Prove that $\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x=-\frac{\pi}{2}\,\ln(2)$. I have discovered via contour integration that
$$\int_0^\infty\,\frac{\exp(t\,u)}{\exp(u)+1}\,\text{d}u={\text{csc}(\pi\,t)}\,\left(\frac{\pi}{2}-\int_0^{\frac{\pi}{2}}\,\frac{\sin\big((1-2t)\,y\big)}{\sin(y)}\,\text{d}y\right)\tag{*}$$
for all $t\in\mathbb{C}\setminus\mathbb{Z}$ such that $\text{Re}(t)<1$.  By taking $t\to 0$, I deduce that
$$\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u=\frac{2}{\pi}\,\int_0^{\frac{\pi}{2}}\,y\,\cot(y)\,\text{d}y\,.$$
With a step of integration by parts, I obtain
$$\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u=-\frac{2}{\pi}\,\int_0^{\frac{\pi}{2}}\,\ln\big(\sin(y)\big)\,\text{d}y\,.$$
Setting $x:=\sin(y)$, I get
$$\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u=-\frac{2}{\pi}\,\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x\,.$$
This shows that
$$\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x=-\frac{\pi}{2}\,\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u\,.$$
The integral $\displaystyle\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u$ can be easily obtained since
$$\int\,\frac{1}{\exp(u)+1}\,\text{d}u=u-\ln\big(\exp(u)+1\big)+\text{constant}\,.$$
That is, I have
$$\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x=-\frac{\pi}{2}\,\ln(2)\,.\tag{#}$$
However, this proof is a very roundabout way to verify the equality above.  Is there a more direct way to prove that (#) is true?  Any technique is appreciated.

A nice consequence of (*) is that
$$\int_0^\infty\,\frac{\sinh(t\,u)}{\exp(u)+1}\,\text{d}u=\frac{\pi}{2}\,\text{csc}(\pi\,t)-\frac{1}{2\,t}$$
for all $t\in\mathbb{C}\setminus\{0\}$ such that $\big|\text{Re}(t)\big|<1$.  This provides a proof that
$$\eta(2r)=\frac{1}{(2r-1)!}\,\int_0^\infty\,\frac{u^{2r-1}}{\exp(u)+1}\,\text{d}u=\frac{\pi^{2r}}{2}\,\Biggl(\left[t^{2r-1}\right]\Big(\text{csc}(t)\Big)\Biggr)$$
for $r=1,2,3,\ldots$.  Here, $\eta$ is the Dirichtlet eta function.  In addition, $[t^k]\big(g(t)\big)$ denotes the coefficient of $t^k$ in the Laurent expansion of $g(t)$ about $t=0$.  This also justifies the well known results that
$$\eta(2r)=\frac{\left(2^{2r-1}-1\right)\,\big|B_{2r}\big|\,\pi^{2r}}{(2r)!}\text{ and }\zeta(2r)=\frac{2^{2r-1}\,\big|B_{2r}\big|\,\pi^{2r}}{(2r)!}$$
for $r=1,2,3,\ldots$, where $\left(B_j\right)_{j\in\mathbb{Z}_{\geq0}}$ is the sequence of Bernoulli numbers and $\zeta$ is the Riemann zeta function.

Similarly,
$$\begin{align}\int_0^\infty\,\frac{\exp(t\,u)-1}{\exp(u)-1}\,\text{d}u&=\ln(2)+2\,\int_0^{\frac{\pi}{2}}\,\frac{\sin\big((1-t)\,y)\,\sin(t\,y)}{\sin(y)}\,\text{d}y\\
&\phantom{aaaaa}-\cot(\pi\,t)\,\left(\frac{\pi}{2}-\int_0^{\frac{\pi}{2}}\,\frac{\sin\big((1-2t)\,y\big)}{\sin(y)}\,\text{d}y\right)\,,\end{align}$$
for all $t\in\mathbb{C}\setminus\mathbb{Z}$ such that $\text{Re}(t)<1$.
This gives
$$\int_0^\infty\,\frac{\sinh(t\,u)}{\exp(u)-1}\,\text{d}u=\frac{1}{2\,t}-\frac{\pi}{2}\,\cot(\pi\,t)$$
for all $t\in\mathbb{C}\setminus\{0\}$ such that $\big|\text{Re}(t)\big|<1$.

