TITLE: Contour Integral $ \int_{0}^1 \frac{\ln{x}}{\sqrt{1-x^2}} \mathrm dx$
QUESTION [8 upvotes]: I need help evaluating this with contour integration
$$
\int_{0}^{1}{\ln\left(\,x\,\right)\over
\,\sqrt{\vphantom{\large A}\,1 - x^{2}\,}}\,{\rm d}x
$$
I am not sure as to how to work with the branch cuts of both
$\ln\left(\,x\,\right)$ and $\sqrt{1 - x^{2}}$
Second part is to evaluate
$$ \int_{0}^1 \frac{\sqrt{\,\vphantom{\large a}\ln\left(\,x\,\right)}}
{\sqrt{\vphantom{\large A}\,1 - x^{2}\,}} \mathrm dx$$

REPLY [3 votes]: For your second integral, we can use Ramanujan's method which I mentioned here. That is, let $x = e^{-t}$. Then we have
\begin{align}
\int_0^1\frac{\log^{1/2}(x)}{\sqrt{1-x^2}}dx &= \int_0^{\infty}\frac{ie^{-t}\sqrt{t}}{\sqrt{1-e^{-2t}}}dt\\
&=i\int_0^{\infty}\sqrt{t}\sum_{n=0}^{\infty}\frac{(1/2)_ne^{-(2n+1)t}}{n!}dt\tag{1}\\
&=i\sum_{n=0}^{\infty}\frac{(1/2)_n}{n!}\int_0^{\infty}t^{1/2}e^{-(2n+1)t}dt\tag{2}\\
&=\frac{i\sqrt{\pi}}{2}\sum_{n=0}^{\infty}\frac{(1/2)_n}{n!(2n+1)^{3/2}}
\end{align}
$(1)$ occurs because $\sum_{n=0}^{\infty}\frac{(1/2)_ne^{-(2n+1)t}}{n!}=\frac{e^{-t}}{\sqrt{1-e^{-2t}}}$ and $(1/2)_n$ is the rising Pochhammer; that is, $(1/2)_0 = 1$ and $(1/2)_n = 1/2(1/2+1)\cdots(1/2+n-1)$. For $(2)$, we have that
$$
\int_0^{\infty}t^{1/2}e^{-(2n+1)t}dt = \frac{\Gamma(3/2)}{(2n+1)^{3/2}}=\frac{\sqrt{\pi}}{2(2n+1)^{3/2}}
$$