A definite integral involving a logarithm and trigonometric functions.

Question

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}

Answer 1

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.