A definite integral involving a logarithm and trigonometric functions. The question is to compute a definite integral below. I have expressed it as a series expansion in powers of $r$ by expanding the integrand in a series and integrating term by term. Then I input the series into Mathematica and Mathematica returned the neat closed form expression in the last line.
I do not understand how this last closed form expression is obtained. Could anybody enlighten me on that? Thanks.
\begin{eqnarray}
\int\limits_0^{2 \pi} \log\left( 1+r \sin(\phi)\right) d \phi = \\
\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n} r^n \left(1+(-1)^n\right) \frac{\Gamma(1/2) \Gamma(\frac{1+n}{2})}{\Gamma(1+\frac{n}{2})}=\\
2 \pi \log\left[\frac{1}{2} \left(1+\sqrt{1-r^2}\right)\right]
\end{eqnarray}
 A: Note first that if $n$ is odd, then the summand is zero. So this is effectively equivalent to the following sum over even numbers
\begin{eqnarray}
-\Gamma(1/2)\sum\limits_{n=1}^\infty \frac{1}{n} r^{2n}  \frac{ \Gamma(1/2+n)}{\Gamma(1+n)}
\end{eqnarray}
Next use
$$
\Gamma(n+1/2)=\frac{\sqrt{\pi } (2 n)!}{4^n n!}
$$
and $\Gamma(n+1)=n!$
to deduce that your sum equals
$$
-\Gamma(1/2)\sum\limits_{n=1}^\infty \frac{1}{n} (r/2)^{2n}  \sqrt{\pi} \frac{(2n)!}{(n!)^2}\qquad (\star)\ .
$$
Then use the known Taylor series for the arcsine
$$
\arcsin(x) = \sum\limits_{n=0}^\infty \cfrac{(2n!)x^{2n+1}}{(2^nn!)^2(2n+1)}
$$
and differentiate both sides
$$
\frac{1}{\sqrt{1-x^2}}=\sum\limits_{n=0}^\infty \cfrac{(2n!)x^{2n}}{(2^nn!)^2}\qquad (\star\star) .
$$
Then use the identity
$$
\frac{1}{n}=\int_0^\infty ds\ e^{-ns}\qquad n>0\ ,
$$
in $(\star)$ to write for your series
$$
-\Gamma(1/2)\sqrt{\pi}\int_0^\infty ds\sum\limits_{n=1}^\infty (r e^{-s/2}/2)^{2n}   \frac{(2n)!}{(n!)^2}
$$
and equating
$$
\frac{x}{2}=r e^{-s/2}/2\ ,
$$
we recognize the sum $(\star\star)$ inside the integrand. Therefore your sum equals
$$
-\Gamma(1/2)\sqrt{\pi}\int_0^\infty ds\left[\frac{1}{\sqrt{1-(r e^{-s/2})^2}}-1\right]=\pi[2 \log \left(\sqrt{1-r^2}+1\right)-\log(4)]\ ,
$$
which is equivalent to your final expression.
