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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 ?

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  • $\begingroup$ Differentiation under the integral $\endgroup$ – Aditya Garg Mar 26 '19 at 20:23
  • $\begingroup$ could you try letting $1=\sin^2(x)+\cos^2(x)$ and seeing if this helps to factorise at all? $\endgroup$ – Henry Lee Mar 26 '19 at 21:36
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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 $$\int_0^\pi\log(1-2a\cos x+a^2)\cos nx\, dx=-\frac\pi n\times\begin{cases} a^n,& |a|\le 1,\\ a^{-n},& |a|>1.\\ \end{cases} $$

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  • $\begingroup$ Quick and easy +1. $\endgroup$ – clathratus Mar 27 '19 at 3:56

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