Integrate $\ln^3(\sin(x))$ using Fourier series

Question

Show that $$\int_0^{\frac\pi2} \ln^3(\sin(x)) \, dx = -\frac\pi2 \ln^3(2) - \frac{\pi^3}8 \ln(2) - \frac{3\pi}4 \zeta(3)$$

I have seen a method for this elsewhere, but I would specifically like to reproduce this result in the same way I've computed the similar integrals of $\ln(\cos(x))\ln(\sin(x))$ and $\ln^2(\sin(x))$, as shown here and here - which involves a "complicated series", as Jack puts it.

In particular, I want to use the Fourier series

$$f(x) = \ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k$$

Expanding the integrand yields

$$\begin{align*} -f(x)^3 &= \ln^3(2) + 3 \ln^2(2) \sum_{a=1}^\infty \frac{\cos(2ax)}a \\ &\quad + 3\ln(2) \left(\sum_{a=1}^\infty \frac{\cos^2(2ax)}{a^2} + 2 \sum_{a\neq b} \frac{\cos(2ax) \cos(2bx)}{ab}\right) \\ & \quad + \left(\sum_{a=1}^\infty \frac{\cos^3(2ax)}{a^3} + 3 \sum_{a\neq b} \frac{\cos^2(2ax) \cos(2bx)}{a^2b} + 6 \sum_{a\neq b\neq c} \frac{\cos(2ax) \cos(2bx) \cos(2cx)}{abc}\right) \end{align*}$$

In the integral, the series with $\cos(2ax)$ vanishes; by orthogonality, $\cos(2ax)\cos(2bx)$ vanishes; in combination of both of these facts, $\cos^2(2ax)\cos(2bx)$ and $\cos^3(2ax)$ also vanish. So the integral reduces to

$$\int_0^{\frac\pi2} f(x)^3 \, dx = -\frac\pi2 \ln^3(2) - \frac{\pi^3}8 \ln(2) - 6 \sum_{a\neq b\neq c} \frac1{abc} \int_0^{\frac\pi2} \cos(2ax) \cos(2bx) \cos(2cx) \, dx$$

In the remaining integral, I'm fairly sure that most of the terms integrate to $0$ using the orthogonality argument, except in the case of $a+b=c$,

$$\int_0^{\frac\pi2} \cos(2ax) \cos(2bx) \cos(2(a+b)x) \, dx \\ = \int_0^{\frac\pi2} \frac{\cos(2(a-b)x)\cos(2(a+b)x) + \cos^2(2(a+b)x)}2 \, dx = \frac\pi8$$

ETA: If there are no other triples, then I should end up with

$$-6 \sum_{a\neq b\neq c} \frac1{abc} \int_0^{\frac\pi2} \cos(2ax) \cos(2bx) \cos(2cx) \, dx = -\frac{3\pi}4 \sum_{a+b=c} \frac1{abc} = -\frac{3\pi}4 \zeta(3) \\ \implies \sum_{a\neq b} \frac1{ab(a+b)} = \zeta(3)$$

Now,

$$\begin{align*} \sum_{a\neq b} \frac1{ab(a+b)} &= \sum_{(a,b)\in\Bbb N^2} \frac1{ab(a+b)} - \frac12 \sum_{a=1}^\infty \frac1{a^3} \\[1ex] &= 2 \sum_{a<b} \frac1{ab(a+b)} - \frac12 \zeta(3) \\[2ex] \sum_{a<b} \frac1{ab(a+b)} &= \sum_{b=2}^\infty \frac1{b(b+1)} + \sum_{b=3}^\infty \frac1{2b(b+2)} + \sum_{b=4}^\infty \frac1{3b(b+3)} + \cdots \\[1ex] &= \sum_{b=2}^\infty \left(\frac1b - \frac1{b+1}\right) + \frac14 \sum_{b=3}^\infty \left(\frac1b - \frac1{b+2}\right) + \frac19 \sum_{b=4}^\infty \left(\frac1b - \frac1{b+3}\right) + \cdots \\[1ex] &= (H_2 - H_1) + \frac{H_4 - H_2}4 + \frac{H_6 - H_3}9 + \cdots \\[1ex] &= \sum_{n=1}^\infty \frac{H_{2n} - H_n}{n^2} \\[1ex] &= \sum_{n=1}^\infty \frac{H_{2n}}{n^2} - 2\zeta(3) \end{align*}$$

