How to prove this identity involving Lerch function? Trying to find another way to answer this question, I arrived at a point where I cannot simplify anymore.
The solution being
$$I_1=\int_0^{\frac \pi 2}\csc (x) \log \left(\frac{1+b \sin (x)}{1-b \sin (x)}\right)=\pi  \sin ^{-1}(b)$$
I am stuck with
$$I_2=\frac 12\Bigg[a_+ \Phi
   \left(a_+^2,2,\frac{1}{2}\right)+a_- \Phi
   \left(a_-^2,2,\frac{1}{2}\right)\Bigg]\quad \text{where}\quad a_\pm=b\pm\sqrt{b^2-1}$$ $\Phi(.)$ being the Lerch transcendent function.
How to prove that, for $\color{red}{-1 \le b \le 1}$, $\color{red}{I_2=I_1}$ ?
 A: As $-1\le b\le 1$, we define
\begin{equation}
 b=\cos\beta
\end{equation}
with $0\le\beta\le\pi$. Then
\begin{align}
 a_{\pm}&=b\pm\sqrt{b^2-1}\\
 &=e^{\pm i\beta}
\end{align}
From the definition of the Lerch transcendent function,
\begin{equation}
\Phi(z,s,a)=\sum_{k=0}^\infty\frac{z^k}{\left[(a+k)^2\right]^{s/2}} 
\end{equation}
for $|z|<1 \bigvee \left(|z|=1\bigwedge \Re s>1\right)\bigwedge -a\notin \mathbb{N}$,
we have
\begin{align}
 I_2&=\frac 12\left[a_+ \Phi
   \left(a_+^2,2,\frac{1}{2}\right)+a_- \Phi
   \left(a_-^2,2,\frac{1}{2}\right)\right]\\
   &=\frac12\left[e^{i\beta}\Phi\left(e^{2i\beta},2,\frac12\right)+e^{-i\beta}\Phi\left(e^{-2i\beta},2,\frac12\right)\right]\\
   &=4\sum_{k=0}^\infty\frac{\cos\left((2k+1)\beta\right)}{(2k+1)^2}
\end{align}
Using the classical Fourier series,
\begin{equation}
 |\beta|=\frac\pi2-\frac{4}{\pi}\sum_{k=0}^\infty\frac{\cos\left((2k+1)\beta\right)}{(2k+1)^2}
\end{equation}
we deduce
\begin{align}
 I_2&=\pi\left( \frac\pi2-\beta \right)\\
 &=\pi\sin^{-1}(b)
\end{align}
as expected.
