Prove that $\frac12 \int_0^1 \log(\sin \pi x) dx = \frac{1}{2π} \int_0^\pi \log(\sin x) dx$

Question

I posted a question earlier but it was closed as duplicate of this question.

I see the answer and it was pretty much explained. But I got stuck on a step where the answerer put $\frac12 \int_0^1 \log(\sin \pi x) dx = \frac{1}{2π} \int_0^\pi \log(\sin x) dx$

I didn't understand this step. Can anyone prove it? It there any formula? Please don't close my question it's really important for me, I'm trying to solve it from 3 hours.

See reverse chain rule substitution method for integration. – Тyma Gaidash Oct 13 '21 at 18:57 — Тyma Gaidash, Oct 13 '21 at 18:57

score 2 · Accepted Answer · answered Oct 13 '21 at 18:52

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let $t = \pi x$ , $x$ is a dummy variable here, so after you perform the substitution change the $t$ back to $x$.

answered Oct 13 '21 at 18:52

mathforfun223

145

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How the upper bond of integral changed to pi? – Oct 13 '21 at 18:55
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when you introduce a substitution you must also accordingly change the bounds. When $t=\pi x$, plug in the lower bound and upper bounds for the first integral into the substitution. $x=0 \rightarrow t=0 \pi=0 $ and $x=\pi \rightarrow t=1\pi=\pi$. – mathforfun223 Oct 13 '21 at 18:56
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Sorry, I'm a beginner to integrals and I know only few rules. – Oct 13 '21 at 18:58
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i edited my comment. does this help? – mathforfun223 Oct 13 '21 at 19:00
Yeah, thank you so much! – Oct 13 '21 at 19:04
@I_am If you are a beginner, first focus on developing your basics. These are advanced integrals and you are definitely not ready for them. I can suggest a book or two, if you want. – Laxmi Narayan Bhandari Oct 14 '21 at 02:39
@LaxmiNarayanBhandari Yes please – Oct 14 '21 at 04:01
@I_am A good place to start will be Spivak's Calculus. If you find that one too hard, then Calculus for Dummies. – Laxmi Narayan Bhandari Oct 14 '21 at 04:07

score 0 · Answer 2 · answered Feb 10 '22 at 01:47

For a beginner, I want to give a complete solution to a him/her.

Let $y=\pi x$, then $d x=\frac{d y}{\pi}$. $$ \begin{aligned} \frac{1}{2} \int_{0}^{1} \ln (\sin \pi x) d x &=\frac{1}{2} \int_{0}^{\pi} \ln (\sin y)\left(\frac{d y}{\pi}\right) \\ &=\frac{1}{2 \pi} \int_{0}^{\pi} \ln (\sin y) d y \\ &=\frac{1}{2 \pi} \int_{0}^{\pi} \ln (\sin x) d x \end{aligned} $$

Prove that $\frac12 \int_0^1 \log(\sin \pi x) dx = \frac{1}{2π} \int_0^\pi \log(\sin x) dx$

2 Answers2