where the last equality is due to (31), and it remains to show

$$\sum_{n=1}^\infty \frac{H_{2n}}{n^2} = \frac{11}4 \zeta(3)$$

Rewriting the sum as follows leads me to think there may be a hidden Cauchy product, but I have not been able to find a decomposition.

$$H_{2n}-H_n = \sum_{k=1}^{2n} \frac1k - \sum_{k=1}^n \frac1k = \sum_{k=n+1}^{2n} \frac1k = \sum_{k=1}^n \frac1{n+k} \\ \implies\sum_{n=1}^\infty \frac{H_{2n}-H_n}{n^2} = \sum_{n=1}^\infty \sum_{m=1}^n \frac1{n^2(n+m)}$$

Answer 1

Use a general method in this post:

$$ \begin{align} &\int_0^{\pi/2}\log^a(\sin(x))\log^b(\cos(x))\,\mathrm{d}x\\[3pt] =&\int_0^1\frac{\log^a(x)\log^b\left(\sqrt{1-x^2}\right)}{\sqrt{1-x^2}}\,\mathrm{d}x\tag{6a}\\[3pt] =&\frac1{2^{a+b+1}}\int_0^1\frac{\log^a(x)\log^b(1-x)}{\sqrt{x(1-x)}}\,\mathrm{d}x\tag{6b}\\ =&\frac{\partial_1^a\partial_2^b\mathrm{B}\!\left(\frac12,\frac12\right)}{2^{a+b+1}}\tag{6c} \end{align} $$

Rules:

$$ \begin{align} \partial_x\mathrm{B}\!\left(x+\tfrac12,y+\tfrac12\right) &=A_{1,0}(x,y)\,\mathrm{B}\!\left(x+\tfrac12,y+\tfrac12\right)\tag{4a}\\[9pt] \partial_y\mathrm{B}\!\left(x+\tfrac12,y+\tfrac12\right) &=A_{0,1}(x,y)\mathrm{B}\!\left(x+\tfrac12,y+\tfrac12\right)\tag{4b}\\[9pt] \partial_xA_{n,m}(x,y) &=A_{n+1,m}(x,y)\tag{4c}\\[9pt] \partial_yA_{n,m}(x,y) &=A_{n,m+1}(x,y)\tag{4d} \end{align} $$

Coefficients:

$$ \begin{align} \mathrm{B}\!\left(\tfrac12,\tfrac12\right) &=\pi\tag{5a}\\[9pt] A_{1,0}(0,0)&=A_{0,1}(0,0) =-2\log(2)\tag{5b}\\[6pt] A_{n,0}(0,0)&=A_{0,n}(0,0) =\frac{(n-1)!}{(-1)^{n-1}}\left(2-2^n\right)\zeta(n)&\text{for }n\ge2\tag{5c}\\ A_{n,m}(0,0) &=\frac{(m+n-1)!}{(-1)^{m+n-1}}\zeta(m+n)&\text{for }m,n\ge1\tag{5d} \end{align} $$

For simple, we use abbreviated notations, use $\mathrm{B}$ for $\mathrm{B}\!\left(\frac12,\frac12\right)$, and use $A_{n,m}$ for $A_{n,m}(0,0)$.

