# How to calculate $\int_0^\pi \ln(1+\sin x)\mathrm dx$

How to calculate this integral $$\int_0^\pi \ln(1+\sin x)\mathrm dx$$ I didn't find this question in the previous questions. With the help of Wolframalpha I got an answer $-\pi \ln 2+4\mathbf{G}$, where $\mathbf{G}$ denotes Catalan's Constant.

First we rewrite $\log(1+\sin(x))=\log(2)+2\log(\sin(x/2+\pi/4))$. This yields, after shifting $x/2+\pi/4\rightarrow y,dx\rightarrow 2 dy$

$$I=\pi\log(2)+4\underbrace{\int_{\pi/4}^{3\pi/4}\log(\sin(y))dy}_{J}$$

Now we might employ the Fourier series of $\log(\sin(x))$ to calculate $J$

$$J=-\log(2)\int_{\pi/4}^{3\pi/4}dy-\sum_{n=1}^{\infty}\frac{1}{k}\int_{\pi/4}^{3\pi/4}\cos(2yk)dy=\\ -\frac{\pi}{2}\log(2)+\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}$$

where we used $\sin(\frac{3\pi}{2} k)-\sin(\frac{\pi}{2} k)=(-1)^{k}-(-1)^{k+1}=2(-1)^k$ in the second line.

Employing the series representation of the Catalan constant we find

$$J=-\frac{\pi}{2}\log(2)+G$$

and therefore

$$I=\pi \log(2)+J=-\pi \log(2)+4G \quad (*)$$

Just for fun, lets see what contour integration can do. We again use $J$ rewriting it by the help of the identity $\log(\sin(x))=\log(i/2)-ix+\log(1-e^{2 i x})$. The integral over the first two terms is trivial leaving us with

$$J=-\frac{\pi}{2}\log(2)+\underbrace{\int_{\pi/4}^{3\pi/4}\log(1-e^{2 i x})}_{K}$$

to evaluate $K$ we integrate the complex valued function

$$f(z)=\log(1-e^{2 i z})$$

over an rectangle $C$ in the complex plane with verticies $(\pi/4,3\pi/4,3\pi/4+i R,\pi/4+i R)$ where in the end we want to take the limit $R\rightarrow +\infty$. As one can easily check, the contribution from the top of the rectangle vanishs (the integrand vanishs as $\mathcal{O}(e^{-2R})$ for big $R$). Furthermore the integrand is holomorphic inside the contour of integration and therefore we can express the integral of interest in terms of the vertical pieces of the contour

$$\int_C f(z)dz=K+i\int_{0}^{\infty}\log(1+ie^{-2y})dy-i\int_{0}^{\infty}\log(1-ie^{-2y})dy=0$$

this can be simplified to

$$K=2\int_0^{\infty}\arctan(e^{-2y})dy=\int_0^{\infty}\arctan(e^{-y})dy$$

using the series repesentation of arctan, we may easily conclude that this integral is equal to $G$ and therefore

$$K=G$$

from which it follows that

$$J=-\frac{\pi}{2}\log(2)+K=-\frac{\pi}{2}\log(2)+G$$

from which (*) follows immediatly

• The value of $J$ is $-\frac{\pi}{2}\ln 2+G$. And maybe the second line of your answer is not correct. – Renascence_5. Feb 25 '16 at 17:41
• @EvilNebula. Ok, was a tiny bit too fast in the beginning...now everything should be fine! Thx – tired Feb 25 '16 at 17:52
• @EvilNebula there is also a possibiliy to integrate this by contour methods, just in case u are interested – tired Feb 25 '16 at 18:14
• I tried some other approaches, but this one was the more efficient than any other one. So, a big +1 ... should be a +2 ... Mark – Mark Viola Feb 25 '16 at 20:40
• Tired. For the contour integral, the initial limits are from $\pi/4$ to $3\pi/4$, yet you had written $0$ to $\infty$. I hope you don't mind my editing it. - Mark – Mark Viola Feb 25 '16 at 22:44