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$$\int_0^{\frac \pi 2} \log(\sin x)dx$$ I tried to use the following substitution, where $I(1)$ is the target integral: $$I(b) = \int_0^{\frac \pi 2} \log(b\sin x)dx \implies I'(b) = \frac \pi {2b}$$ Doing a definite integral on $I'(b)$, I get $$I(1)-I(l) = -\frac \pi 2 \log(l) \implies I(1) = I(l) - \frac \pi 2 \log(l)$$ where $l$ is the lower limit of the integral. I am unable to find a suitable value of $I(l)$ to carry out this substitution. Any help is appreciated!

Aniruddha Deb
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  • $$I(1) = I(l) - \frac \pi 2 \log(l)$$ is trivial so you won't find a suitable value for $\ell$ – Mohammed M. Zerrak Mar 10 '20 at 10:09
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    Feynman's trick is unsuitable for this choice of variable and integral because $$\log(b\sin x) = \log(b) + \log(\sin x)$$ which decouples the constant from whatever complicated function was inside the log in the first place. Generally the point of a Feynman trick manipulation is not to decouple the complicated stuff, but to multiply it or divide it elsewhere. – Ninad Munshi Mar 10 '20 at 10:13
  • By the way there are so many of these, see here – Maximilian Janisch Mar 10 '20 at 11:13
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    @MaximilianJanisch None of the answers posted there use Feynman's trick... I specifically wanted to see how to use that method here. – Aniruddha Deb Mar 10 '20 at 13:49

1 Answers1

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One way is to first integrate by parts to find $$\int_0^{\pi/2} \log \sin x \, dx = - \int_0^{\pi/2} \frac{x}{\tan x} dx.$$

The idea behind Feynman's trick, as pointed out in the comments, is to introduce the parameter $b$ in a non-trivial way so that the recovery of $I(1)$ from $I'(b)$ doesn't reduce to an identity. To accommodate the $\tan x$ in the denominator you can write $x = \arctan(\tan x)$ (valid since $x \in [0,\pi/2)$) and define for $b \ge 0$ $$I(b) = \int_0^{\pi/2} \frac{\arctan(b \tan x)}{\tan x} \, dx$$ which gives you $$I'(b) = \int_0^{\pi/2} \frac{1}{(b\tan x)^2 + 1} \, dx.$$ This integral can be evaluated with standard techniques to $$I'(b) = \frac{\pi}{2(b+1)}$$ so that $$I(1) = I(0) + \int_0^{1} \frac{\pi}{2(b+1)} \, db = \frac{\pi}2 \log 2.$$

Umberto P.
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