# Integral ${\large\int}_0^{\pi/2}\frac{\sin\left(x-a\ln2\cdot\tan x\right)}{\left(e^{2\pi a\tan x}-1\right)\cdot\cos x}\,dx$

Analyzing results Mathematica returned for several integrals and series, and putting pieces together, I came up with this conjecture:
$$\int_0^{\pi/2}\frac{\sin\left(x-a\ln2\cdot\tan x\right)}{\left(e^{2\pi a\tan x}-1\right)\cdot\cos x}\,dx\stackrel{\color{gray}?}=\frac{\operatorname{li}\left(2^a\right)}{2^{a+1}}+\frac{2\,a-1+a\cdot\Re\,{_2F_1}(1,1;a;2)}{4\,a\,(1-a)}\color{gray}{,\,a>0\land a\ne1 }$$
Can we prove this result? It is possible to generalize it to complex values of the parameter $a$?

#### Solutions Collecting From Web of "Integral ${\large\int}_0^{\pi/2}\frac{\sin\left(x-a\ln2\cdot\tan x\right)}{\left(e^{2\pi a\tan x}-1\right)\cdot\cos x}\,dx$"

I only managed to arrive at a closed form for the integral in a different form from that which was proposed in the question.

Preliminary Manipulations:

\begin{align}
\int^\frac{\pi}{2}_0\frac{\sin(x-a\ln{2}\tan{x})}{(e^{2\pi a\tan{x}}-1)\cos{x}}\ {\rm d}x
&=\Im\int^\infty_0\frac{e^{-i(a\ln{2})x}}{(e^{2\pi ax}-1)(1-ix)}\ {\rm d}x\tag1\\
&=\Im\int^0_{-\infty}\frac{e^{-i(a\ln{2})x}}{1-ix}\left(1+\frac{1}{e^{2\pi ax}-1}\right)\ {\rm d}x\tag2\\
&\color{blue}{=\frac{1}{2}\Im\int^\infty_0\frac{e^{i(a\ln{2})x}}{1+ix}\ {\rm d}x}+\color{red}{\frac{1}{2}\Im\int^\infty_{-\infty}\frac{e^{-i(a\ln{2})x}}{(e^{2\pi ax}-1)(1-ix)}\ {\rm d}x}\tag3
\end{align}

Explanation:
$(1)$: Substituted $x\mapsto\arctan{x}$
$(2)$: Substituted $x\mapsto-x$.
$(3)$: Averaged $(1)$ and $(2)$.

Contour Integration: The First Integral

Integrating $\displaystyle f(z)=\frac{1}{2}\frac{e^{i\alpha z}}{1+iz}$ along the contour $[0,R]\cup Re^{i[0,\pi/2]}\cup i[R,0]$,
\begin{align}
\color{blue}{\frac{1}{2}\Im\int^\infty_0\frac{e^{i(a\ln{2})x}}{1+ix}\ {\rm d}x}
&=\Im\left(\color{grey}{\pi i\operatorname*{Res}_{z=i}f(z)}+\frac{i}{2}PV\int^\infty_0\frac{e^{-(a\ln{2})y}}{1-y}\ {\rm d}y\right)\\
&=\left.-\frac{1}{2}e^{-a\ln{2}}\operatorname{Ei}(a\ln{2}(1-y))\right|^\infty_{y=0}=\frac{1}{2}\frac{\operatorname{Ei}(a\ln{2})}{e^{a\ln{2}}}\\
&\color{blue}{=\frac{\operatorname{li}(2^a)}{2^{a+1}}}
\end{align}