Another consequence of (*) is that
$$\int_0^{\frac{\pi}{2}}\,\frac{\sin(k\,y)}{\sin(y)}\,\text{d}y=\frac{\pi}{2}\,\text{sign}(k)$$
for all odd integers $k$.  It is an interesting challenge to determine the integral $\displaystyle \int_0^{\frac{\pi}{2}}\,\frac{\sin(k\,y)}{\sin(y)}\,\text{d}y$ for all even integers $k$.
 A: By letting $x=\sin(t)$, and by using the symmetry $\sin(\pi/2-t)=\cos(t)$, we get
$$\begin{align}I&=\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,dx=\int_0^{\pi/2} \ln (\sin t)\, dt
=\frac{1}{2}\left(\int_0^{\pi/2} \ln (\sin t)\, dt+\int_0^{\pi/2} \ln (\cos t)\, dt\right)\\&=\frac{1}{2}\left(\int_0^{\pi/2} \ln(\sin(2t))dt - \int_0^{\pi/2} \ln(2)dt\right)=\frac{I}{2}-\dfrac{\pi}4 \ln(2)\end{align}$$
and the result easily follows.
A: I would start with
$$ J(a) = \int_0^1 \frac{x^a \; dx}{\sqrt{1-x^2}}, \ a > -1$$
The substitution $x = \sqrt{t}$ gives you
$$ J(a) = \frac{1}{2} \int_0^1 t^{(a-1)/2} (1-t)^{-1/2}\; dt = \frac{B((a+1)/2,1/2)}{2} = \frac{\Gamma(a+1/2) \Gamma(1/2)}{\Gamma(a+1)} $$
where $B$ is the Beta function.
Then your integral is $$J'(0) = - \frac{-\pi \ln(2)}{2}$$ 
A: Take $x = \sin t$ then $dx = \cos t \ dt $ so
\begin{equation}
 \frac{\ln (x) }{\sqrt{1 - x^2}} \ dx = \frac{\ln (\sin t) }{\cos t} \cos t \ dt = \ln (\sin t) \ dt
\end{equation}
SO the integral becomes
\begin{equation}
 A
 =
 \int\limits_{0}^{\frac{\pi}{2}}
 \ln (\sin t) \ dt
\end{equation}
or just name $x$ instead of $t$
\begin{equation}
 A
 =
 \int\limits_{0}^{\frac{\pi}{2}}
 \ln (\sin x) \ dx
\end{equation}
Now use the following change of variable
\begin{equation}
 t = \frac{\pi}{2} - x
\end{equation}
We get
\begin{equation}
 A 
 =
 -
  \int\limits_{\frac{\pi}{2}}^{0}
 \ln (\sin (\frac{\pi}{2} - t)) \ dt
 =
\int\limits_{0}^{\frac{\pi}{2}}
 \ln (\cos t) \ dt
\end{equation}
This means 
\begin{equation}
 2A = 
 \int\limits_{0}^{\frac{\pi}{2}}
 \ln (\sin x) \ dx
 +
 \int\limits_{0}^{\frac{\pi}{2}}
 \ln (\cos x) \ dx
 =
 \int\limits_{0}^{\frac{\pi}{2}}
 [
 \ln (\sin x) \ dx
 +
 \ln (\cos x) \ dx
 ] 
\end{equation}
But $\ln ab = \ln a + \ln b$ so 
\begin{equation}
 2A
 =
 \int\limits_{0}^{\frac{\pi}{2}}
 \ln (\sin x \cos x) \ dx
 =
 \int\limits_{0}^{\frac{\pi}{2}}
 \ln (\frac{1}{2} \sin 2x) \ dx =
 \int\limits_{0}^{\frac{\pi}{2}}
 \ln (\frac{1}{2}) + \ln( \sin 2x) \ dx
\end{equation}
this means that
\begin{equation}
 2A
 =
 \frac{\pi}{2}
 \ln \frac{1}{2}
 +
 B
\end{equation}
where $B =\int\limits_{0}^{\frac{\pi}{2}}
 \ln( \sin 2x) \ dx $
\begin{equation}
 B
 =
\int\limits_{0}^{\frac{\pi}{4}}
 \ln( \sin 2x) \ dx
 +
\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}
