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 $\frac{\infty}{\infty}$.

  • 9
    $\begingroup$ How in the denominator do you handle the logarithm's argument changing sign at $\pi/\sqrt{2}$? $\endgroup$ – J.G. May 20 at 16:11
  • $\begingroup$ Where did you get this integral? Is it a part of some larger problem? $\endgroup$ – Yuriy S May 21 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. $\endgroup$ – RandomAspirant May 21 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 at 21:02
  • $\begingroup$ @Threesidedcoin I am so so so so so sorry. I'll make sure this doesn't happen in the future $\endgroup$ – RandomAspirant May 23 at 6:10

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

We can take the upper integral and perform Emperor's rule $(\pi-x\to x$ and sum a $0$ in the end): $$I=\int_0^\pi (\pi^3-3\pi^2x+3\pi x^2-x^3)\ln(\sin x)dx+ 3\pi(\underbrace{\ln \sqrt 2-\ln \sqrt 2}_{=0})\int_0^\pi x^2 dx$$ $$\small=\pi^3 \int_0^\pi \ln(\sin x)dx-3\pi^2 \int_0^\pi x\ln(\sin x)dx+3\pi\int_0^\pi x^2(\ln(\sin x)+\ln\sqrt 2)dx-I-{\pi^4}\ln \sqrt 2$$ $$\small 2I=\left(\pi^3-\frac{3\pi^3}{2}\right)\int_0^\pi \ln(\sin x)dx-{\pi^4}\ln \sqrt 2+3\pi\int_0^\pi x^2\ln(\sqrt2 \sin x)dx$$ $$\require{cancel} 2I=\cancel{\frac{\pi^3}{2}\cdot 2\pi \ln \sqrt 2}-\cancel{\pi^4 \ln \sqrt 2}+3\pi J\Rightarrow I=\frac{3\pi}2J$$

Things used above: $$K=\int_0^\pi x\ln(\sin x)dx=\int_0^\pi (\pi-x)\ln(\sin x)dx$$ $$2K=\int_0^\pi (x+\pi-x)\ln(\sin x)dx\Rightarrow K=\frac{\pi}{2}\int_0^\pi \ln(\sin x)dx$$ Split into two parts the latter integral in the point $\frac{\pi}{2}$ and reduce both bounds to $0\to \frac{\pi}{2}$. $$L=\int_0^\pi \ln(\sin x)dx=\int_0^\frac{\pi}{2} \ln(\sin x)dx+\int_0^\frac{\pi}{2} \ln(\cos x)dx$$ $$=\int_0^\pi \ln\left(\frac22\sin x\cos x\right)=\int_0^\frac{\pi}{2} \ln(\sin 2x)dx-\int_0^\frac{\pi}{2} \ln 2dx$$ $$=\frac12 \int_0^\pi \ln(\sin x) dx-\ln\sqrt 2 \int_0^{\pi} dx\Rightarrow L=-2\pi \ln\sqrt 2$$

  • 1
    $\begingroup$ Thanks a lot mate. It was really helpful $\endgroup$ – RandomAspirant May 23 at 9:41

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 $$\frac{p}{q}=\frac{\frac{\pi^4}{4}\ln2+\frac{3\pi^2}4\zeta(3)}{\frac{\pi^3}{6}\ln2+\frac\pi2\zeta(3)}=\frac32\pi$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.