Start:

$$\begin{align} \int_0^{\frac{\pi}{2}} \ln^3(\sin(x)) dx &=\frac{1}{2^{3+1}}\cdot\partial_1^3 \mathrm{B}\\ &=\frac{1}{16}\cdot\partial_1^2 (A_{1,0}\mathrm{B})\\ &=\frac{1}{16}\cdot\partial_1 (A_{2,0}\mathrm{B}+A_{1,0}A_{1,0}\mathrm{B})\\ &=\frac{1}{16}\cdot\partial_1 (A_{2,0}\mathrm{B}+A^2_{1,0}\mathrm{B})\\ &=\frac{1}{16}\cdot(A_{3,0}\mathrm{B}+A_{2,0}A_{1,0}\mathrm{B}+2A_{1,0}A_{2,0}\mathrm{B}+A^2_{1,0}A_{1,0}\mathrm{B})\\ &=\frac{1}{16}\cdot(A_{3,0}+3A_{1,0}A_{2,0}+A^3_{1,0})\mathrm{B} \end{align}$$

Plug in values: $$\mathrm{B}=\mathrm{B}\!\left(\frac12,\frac12\right)=\pi,~~A_{1,0}=-2\ln(2),~~A_{2,0}=2\zeta(2)=\frac{\pi^2}{3},~~A_{3,0}=-12\zeta(3)$$

$$\begin{align} \int_0^{\frac{\pi}{2}} \ln^3(\sin(x)) dx&=\frac{\pi}{16}(-12\zeta(3)-2\pi^2\ln(2)-8\ln^3(2))\\ \int_0^{\frac{\pi}{2}} \ln^3(\sin(x)) dx&=-\frac{3\pi}{4}\zeta(3)-\frac{\pi^3}{8}\ln(2)-\frac{\pi}{2}\ln^3(2) \end{align}$$

Answer 2

$$\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n^2-H_n^{(2)}}{n+1}\cos(2(n+1)x)\\ =\Re\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n+1}(-e^{2ix})^{n+1}$$

$$=\Re\left\{-\frac13 \ln^3(1+e^{2ix})\right\}$$

$$=x^2\ln(2\cos x)-\frac13\ln^3(2\cos x)\tag{*}$$

where we used the generating function

$$\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n+1}x^{n+1}=-\frac13\ln^3(1-x)$$

which follows from integrating

$$\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n}=\frac{\ln^2(1-x)}{1-x}.$$

Integrate both sides of $(*)$ from $x=0\to \pi/2$, $$\int_0^{\pi/2}\left[x^2\ln(2\cos x)-\frac13\ln^3(2\cos x)\right]dx$$

$$=\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n^2-H_n^{(2)}}{n+1}\int_0^{\pi/2}\cos(2(n+1)x)dx$$

$$=\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n^2-H_n^{(2)}}{n+1}\left(-\frac{\sin(n\pi)}{2(n+1)}\right)=0$$

so $$\int_0^{\pi/2}\ln^3(2\cos x)dx=3\int_0^{\pi/2}x^2\ln(2\cos x)dx=-\frac34\pi\zeta(3).$$

Note that $\int_0^{\pi/2}\ln^3(2\sin x)dx=\int_0^{\pi/2}\ln^3(2\cos x)dx.$

Answer 3

Some of the last several steps I took in the OP are unnecessary. The last obstacle is the sum

$$\sum_{(a,b)\in\Bbb N^2} \frac1{ab(a+b)} = \sum_{a=1}^\infty \sum_{b=1}^\infty \frac1{ab(a+b)} = \Gamma(3)\zeta(3) = 2\zeta(3)$$

Let

$$f(x) = \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{x^{a+b}}{ab(a+b)}$$

Then for $|x|<1$,

$$\begin{align*} f'(x) &= \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{x^{a+b-1}}{ab} \\[1ex] x f'(x) &= \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{x^{a+b}}{ab} \\[1ex] &= \sum_{a=1}^\infty \left(\frac{x^a}a \sum_{b=1}^\infty \frac{x^b}b\right) \\[1ex] &= \ln^2(1-x) \\[2ex] f'(x) &= \frac{\ln^2(1-x)}x \\[2ex] f(x) &= \int_0^x \frac{\ln^2(1-y)}y \, dy \end{align*}$$

and the result follows. (Incidentally, this is a proof I had come across on this site several months ago but am unable to track it down.)