How to solve $\int_0^{\frac{\pi}{2}}\frac{x^2\cdot\log\sin x}{\sin^2 x}dx$ using a very cute way?

Question

Few days ago my friend gave me this integral and i cant get how to solve this. The integral is:$$\int_0^{\large \frac{\pi}{2}}\frac{x^2\cdot\log{{\sin{x}}}}{\sin^2{x}}dx$$

What do you mean "a very cute way"? – Rory Daulton Apr 22 '15 at 22:09 — Rory Daulton, Apr 22 '15 at 22:09

Olivier Oloa · Answer 1 · 2015-04-12T08:21:23.130

13

Hint. You may observe that $$ \frac{\log{{\sin{x}}}}{\sin^2{x}}=\left( -\frac{x^2}{2}-\log \sin x-\frac{1}{2} \log^2 \sin x\right)'' $$ thus integrating by parts twice, using $(x^2)''=2$, leads to $$ \begin{align} \int_0^{\pi/2}\frac{x^2\cdot\log{{\sin{x}}}}{\sin^2{x}}dx&=2\int_0^{\pi/2}\left( -\frac{x^2}{2}-\log \sin x-\frac{1}{2} \log^2 \sin x\right)dx\\\\ &=-\int_0^{\pi/2}x^2dx-2\int_0^{\pi/2}\log \sin x dx-\int_0^{\pi/2}\log^2 \sin x dx\\\\ &=-\frac{\pi ^3}{24}+\pi \ln 2-\frac{1}{24} \pi \left(\pi ^2+12 \ln^2 2\right) \end{align} $$ that is

$$ \int_0^{\pi/2}\frac{x^2\cdot\log{{\sin{x}}}}{\sin^2{x}}dx=-\frac{\pi ^3}{12}+\pi \ln 2-\frac{1}{2} \pi \ln^2 2, $$

where we have used standard evaluations for the last integrals.

edited Apr 12 '15 at 08:21

answered Apr 12 '15 at 08:11

Olivier Oloa

120,989

Did you check that $[x^2f']_0^{2/\pi}$=0? and also $[2xf]_0^{2/\pi}$=0? – user128766 Apr 12 '15 at 08:27
The first step seem very hard to "guess". Is there any trick to figure this step out? – MonkeyKing Apr 12 '15 at 08:29
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This is a very good catch ! Thanks for such a nice answer. – Claude Leibovici Apr 12 '15 at 08:31
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(+1) I bow to your clever observation. I also finished my calculation and obtained exactly the same answer, so I guess your solution is among the shortest ones. – Sangchul Lee Apr 12 '15 at 08:31
@user128766 Not each term is $0$, but the sum of both. Thanks. – Olivier Oloa Apr 12 '15 at 08:35
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Thanks for the answer @OlivierOloa.....a great answer – Apr 12 '15 at 08:40
@Chris'ssis Please, which one is ridiculous? – Olivier Oloa Apr 12 '15 at 10:41
@Chris'ssis The OP wanted a cute way... – Olivier Oloa Apr 12 '15 at 10:43
@OlivierOloa anyway – user 1591719 Apr 12 '15 at 10:44

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

This turns out to be similar to Olivier Oloa's, but perhaps with a bit more detail or motivation.

Since $\frac{\mathrm{d}}{\mathrm{d}x}\cot(x)=-\csc^2(x)$, we have $$ \begin{align} \int\frac{\log(\sin(x))}{\sin^2(x)}\,\mathrm{d}x &=-\int\log(\sin(x))\,\mathrm{d}\cot(x)\\ &=-\cot(x)\log(\sin(x))+\int(\csc^2(x)-1)\,\mathrm{d}x\\[6pt] &=-\cot(x)\log(\sin(x))-\cot(x)-x\tag{1} \end{align} $$ Then since $\int\cot(x)\,\mathrm{d}x=\log(\sin(x))$ we have $$ \begin{align} &-\int(\cot(x)\log(\sin(x))+\cot(x)+x)\,\mathrm{d}x\\ &=-\frac12\log(\sin(x))^2-\log(\sin(x))-\frac12x^2\tag{2} \end{align} $$ Therefore, since $\int x^2f''(x)\,\mathrm{d}x=\color{#C00000}{x^2f'(x)}\color{#00A000}{-2xf(x)}+2\int f(x)\,\mathrm{d}x$ $$ \begin{align} &\int_0^{\pi/2} x^2\frac{\log(\sin(x))}{\sin^2(x)}\,\mathrm{d}x\\ &=\color{#C00000}{-\frac{\pi^3}8}\color{#00A000}{+\frac{\pi^3}8}-\int_0^{\pi/2}\left(\log(\sin(x))^2+2\log(\sin(x))+x^2\right)\,\mathrm{d}x\tag{3} \end{align} $$ Using $(2)$ and $(9)$ from this answer, we get $$ \begin{align} \int_0^{\pi/2} x^2\frac{\log(\sin(x))}{\sin^2(x)}\,\mathrm{d}x &=-\left(\frac{\pi^3}{24}+\frac\pi2\log(2)^2\right)+2\left(\frac\pi2\log(2)\right)-\frac{\pi^3}{24}\\ &=\bbox[5px,border:2px solid #C0A000]{\pi\log(2)-\frac\pi2\log(2)^2-\frac{\pi^3}{12}}\tag{4} \end{align} $$

score 1 · Answer 3 · answered Apr 12 '15 at 08:09

I am too sleepy to finish my calculation, but you can check that

$$ \int_{0}^{\frac{\pi}{2}} \frac{x^2 \log\sin x}{\sin^2 x} \, dx = -\pi \int_{0}^{1} \frac{\log x \log(\frac{1+x}{2})}{(1+x)^2}\,dx.$$

I guess this may be an alternate starting point of a further calculation.

How to solve $\int_0^{\frac{\pi}{2}}\frac{x^2\cdot\log\sin x}{\sin^2 x}dx$ using a very cute way?

3 Answers3

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