Integral through complex methods $\int_{0}^{2\pi}\log \sin^{2}2\theta d\theta=4\int_{0}^{\pi}\log \sin\theta d\theta=-4\pi \log 2$

Asked Apr 13 '19 at 23:20

Active Apr 13 '19 at 23:49

Viewed 92 times

Using residue theorem show that: $$\int_{0}^{2\pi}\log \sin^{2}2\theta d\theta=4\int_{0}^{\pi}\log \sin\theta d\theta=-4\pi \log 2$$

I cannot show even the first identity let alone the second one.

Can anyone show please how to calculate this integral using residue and the first identity?

P.S. I have solved bunch of integrals but have problems with this one.

asked Apr 13 '19 at 23:20

RFZ

1

The first equality is just a simple substitution $\theta=2t$. Afterwards, replace $t$ with $\theta$. – John Wayland Bales Apr 13 '19 at 23:43
1

It's a bit of a nuanced argument. See, for example, page 160 of Ahlfors 3rd Ed. An easier alternative to residue theory is to use Jensen's Theorem. – Brevan Ellefsen Apr 14 '19 at 01:38

0 Answers0