Finding value of $\displaystyle \int^{\pi}_{0}\ln(1+k\cos x)dx$ for $0<k<1$
what I try
Let $\displaystyle I =\int^{\pi}_{0}\ln(1+k\cos x)dx$
put $\displaystyle x\rightarrow \frac{\pi}{2}-x$
$\displaystyle I=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\ln(1-k\sin x)dx$
$\displaystyle I =\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\bigg[k\sin x-\frac{k^2\sin^2 x}{2}+\frac{k^3\sin^3 x}{3}-\cdots \bigg]dx$
$\displaystyle I =-2\int^{\frac{\pi}{2}}_{0}\bigg[\frac{k^2\sin^2 x}{2}+\frac{k^4\sin^4 x}{4}+\cdots \bigg]dx$
$\displaystyle I =-\pi\bigg[\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots \bigg]$
How can I find sum of that series. Help me please.
Best Answer
For any $R\geq 1$,
$$(R+e^{i\theta})(R+e^{-i\theta}) = (R^2+1)+2R\cos\theta \tag{1}$$ $$ 1+\frac{2R}{R^2+1}\cos\theta = \frac{R^2}{R^2+1}\left(1+\frac{e^{i\theta}}{R}\right)\left(1+\frac{e^{-i\theta}}{R}\right)\tag{2}$$ $$ \log\left(1+\tfrac{2R}{R^2+1}\cos\theta\right) = \log\left(\tfrac{R^2}{R^2+1}\right)+2\sum_{n\geq 1}\frac{(-1)^{n+1}\cos(n\theta)}{n R^n}\tag{3} $$ and for any $n\in\mathbb{N}^+$ we have $\int_{0}^{\pi}\cos(n\theta)\,d\theta=0$, therefore $$ \int_{0}^{\pi}\log\left(1+\tfrac{2R}{R^2+1}\cos\theta\right)\,d\theta = \pi\log\left(\tfrac{R^2}{R^2+1}\right).\tag{4}$$ Now it is enough to enforce the substitution $\frac{2R}{R^2+1}=\kappa$.