MHB Definite integral challenge ∫ln(2−2cosx)dx=0

[lfdahl]

Prove, that the definite integral

$$\int_{0}^{\pi}\ln (2-2\cos x)dx = 0.$$

 

[MountEvariste]

Spoiler

Call the integral $I$. Let $x \mapsto \pi-x$ then we get $\displaystyle \int_0^{\pi}\ln(2+2\cos{x})\,{dx}$.

Then $\displaystyle 2I = \int_0^{\pi}\log(4\sin^2{x})\,{dx} \implies I = \int_0^{\pi}\log(\sin{x}) +\log(2)\,{dx}$

But $\displaystyle\log (\sin x)=-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}-\log( 2)$. Using this we end up with

$\displaystyle I = -\int_0^{\pi} \sum_{j=1}^{\infty} \frac{1}{j} \cos{2jx} \,{dx}= -\sum_{j=1}^{\infty} \frac{1}{j} \int_0^{\pi} \cos(2jx) \,{dx} = 0$ as required.

 

[lfdahl]

Spoiler

Call the integral $I$. Let $x \mapsto \pi-x$ then we get $\displaystyle \int_0^{\pi}\ln(2+2\cos{x})\,{dx}$.

Then $\displaystyle 2I = \int_0^{\pi}\log(4\sin^2{x})\,{dx} \implies I = \int_0^{\pi}\log(\sin{x}) +\log(2)\,{dx}$

But $\displaystyle\log (\sin x)=-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}-\log( 2)$. Using this we end up with

$\displaystyle I = -\int_0^{\pi} \sum_{j=1}^{\infty} \frac{1}{j} \cos{2jx} \,{dx}= -\sum_{j=1}^{\infty} \frac{1}{j} \int_0^{\pi} \cos(2jx) \,{dx} = 0$ as required.

Welcome, June29! - to the challenge forum. What a nice entry you make with this fine solution of yours!
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