Quotient of two integrals $\frac{\int_0^\pi x^3\ln(\sin x)\,dx}{\int_0^\pi x^2\ln(\sqrt{2}(\sin x))\,dx}$

Solution 1:

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

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

Things used above: $$K=\int_0^\pi x\ln(\sin x)dx=\int_0^\pi (\pi-x)\ln(\sin x)dx$$ $$\Rightarrow 2K=\int_0^\pi (x+\pi-x)\ln(\sin x)dx\Rightarrow K=\frac{\pi}{2}\int_0^\pi \ln(\sin x)dx$$ $$L=\int_0^\pi \ln(\sin x)dx=\int_0^\frac{\pi}{2} \ln(\sin x)dx+\int_0^\frac{\pi}{2} \ln(\cos x)dx$$ $$=\int_0^\pi \ln\left(\frac22\sin x\cos x\right)=\int_0^\frac{\pi}{2} \ln(\sin 2x)dx-\int_0^\frac{\pi}{2} \ln 2dx$$ $$=\frac12 \int_0^\pi \ln(\sin x) dx-\ln\sqrt 2 \int_0^{\pi} dx\Rightarrow L=-2\pi \ln\sqrt 2$$

Solution 2:

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