 \ln( \sin 2x) \ dx 
\end{equation}
Let $ t = 2x - \frac{\pi}{2}$ so 
\begin{equation}
 B
 =
\int\limits_{0}^{\frac{\pi}{4}}
 \ln( \sin 2t) \ dx
 +
\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}
 \ln( \sin 2(t + \frac{\pi}{2})) \ dx 
 =
 \frac{1}{2}
 \int\limits_{0}^{\frac{\pi}{2}}
 \ln(\sin(t)) \ dt 
 +
 \frac{1}{2}
 \int\limits_{0}^{\frac{\pi}{2}}
 \ln(\cos(t)) \ dt  
\end{equation}
which is
\begin{equation}
 B
  =
 \frac{1}{2}
 A
 +
 \frac{1}{2}
 A
 =
 A
\end{equation}
So
\begin{equation}
 2A = \frac{\pi}{2}\ln \frac{1}{2}
 +
 A
\end{equation}
So
\begin{equation}
 A = \frac{\pi}{2}\ln \frac{1}{2}
\end{equation}
A: \begin{align}
\int_0^1 x^\alpha\ dx &= \dfrac{x^{\alpha+1}}{\alpha+1}\\
\dfrac{d}{d\alpha}\int_0^1 x^\alpha\ dx &=\dfrac{d}{d\alpha}\dfrac{x^{\alpha+1}}{\alpha+1}\\
\int_0^1 x^\alpha\ln x\ dx &=\dfrac{x^{\alpha+1}((\alpha+1)\ln x-1)}{(\alpha+1)^2}\Big|_0^1=\dfrac{-1}{(\alpha+1)^2}
\end{align}
\begin{align}
I
&= \int_{0}^{1} \frac{\ln x}{\sqrt {1-x^2}}dx \\
&= \int_0^1\ln x\sum_{n=0}^\infty{2n\choose n}\dfrac{1}{4^n}x^{2n} dx\\
&= \sum_{n=0}^\infty{2n\choose n}\dfrac{1}{4^n}\dfrac{-1}{(2n+1)^2} \\
&= \color{blue}{-\dfrac{\pi}{2}\ln2}
\end{align}
using generator function
$$\sum_{n=0}^\infty{2n\choose n}\dfrac{1}{4^n}\dfrac{x^{2n+1}}{(2n+1)^2}=\ln\dfrac{1+\sqrt{1-x^2}}{x}-\sqrt{1-x^2}$$
A: Riemann sums are enough. For any $n\in\mathbb{N}^+$ we have
$$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n},\tag{1} $$
hence
$$ \int_{0}^{\pi/2}\log\sin\theta\,d\theta=\frac{1}{2}\int_{0}^{\pi}\log\sin\theta\,d\theta = \frac{\pi}{2}\lim_{n\to +\infty}\frac{\log(2n)-n\log 2}{n} = \color{red}{-\frac{\pi}{2}\log 2}.\tag{2}$$
A: The antiderivative
$$I=\int\,\frac{\ln(x)}{\sqrt{1-x^2}}\,dx$$ can be computed (a CAS gave it).
For $\color{red}{2i I}$, the obtained result is
$$\text{Li}_2\left(e^{-2 i \sin ^{-1}(x)}\right)+2 \log (x) \log \left(\sqrt{1-x^2}+i
   x\right)+\sin ^{-1}(x)^2-2 i \sin ^{-1}(x) \log \left(1-e^{-2 i \sin
   ^{-1}(x)}\right)$$
$$\lim_{x\to 1} \, I=-\frac{i \pi }{12}  (\pi -6 i \log (2))$$
$$\lim_{x\to 0} \, I=-\frac{i \pi ^2}{12}$$
A: \begin{align}
I=\int_0^1 \frac{\ln x}{\sqrt{1-x^2}} dx
& \overset{x^2\to x}=\frac14 \int_0^1 \frac{\ln x}{\sqrt{x(1-x)}} dx
\overset{x\to 1-x}= \frac14\int_0^1 \frac{\ln (1-x)}{\sqrt{x(1-x)}} dx\\
&= \frac18 \int_0^1 \frac{\ln (x(1-x)) }{\sqrt{x(1-x)}} dx \overset{symmetry}=\frac14 \int_0^{\frac12} \frac{\ln (x(1-x)) }{\sqrt{x(1-x)}} dx \\
&\overset{x(1-x)=\frac t4}=\frac12I -\frac{\ln2}4\int_0^1 \frac{dt}{\sqrt{t(1-t)}}= \frac12I- \frac\pi4 \ln2\\
\end{align}
which yields $I=-\frac{\pi}{2}\ln 2$.