Contour Integration: The Second Integral

$\text{Case}\ 1$: $a\in\mathbb{R}^+\backslash\mathbb{Z}^+$

Integrate $\displaystyle g(z)=\frac{1}{2}\frac{e^{-i(a\ln{2})z}}{(e^{2\pi az}-1)(1-iz)}$ along the contour $[-R,R]\cup Re^{i[2\pi,\pi]}$. This (clockwise) contour encloses only simple poles at $z=-i$ and $z=-\frac{in}{a}$. Summing up the residues,
\begin{align}
\color{red}{\frac{1}{2}\Im\int^\infty_{-\infty}\frac{e^{-i(a\ln{2})x}}{(e^{2\pi ax}-1)(1-ix)}\ {\rm d}x}
&=\color{red}{-}\frac{1}{2}\Im\left(\pi i\operatorname*{Res}_{z=0}g(z)+2\pi i\operatorname*{Res}_{z=-i}g(z)+2\pi i\sum^\infty_{n=1}\operatorname*{Res}_{z=-in/a}g(z)\right)\\
&=-\frac{1}{2}\Im\left(\pi i\cdot\frac{1}{2\pi a}+2\pi i\cdot\frac{i}{2^a}\frac{1}{e^{-2\pi ia}-1}+2\pi i\sum^\infty_{n=1}\frac{1}{2^{n+1}\pi(a-n)}\right)\\
&=-\frac{1}{4a}+\frac{\pi\cot(\pi a)}{2^{a+1}}+\frac{1}{2}\sum^\infty_{n=1}\frac{1}{2^n(n-a)}\\
&\color{red}{=\frac{1}{4a}+\frac{\pi\cot(\pi a)}{2^{a+1}}+\frac{1}{2}\Phi\left(\frac{1}{2},1,-a\right)}
\end{align}
where $\Phi$ is the lerch transcendent defined as $\displaystyle\Phi(z,s,\alpha)=\sum^\infty_{n=0}\frac{z^n}{(n+\alpha)^s}$ for $|z|<1$ and $\alpha\notin\mathbb{Z}^-$. (It is implemented in Mathematica or Wolfram as $\text{HurwitzLerchPhi[z,s,a]}$, not $\text{LerchPhi[z,s,a]}$).

$\text{Case}\ 2$: $a\in\mathbb{Z}^+$

In this case, the pole at $z=-i$ is order $2$. Applying the residue theorem once again,
\begin{align}
\color{red}{\frac{1}{2}\Im\int^\infty_{-\infty}\frac{e^{-i(a\ln{2})x}}{(e^{2\pi ax}-1)(1-ix)}\ {\rm d}x}
&=-\frac{1}{2}\Im\left(\pi i\cdot\frac{1}{2\pi a}+2\pi i\cdot\frac{\ln{2}-\pi i}{2^{a+1}\pi}+2\pi i\sum^\infty_{\substack{n=1\\n\neq a}}\frac{1}{2^{n+1}\pi(a-n)}\right)\\
&=\frac{1}{4a}-\frac{\ln{2}}{2^{a+1}}+\frac{1}{2}\sum^\infty_{\substack{n=0\\n\neq a}}\frac{1}{2^{n}(n-a)}\\
&=\frac{1}{4a}\color{grey}{-\frac{\ln{2}}{2^{a+1}}}+\frac{1}{2}\sum^{a-1}_{n=0}\frac{1}{2^{n}(n-a)}+\color{grey}{\frac{1}{2}\sum^\infty_{n=a+1-(a+1)}\frac{1}{2^{n+(a+1)}(n+(a+1)-a)}}\\
&\color{red}{=\frac{1}{4a}+\frac{1}{2}\sum^{a-1}_{n=0}\frac{1}{2^{n}(n-a)}}
\end{align}

The Final Result:

Combining both blue and red integrals gives
\begin{align}
\int^\frac{\pi}{2}_0\frac{\sin(x-a\ln{2}\tan{x})}{(e^{2\pi a\tan{x}}-1)\cos{x}}\ {\rm d}x
=
\begin{cases}
\displaystyle\small{\frac{\mathrm{li}(2^a)}{2^{a+1}}+\frac{1}{4a}+\frac{\pi\cot(\pi a)}{2^{a+1}}+\frac{1}{2}\Phi\left(\frac{1}{2},1,-a\right)}, &\text{if } a\in\mathbb{R}^+\backslash\mathbb{Z}^+\\
\displaystyle\small{\frac{\mathrm{li}(2^a)}{2^{a+1}}+\frac{1}{4a}+\frac{1}{2}\sum^{a-1}_{n=0}\frac{1}{2^{n}(n-a)}}, &\text{if } a\in\mathbb{Z}^+
\end{cases}
\end{align}