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Calculate $$\frac{\int_0^\pi x^3\ln(\sin x)\,dx}{\int_0^\pi x^2\ln(\sqrt{2}(\sin x))\,dx}$$

In this problem, I'm unable to understand how to start.

I tried applying integration by parts but I couldn't solve it. I also tried the various properties of definite integration but they were of no use. Maybe applying integration by parts (or DI method) successively may work but it leads to a form of $\frac{\infty}{\infty}$.

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    $\begingroup$ How in the denominator do you handle the logarithm's argument changing sign at $\pi/\sqrt{2}$? $\endgroup$
    – J.G.
    May 20, 2019 at 16:11
  • $\begingroup$ Where did you get this integral? Is it a part of some larger problem? $\endgroup$
    – Yuriy S
    May 21, 2019 at 13:49
  • $\begingroup$ No , It is from a hard book for JEE Advanced that my coaching school has given me. It is a standalone question , I have been asked that this equates to kπ , and we have to find the value of k , The answer provided is 1.5 , but no solution has been given. $\endgroup$ May 21, 2019 at 13:51
  • $\begingroup$ I highly doubt that either integral can be computed in closed form without knowledge of the Riemann zeta function at least. Perhaps you miscoppied the question, and wrote $$\ln(\sin(x\sqrt{2}))$$ instead of $$\ln(\sqrt{2}\sin x)$$ . I am not even sure if the integral on the denominator even exists. It would also help if you added more context to your question. $\endgroup$
    – clathratus
    May 22, 2019 at 21:02
  • $\begingroup$ @Threesidedcoin I am so so so so so sorry. I'll make sure this doesn't happen in the future $\endgroup$ May 23, 2019 at 6:10

2 Answers 2

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We want to prove that: $$\frac{I}{J}=\frac{\int_0^\pi x^3\ln(\sin x)dx} {\int_0^\pi x^2\ln\left(\sqrt 2\sin x\right)dx}=\frac{3\pi}2$$

First, let's take the $I$ integral and perform the $x\to \pi-x$ substitution:

$$I=\int_0^\pi x^3\ln(\sin x)dx\overset{x\to\pi-x}=\int_0^\pi (\pi^3-3\pi^2x+3\pi x^2-x^3)\ln(\sin x)dx$$

In the $J$ integral we had an additional $\ln \sqrt 2$ for the $x^2$ term, thus we can also add it here:

$$\Rightarrow I=\int_0^\pi (\pi^3-3\pi^2x+3\pi x^2-x^3)\ln(\sin x)dx+ 3\pi(\underbrace{\ln \sqrt 2-\ln \sqrt 2}_{=0})\int_0^\pi x^2 dx$$

$$=\pi^3 \underbrace{\int_0^\pi \ln(\sin x)dx}_{=\mathcal L}-3\pi^2 \underbrace{\int_0^\pi x\ln(\sin x)dx}_{=\mathcal K}+3\pi J-I-{\pi^4}\ln \sqrt 2$$

$$\Rightarrow 2I=\left(\pi^3-\frac{3\pi^3}{2}\right)\int_0^\pi \ln(\sin x)dx+3\pi J-{\pi^4}\ln \sqrt 2$$

$$\require{cancel} 2I=\cancel{\frac{\pi^3}{2}\cdot 2\pi \ln \sqrt 2}+3\pi J-\cancel{\pi^4 \ln \sqrt 2}\Rightarrow I=\frac{3\pi}2J$$


Note that above we used: $$\mathcal K=\int_0^\pi x\ln(\sin x)dx\overset{x\to \pi-x}=\int_0^\pi (\pi-x)\ln(\sin x)dx$$

$$\Rightarrow \mathcal K=\frac{\pi}{2}\underbrace{\int_0^\pi \ln(\sin x)dx}_{=\mathcal L}=\frac{\pi}{2}\mathcal L$$

$$\mathcal L=\int_0^\pi \ln(\sin x)dx=\int_0^\frac{\pi}{2} \ln(\sin x)dx+\int_0^\frac{\pi}{2} \ln(\cos x)dx$$

$$=\int_0^\frac{\pi}{2} \ln\left(\sin x\cos x\right)=\int_0^\frac{\pi}{2} \ln(\sin 2x)dx-\int_0^\frac{\pi}{2} \ln 2dx$$

$$=\frac12 \underbrace{\int_0^\pi \ln(\sin x) dx}_{=\mathcal L}-\ln\sqrt 2 \int_0^{\pi} dx\Rightarrow \mathcal L=-2\pi \ln\sqrt 2$$

