right)\right)}

Ricardo Albahari

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