TITLE: Fourier series for a logarithm
QUESTION [0 upvotes]: Is there an explicit Fourier sine series for the function $f$ 
defined below (valid for $x\in[0,\pi]$) ?
$$f(x) := \ln\big(\sqrt{1 + \sin x} + \sqrt{\sin x}\big)$$
In case this is well known,  a reference would be
more than enough.
Thanks very much.

REPLY [4 votes]: Since $f$ is only defined over the interval $[0,\pi]$, we must modify the standard Fourier series integral to cover only that range:
$$
f(x)=\sum_k b_k\sin(kx) \\
b_k=\frac 2\pi\int_0^\pi f(x)\sin(kx)\ dx
$$
The factor of 2 comes from the fact that we are effectively integrating over only half the period.
$$
b_k=\frac 2\pi\int_0^\pi\log\left(\sqrt{1+\sin x}+\sqrt{\sin x}\right)\sin(kx)\ dx
$$
I'll start with integration by parts:
$$
\begin{align}
u&= \log\left(\sqrt{1+\sin x}+\sqrt{\sin x}\right) & du&=\frac{\cos x}{2\sqrt{\sin x\left(1+\sin x\right)}}dx\\
dv&= \sin(kx)\ dx & v&=-\cos(kx)/k
\end{align}
$$
$$
b_k=\frac 2\pi\left(\left.-\frac{\log\left(\sqrt{1+\sin x}+\sqrt{\sin x}\right)\cos(kx)}k\right|_{x=0}^\pi+\int_0^\pi\frac{\cos x\cos(kx)}{2k\sqrt{\sin x(1+\sin x)}}dx\right) \\
=0+\frac 1{\pi k}\int_0^\pi\frac{\cos x\cos(kx)}{\sqrt{\sin x(1+\sin x)}}dx
$$
Note that this integral is symmetric around $\pi/2$ if $k$ is odd, and antisymmetric (and therefore $0$) if $k$ is even.
$$
b_{k\textrm{ (even)}}=0\\
b_{k\textrm{ (odd)}}=\frac 2{\pi k}\int_0^{\pi/2}\frac{\cos x\cos(kx)}{\sqrt{\sin x(1+\sin x)}}dx
$$
Although I can't prove it, I found that this evaluates to:
$$
b_{k\textrm{ (odd)}}=\frac{\Gamma(k/2)}{\sqrt\pi k\ \Gamma((k+1)/2)}=\frac{2^{1-k}\ \Gamma(k)}{k\ \Gamma^2((k+1)/2)}
$$
So your series is:
$$
\log\left(\sqrt{1+\sin x}+\sqrt{\sin x}\right)=\sum_{i=0}^\infty\frac{4^{-i}(2i)!}{(2i+1)(i!)^2}\sin((2i+1)x)
$$
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