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  • $\begingroup$ Hi. How did you write $L=\int_0^\pi \ln(\sin x)dx=\int_0^\frac{\pi}{2} \ln(\sin x)dx+\int_0^\frac{\pi}{2} \ln(\cos x)dx$? $\endgroup$
    – aarbee
    Jun 13, 2021 at 14:32
  • $\begingroup$ @aarbee Split the LHS integral as a sum of two integrals with the bounds $[0,\frac{\pi}{2}]$ and $[\frac{\pi}{2},\pi]$ respectively. Then for the second integral, with the bounds $[\frac{\pi}{2},\pi]$, let $t=\frac{\pi}{2}-x$ and arrive at the integral with $\cos x$ in it. $\endgroup$
    – Zacky
    Jun 14, 2021 at 0:01
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    $\begingroup$ Ohkay, thanks, got it. $\endgroup$
    – aarbee
    Jun 14, 2021 at 5:16
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For those interest in the overkill approach, I will be providing closed forms for each integral with the use of special functions just for the hell of it.


We define $$p=\int_0^\pi x^3\ln\sin x\,dx$$ We recall the definition of the Clausen function of order $2$: $$\mathrm{Cl}_2(x)=-\int_0^x \ln\left|2\sin\frac{t}2\right|\,dt=\sum_{k\geq1}\frac{\sin kx}{k^2}$$ so $$-\ln\left(2\sin \frac{x}2\right)=\mathrm{Cl}_1(x)=\sum_{k\geq1}\frac{\cos kx}{k}$$ and thus $$\ln\sin x=-\ln2-\sum_{k\geq1}\frac{\cos2kx}{k}$$ then $$\begin{align} p&=-\int_0^\pi x^3\left(\ln2+\sum_{k\geq1}\frac{\cos2kx}{k}\right)dx\\ &=-\frac{\pi^4}4\ln2-\frac1{16}\sum_{k\geq1}\frac1{k^5}\int_0^{2k\pi}x^3\cos x\,dx \end{align}$$ We can use IBP to show that $$\int_0^{2k\pi}x^3\cos x\,dx=12\pi^2k^2$$ Which I leave to you as a challenge.

Long story short, $$p=-\frac{\pi^4}{4}\ln2-\frac{3\pi^2}4\zeta(3)$$ Where $\zeta(3)=\sum_{k\geq1}k^{-3}$ is Apery's Constant. And $\zeta(s)=\sum_{k\geq1}k^{-s}$ is the Riemann Zeta function.


Next up: $$q=\int_0^\pi x^2\ln(\sqrt{2}\sin x)\,dx=\frac{\pi^3}{6}\ln2+\int_0^\pi x^2\ln\sin x\,dx$$ Using the same series as last time, $$\begin{align} \int_0^\pi x^2\ln\sin x\,dx&=-\frac{\pi^3}{3}\ln2-\frac18\sum_{k\geq1}\frac1{k^4}\int_0^{2k\pi}x^2\cos x\,dx \end{align}$$ IBP shows that $$\int_0^{2k\pi}x^2\cos x\,dx=4\pi k$$ So of course $$\int_0^\pi x^2\ln\sin x\,dx=-\frac{\pi^3}{3}\ln2-\frac\pi2\zeta(3)$$ Hence $$q=-\frac{\pi^3}{6}\ln2-\frac\pi2\zeta(3)$$


So the ratio in question is $$\frac{p}{q}=\frac{\frac{\pi^4}{4}\ln2+\frac{3\pi^2}4\zeta(3)}{\frac{\pi^3}{6}\ln2+\frac\pi2\zeta(3)}=\frac32\pi$$

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