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I want to show that $$ \int\limits_0^\pi \ln\left(1-\cos\left(t\right)\right)\text{d}t=-\pi\ln\left(2\right) $$ I wanted to use the integral $\displaystyle \int\limits_0^\pi \ln\sin\left(t\right)\ dt$ which I know the value but I struggle finding a judicious change of variable. Is there a way to do it ?

GNUSupporter 8964民主女神 地下教會
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Atmos
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  • This is equivalent to $\int_{0}^{\pi/2}\log\sin(x),dx = -\frac{\pi}{2}\log(2) $ which has been asked many times. It can be proved via $\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{2n}{2^n}$ and Riemann sums, just symmetry, Fourier series, derivatives of the Beta function and probably many other ways. – Jack D'Aurizio Feb 16 '18 at 18:20
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1 Answers1

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Hint:

As $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx=I$(say),

$$I+I=\int_a^bf(x)+f(a+b-x)\ dx$$

Now use Showing that $\int_0^1 \log(\sin \pi x)dx=-\log2$

lab bhattacharjee
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  • Thanks for the hint but could you provide a more detailed way ? I've shown the equality with $\log(\sin(\pi x))$ but I dont come up with the result – Atmos Feb 16 '18 at 19:14