Solving the integral $\int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s$ For $t \in [0,1]$, let
$$f(t) = \int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s.$$
Is it possible to evaluate this integral to obtain an explicit expression for $f(t)$? Mathematica and Maple both fail to give an answer.
 A: Let us set $t=\sin\varphi$ for $\varphi\in\left[0,\frac{\pi}{2}\right]$ and $s=\sin\theta$ for $\theta\in\left[0,\frac{\pi}{2}\right]$. We have to evaluate
$$ \int_{0}^{\pi/2}\frac{\sin\theta\log\sin\theta-\sin\varphi\log\sin\varphi}{\sin^2\theta-\sin^2\varphi}\,d\theta $$
and we may use the Fourier serie of $\log\sin$
$$ \log\sin(x) = -\log(2)-\sum_{n\geq 1}\frac{\cos(2n x)}{n}, $$
$$\sin(x)\log\sin(x) = -\log(2)\sin(x) - \sum_{n\geq 1}\frac{\sin((1+2n)x)-\sin((2n-1)x)}{2n} $$
to write the previous integral as a reasonably simple series of integrals.  The first term is
$$ -\log(2)\int_{0}^{\pi/2}\frac{d\theta}{\sin\theta+\sin\varphi} = \frac{\log 2}{\cos\varphi}\log\tan\frac{\varphi}{2}=\frac{1}{\sqrt{1-t^2}}\,\log\left(\frac{2t}{1+\sqrt{1-t^2}}\right)$$
and the other terms depend on
$$ J(n,\varphi) = \int_{0}^{\pi/2}\frac{\sin((2n+1)\theta)-\sin((2n+1)\varphi)}{\sin^2\theta-\sin^2\varphi}\,d\theta.$$
A: Split the integral into two
\begin{align}
I=& \int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds=I_1+I_2
\end{align}
where, with $b = \sqrt{1-t^2}$
\begin{align}
I_1= & \int_{0}^{1} \frac{s \ln{s} - s \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds
\overset{x=\sqrt{1-s^2}}=\frac12 \int_{0}^{1}
 \frac{\ln\frac{1-x^2}{1-b^2}}{b^2-x^2}dx\\
=&\ \frac1{4b}\bigg[\text{Li}_2\left(\frac{b-1}{b+1}\right)-\text{Li}_2\left(\frac{b+1}{b-1}  \right)\bigg]\\
\\
I_2= & \int_{0}^{1} \dfrac{s \ln{t} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds\\
=& \int_{0}^{1} \dfrac{\ln{t}}{(s +t)\sqrt{1-s^{2}}}ds=\frac{2\ln t}{\sqrt{1-t^2}}\tanh^{-1}\sqrt{\frac{1-t}{1+t}}\\
\end{align}
Thus
\begin{align}
I=& \frac1{4\sqrt{1-t^2}}\left[ \text{Li}_2\bigg(\frac{\sqrt{1-t^2}-1}{\sqrt{1-t^2} +1}\bigg)-\text{Li}_2\bigg(\frac{\sqrt{1-t^2} +1}{\sqrt{1-t^2}-1}  \bigg)
+ 8\ln t\tanh^{-1}\sqrt{\frac{1-t}{1+t}}\right]
\end{align}
