How to solve the integration

Question

How to find $$ \frac{\int_0^{\pi}x^3\log(\sin x)\,dx}{\int_0^{\pi} x^2 \log(\sqrt{2} \sin x)\,dx} $$

I couldn’t resolve it by using integration by parts.

Answer 1

Using Fourier series, we see that \begin{align} \log(\sin x)= -\log 2-\sum^\infty_{k=1}\frac{\cos(2kx)}{k} \end{align} for $0\le x\le \pi$. Hence we have \begin{align} \int^\pi_0x^3\log(\sin x)\ dx =&\ -\log 2\int^\pi_0x^3\ dx -\sum^\infty_{k=1} \frac{1}{k}\int^\pi_0 x^3\cos(2kx)\ dx\\ =&\ -\frac{\pi^4}{4}\log 2 - \frac{3\pi^2}{4}\sum^\infty_{k=1} \frac{1}{k^3} = -\frac{\pi^4}{4}\log 2- \frac{3\pi^2}{4}\zeta(3). \end{align} On the other hand, we have \begin{align} \int^\pi_0 x^2\log(\sqrt{2}\sin x)\ dx =&\ \int^\pi_0 \frac{1}{2}x^2\log 2+ x^2\log(\sin x)\ dx\\ =&\ \frac{\pi^3}{6}\log 2 + \int^\pi_0 x^2\log(\sin x)\ dx\\ =&\ \frac{\pi^3}{6}\log 2 -\log 2\int^\pi_0x^2\ dx -\sum^\infty_{k=1} \frac{1}{k}\int^\pi_0 x^2\cos(2kx)\ dx\\ =&\ -\frac{\pi^3}{6}\log 2 -\frac{\pi}{2}\sum^\infty_{k=1} \frac{1}{k^3}= -\frac{\pi^3}{6}\log 2 - \frac{\pi}{2}\zeta(3). \end{align}

Then we see that \begin{align} \frac{\int^\pi_0 x^3 \log(\sin x)\ dx}{\int^\pi_0 x^2\log(\sqrt{2}\sin x)\ dx} = \frac{-\frac{\pi^4}{4}\log 2- \frac{3\pi^2}{4}\zeta(3)}{-\frac{\pi^3}{6}\log 2 - \frac{\pi}{2}\zeta(3)} = \frac{3\pi}{2}. \end{align}

Answer 2

$\newcommand{\intd}{\,\mathrm{d}}$ Alternatively, and quite similarly I might add, this can be done using the substitution $u=a+b-x$ where $a$ and $b$ are the bounds of the integration. It will be important to note that $\sin(\pi-x)=\sin x$, looking at the numerator we have $I_1=\int_0^{\pi}x^3\log(\sin x)\,\mathrm{d}x$, making the substitution $u=\pi-x$ and switching the variable back to $x$ we get $$I_1=\int_0^\pi(\pi-x)^3\log(\sin x)\,\mathrm{d}x=\int_0^\pi(\pi^3-3\pi^2x+3\pi x^2-x^3)\log(\sin x)\,\mathrm{d}x$$ Adding $I_1+I_1$ we get $$2I_1=\int_0^\pi(\pi^3-3\pi^2x+3\pi x^2)\log(\sin x)\,\mathrm{d}x$$ We now have 3 resulting integrals which will be computed with the same method. Consider $I_2=\int_0^\pi x\log(\sin x)\intd x$, making the same substitution we have $I_2=\int_0^\pi(\pi-x)\log(\sin x)\intd x$ which implies $2I_2=\int_0^\pi \pi\log(\sin x)\intd x$ so $I_2=\frac\pi2\int_0^\pi\log(\sin x)\intd x$. Rewriting our expression for $I_1$ we obtain \begin{align} \frac{\pi^3}{2}\int_0^\pi\log(\sin x)\intd{x}-\frac{3\pi^2}{2}\int_0^\pi x\log(\sin x)\intd{x}+\frac{3\pi}{2}\int_0^\pi x^2\log(\sin x)\intd x \end{align} Let $I_3=\int_0^\pi\log(\sin x)\intd x$, this implies $I_2=\frac\pi2I_3$ so \begin{align} I_1&=\frac{\pi^3}{2}I_3-\frac{3\pi^2}{2}\cdot\frac\pi2I_3+\frac{3\pi}{2}\int_0^\pi x^2\log(\sin x)\intd x\\ &=-\frac{\pi^3}{4}I_3+\frac{3\pi}{2}\int_0^\pi x^2\log(\sin x)\intd{x} \end{align} By a similar fashion in the denominator we have \begin{align} \int_0^\pi x^2\log(\sqrt2\sin x)\intd x&=\int_0^\pi \frac{x^2}{2}\log 2+x^2\log(\sin x)\intd{x}\\ &=\frac{\pi^3}{6}\log2+\int_0^\pi x^2\log(\sin x)\intd{x} \end{align} All that is left to compute is $I_3$ which will be done by use of the same substitution. We have that since $I_3=\int_0^\pi\log(\sin(\pi-x))\intd{x}$ that $$I_3=2\int_0^{\frac\pi2}\log(\sin x)\intd{x}=2\int_0^{\frac\pi2}\log(\cos x)\intd{x}$$ From this it follows that \begin{align} I_3&=\int_0^\pi\log(\sin x)\intd{x}=2\int_0^{\frac\pi2}\log(\sin x)\intd{x}\\ &=\int_0^{\frac\pi2}\log(\sin x)\intd{x}+\int_0^{\frac\pi2}\log(\cos x)\intd{x}\\ &=\int_0^{\frac\pi2}\log(\sin x\cos x)\intd{x}=\int_0^{\frac\pi2}\log\left(\frac{\sin 2x}{2}\right)\intd{x}\\ &=\int_0^{\frac\pi2}\log(\sin 2x)\intd{x}-\frac\pi2\log2\\ &\overbrace{=}^{u=2x}\frac12\int_0^\pi\ln(\sin u)\intd u-\frac\pi2\log2\\ &=\frac12I_3-\frac\pi2\log2 \end{align} So $I_3=-\pi\log2$. Now we have condensed the expression down to \begin{align*}\frac{\int_0^\pi x^3\log(\sin x)\intd{x}}{\int_0^\pi x^2\log(\sqrt2\sin x)\intd{x}}&=\frac{-\frac{\pi^3}{4}\cdot-\pi\log2+\frac{3\pi}{2}\int_0^\pi x^2\log(\sin x)\intd{x}}{\frac{\pi^3}{6}\log2+\int_0^{\pi}x^2\log(\sin x)\intd{x}}\\ &=\frac{\frac{\pi^4}{4}\log2+\frac{3\pi}{2}\int_0^\pi x^2\log(\sin x)\intd{x}}{\frac{\pi^3}{6}\log2+\int_0^\pi x^2\log(\sin x)\intd{x}} \end{align*} By inspection we see that the numerator is $\frac{3\pi}{2}$ times the denominator so the final answer is $\boxed{\frac{3\pi}{2}}$. Interestingly enough, although my approach did not involve Fourier series nor $\zeta(3)$ the final steps to the answer are remarkably similar. It is also important to note that since $x^2$ is even we could not make the same $\pi-x$ substitution to solve for that integral.