<s>[INST] How to calculate $\int_{-a}^{a} \sqrt{a^2-x^2}\ln(\sqrt{a^2-x^2})\mathrm{dx}$ Well,this is a homework problem.
I need to calculate the differential entropy of random variable 
$X\sim f(x)=\sqrt{a^2-x^2},\quad -a<x<a$ and $0$ otherwise. Just how to calculate
$$
\int_{-a}^a \sqrt{a^2-x^2}\ln(\sqrt{a^2-x^2})\,\mathrm{d}x
$$
I can get the result with Mathematica,but failed to calculate it by hand.Please give me some idea.
 [/INST] This integral can be performed via differentiation under the integral sign. First note that for $|x|\leq1$ we have $\ln \sqrt{1-x^2} = \frac12\ln (1-x^2)$. Moreover, simple application of the chain rule yields
$$ \frac{d}{d\alpha} (1-x^2)^\alpha = (1-x^2)^\alpha \ln(1-x^2) .$$
The remaining integral is a special case of the beta function with $x=1/2$ and $y=\alpha+1$. Thus, we have
$$\int_0^1\!dx\,(1-x^2)^\alpha = 
\frac12\int_0^1\!dy\,y^{1/2} (1-y)^\alpha= \frac{\sqrt{\pi} \Gamma(1+\alpha)}{2 \Gamma(\frac{3}{2} + \alpha)}.$$
The original integral, we obtain by taking the derivative with respect to $\alpha$ and afterwards setting $\alpha=1/2$;
$$
\begin{align}\int_{-a}^a\!dx\,\sqrt{a^2-x^2} \ln\sqrt{a^2-x^2}
&=a^{2} \int_{0}^1\!dx\,\sqrt{1-x^2} [\ln a^2 + \ln(1-x^2)]\\
&= \frac{a^2 \pi \log a}{2}+ a^{2} \frac{d}{d\alpha} \int_0^1\!dx\,(1-x^2)^\alpha \Big|_{\alpha=1/2}\\
&= \frac{a^2 \pi \log a}{2}+ a^{2} \frac{d}{d\alpha} \frac{\sqrt{\pi} \Gamma(1+\alpha)}{2 \Gamma(\frac{3}{2} + \alpha)}  \Big|_{\alpha=1/2} \\
&= \frac{a^2 \pi \log a}{2}+ a^{2}  \frac{\Gamma(1+\alpha)}{\Gamma(3/2+\alpha)} [\psi(1+\alpha)-\psi(\tfrac{3}{2}+\alpha)]\Big|_{\alpha=1/2} \\
&=\frac{a^2 \pi \log a}{2}+  \frac{a^{2}\pi}4(1-2 \ln 2). 
\end{align}$$
Where we used the special values of the $\Gamma$ and the $\psi = (\log \Gamma)'$ at integer and half-integer values.
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