<s>[INST] Integrate $\ln^3(\sin(x))$ using Fourier series 
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.
$$H_{2n}-H_n = \sum_{k=1}^{2n} \frac1k - \sum_{k=1}^n \frac1k = \sum_{k=n+1}^{2n} \frac1k = \sum_{k=1}^n \frac1{n+k} \\ \implies\sum_{n=1}^\infty \frac{H_{2n}-H_n}{n^2} = \sum_{n=1}^\infty \sum_{m=1}^n \frac1{n^2(n+m)}$$
 [/INST] 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.)
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