# $\int_{0}^{\pi }{\ln \left( 1-2a\cos x+{{a}^{2}} \right)\cos \left( nx \right)dx}$

This integral seems to me not easy to figure it out

$$\int_{0}^{\pi }{\ln \left( 1-2a\cos x+{{a}^{2}} \right)\cos \left( nx \right)dx}$$ for $$n=1,2,3,...$$ and $$a\in\mathbb{R}$$. I start thinking in using this $$f(x)=f(\pi-0+x)$$ with integrals for both sides but it never help me at all , then I switch to induction to obtain some formula also failed . As a result, is there any shortcut to crack this exercise ?

• Differentiation under the integral – Aditya Garg Mar 26 '19 at 20:23
• could you try letting $1=\sin^2(x)+\cos^2(x)$ and seeing if this helps to factorise at all? – Henry Lee Mar 26 '19 at 21:36

Assume $$|a|<1$$. Then \begin {align} \log(1-2a\cos x+a^2) &= \log((1-ae^{ix})(1-ae^{-ix}))\\ &= \log{(1-ae^{ix})}+\log(1-ae^{-ix}) \\ &= -\sum_{k=1}^{\infty} \frac{1}{k}a^k (e^{ikx}+e^{-ikx})\\ &= -2\sum_{k=1}^{\infty} \frac{a^k}{k} \cos{kx}.\tag1\end {align}
Observe that by Dirichlet's test the series $$(1)$$ converges for $$|a|=1$$ as well except for the following points: $$\begin{cases} a=1,& x=2n\pi;\\ a=-1,&x=(2n+1)\pi, \end{cases}\quad n\in\mathbb Z.$$ This divergence at a set of a measure $$0$$ however does not matter for the subsequent integration, so that $$|a|\le1$$ will be assumed.
Thus: $$\int_0^\pi\log(1-2a\cos x+a^2)\cos nx\, dx=-2\sum_{k=1}^{\infty} \frac{a^k}{k}\int_0^\pi \cos{kx}\cos nx\, dx=-\frac {a^n}n\pi.\tag2$$
For $$|a|>1$$, we can write: $$\log(1-2a\cos x+a^2)=\log(1-2a^{-1}\cos x+a^{-2})+\log a^2,$$ so that after integration one obtains already known expression $$(2)$$ with $$a$$ replaced by $$a^{-1}$$ since integration over $$\log a^2\cos nx$$ results in $$0$$.
Thus, finally $$\int_0^\pi\log(1-2a\cos x+a^2)\cos nx\, dx=-\frac\pi n\times\begin{cases} a^n,& |a|\le 1,\\ a^{-n},& |a|>1.\\ \end{cases}$$