Integral of $\log|x| \sqrt{4 - x^2}$ How can I prove that $$ \frac{1}{2\pi} \int_{-2}^2 \log|x| \sqrt{4 - x^2} dx = - \frac{1}{2}$$? I have tried the substitution $x = 2\sin \theta$, which leads to $$ \frac{2}{\pi} \int_{-\pi/2}^{\pi/2} \log|2\sin\theta| \cos^2\theta d\theta.$$ But I don't know how to proceed from here.
 A: By parity the given integral equals
$$ \frac{1}{\pi}\int_{0}^{2}\log(x)\sqrt{4-x^2}\,dx\stackrel{x\mapsto 2z}{=}\color{blue}{\frac{4}{\pi}\int_{0}^{1}\log(z)\sqrt{1-z^2}\,dz}+\frac{1}{2}\log 2 $$
and by integration by parts the blue integral turns into
$$ -\frac{2}{\pi}\int_{0}^{1}\left(\sqrt{1-z^2}+\frac{\arcsin z}{z}\right)\,dz =-\frac{2}{\pi}\left(\frac{\pi}{4}+\int_{0}^{\pi/2}\frac{\theta\cos\theta}{\sin\theta}d\theta\right).$$
Since $\int_{0}^{\pi/2}\frac{\sin(2m\theta)\cos\theta}{\sin\theta}\,d\theta=\frac{\pi}{2}$, by expanding $\theta$ as a Fourier sine series we get:
$$ I = \frac{1}{\pi}\int_{0}^{2}\log(x)\sqrt{4-x^2}\,dx = \color{red}{-\frac{1}{2}}$$
as wanted. As an alternative:
$$\begin{eqnarray*} \int_{0}^{1}\log(z)\sqrt{1-z^2}\,dz &\stackrel{z\mapsto\sqrt{t}}{=}&\frac{1}{4}\int_{0}^{1}t^{-1/2}(1-t)^{1/2}\log(t)\,dt\\ &=& \frac{1}{4}\cdot\frac{d}{d\alpha}\left.\left(\frac{\Gamma\left(\frac{3}{2}\right)\,\Gamma\left(\alpha\right)}{\Gamma\left(\frac{3}{2}+\alpha\right)}\right)\right|_{\alpha=1/2}\end{eqnarray*}$$
and since $\frac{df}{d\alpha}=f\cdot\frac{d}{d\alpha}\log f$ the RHS equals
$$ \frac{\pi}{8} \left(\psi\left(\frac{1}{2}\right)-\psi(2)\right)=-\frac{\pi}{8}(1+2\log 2). $$
A: Noting
$$ \int\sqrt{4 - x^2} dx=\frac12x\sqrt{4-x^2}+2\arcsin(\frac x2)+C$$
one has
\begin{eqnarray}
\int_{-2}^2 \log|x| \sqrt{4 - x^2} dx&=&2\int_0^2 \log x \sqrt{4 - x^2} dx\\
&=&2\int_0^2 \log x d\bigg[\frac12x\sqrt{4-x^2}+2\arcsin(\frac x2)\bigg]\\
&=&2\bigg\{\log x \bigg[\frac12x\sqrt{4-x^2}+2\arcsin(\frac x2)\bigg]\bigg|_0^2-\int_0^2 \frac1x \bigg[\frac12x\sqrt{4-x^2}+2\arcsin(\frac x2)\bigg]dx\bigg\}\\
&=&2\bigg\{\pi\log 2 -\int_0^2 \bigg[\frac12\sqrt{4-x^2}+\frac2x\arcsin(\frac x2)\bigg]dx\bigg\}\\
&=&2\bigg\{\pi\log 2 -\frac\pi2-2\int_0^2\frac1x\arcsin(\frac x2)dx\bigg\}\\
&=&2\bigg\{\pi\log 2 -\frac\pi2-2\int_0^1\frac1x\arcsin xdx\bigg\}.
\end{eqnarray}
Since
\begin{eqnarray}
\int_0^1\frac1x\arcsin xdx&=&\int_0^1\arcsin xd\ln x\\
&=&\arcsin x\ln x|_0^1-\int_0^1\frac{\ln x}{\sqrt{1-x^2}}dx\\
&=&-\int_0^{\pi/2}\ln\sin xdx\\
&=&\frac12\pi\ln 2
\end{eqnarray}
from here, it is easy to obtain $I=-\frac12$.
