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.

  • $\begingroup$ $f(x)$ is not a function but a number ! $\endgroup$ – idm Dec 30 '14 at 20:34
  • $\begingroup$ I edited the post so there is now no mistakes, but if you want to understand my remark, you can see the original post. $\endgroup$ – idm Dec 30 '14 at 20:44
  • $\begingroup$ de gustibus non est disputandum $\endgroup$ – user2052 Mar 1 '15 at 20:40

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: $$ \log\left(\sqrt{1+\sin x}+\sqrt{\sin x}\right)=\sum_{i=0}^\infty\frac{4^{-i}(2i)!}{(2i+1)(i!)^2}\sin((2i+1)x) $$

  • 1
    $\begingroup$ thanks for the speedy reply - notice that the right hand side is the imaginary part of arcsin(e^ix) $\endgroup$ – user2052 Dec 31 '14 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.