Simple looking integrals can be very nasty

The details of this can be found in the following: paper .

We are interested in calculating what is referred to as the log-sine integral (evaluated at $\pi$ ): $\mathrm{Ls}_4^{(1)}(\pi) \; = \; -\int_0^\pi \theta \log^2\big(2\sin\tfrac12\theta\big)\,d\theta \; = \; -4\int_0^{\frac12\pi} \theta\big( \log2 + \log(\sin\theta)\big)^2\,d\theta \;,$ so that $\begin{array}{rcl} \displaystyle I \; = \; \int_0^{\frac12\pi} \theta \log(\sin\theta)\,d\theta & = & \displaystyle -\tfrac18\pi^2\log^22 - 2\log 2\int_0^{\frac12\pi} \theta \log(\sin\theta)\,d\theta - \tfrac14\mathrm{Ls}_4^{(1)}(\pi) \\ & = & \tfrac18\pi^2 \log^22 - 2\log 2\left\{ \tfrac{7}{16}\zeta(3) - \tfrac18\pi^2\log2\right\} - \tfrac14\mathrm{Ls}_4^{(1)}(\pi) \\ & = & \tfrac18\pi^2 \log^22 - \tfrac78\log2\zeta(3) - \tfrac14\mathrm{Ls}_4^{(1)}(\pi) \\ & = & \tfrac18\pi^2\log^22 - \tfrac78\log2\zeta(3) - \tfrac{19}{32}\zeta(4) + \tfrac12\lambda_4(\tfrac12) \end{array}$ where we are using the integral (due to Euler) $\int_0^{\frac12\pi} \theta \log(\sin\theta)\,d\theta \; = \; \tfrac{7}{16}\zeta(3) - \tfrac18\pi^2\log2$ and we have introduced a special case of the so-called Kummer polylogarithm $\lambda_4(\tfrac12) \; = \; 2\sum_{k=0}^2\tfrac{1}{k!} \mathrm{Li}_{n-k}(\tfrac12) \log^k2 + \tfrac14\log^42 \;.$ Thus $\begin{array}{rcl} I & = & \tfrac18\pi^2\log^22 - \tfrac78\log2\zeta(3) - \tfrac{19}{32}\zeta(4) + \mathrm{Li}_4(\tfrac12) + \log2 \mathrm{Li}_3(\tfrac12) + \tfrac12\log^22 \mathrm{Li}_2(\tfrac12) + \tfrac18\log^42 \\ & = & \tfrac18\pi^2\log^22 - \tfrac78\log2\zeta(3) - \tfrac{19}{32}\zeta(4) + \mathrm{Li}_4(\tfrac12) \\ & & {} + \log2\left\{ -\tfrac{1}{12}\pi^2\log2 + \tfrac16\log^32 + \tfrac78\zeta(3)\right\} + \tfrac12\log^22 \left\{ \tfrac{1}{12}\pi^2 - \tfrac12\log^22\right\} + \tfrac18\log^42 \\ & = & \mathrm{Li}_4(\tfrac12) - \tfrac{19}{32}\zeta(4) + \tfrac{1}{24}\log^42 + \tfrac{1}{12}\pi^2\log^22 \\ & = & \mathrm{Li}_4(\tfrac12) - \tfrac{19}{2880}\pi^4 + \tfrac{1}{24}\log^42 + \tfrac{1}{12}\pi^2\log^22 \end{array}$ so that the answer is $1+2+19+4+2880 + 4 + 2 + 24 + 2 + 2 + 2 + 12 \,=\, \boxed{2954}$ .

Hint: Using the integral representation of $H_{n}$ , we have

$\displaystyle \sum_{n=1}^{\infty} H_{n} t^n = -\dfrac{\ln (1-t)}{1-t}$

Dividing both sides by $t$ and integrating from $0$ to $x$ , we have,

\displaystyle \sum_{n=1}^{\infty} \dfrac{H_{n} x^n}{n} = \operatorname{Li}_{2}(x) + \dfrac{1}{2} \ln^2(1-x) \tag{*}

Now, in the integrand, write the $\sin x$ term in $\ln^2 (\sin x )$ as $\dfrac{e^{ix}-e^{-ix}}{2i}$ and take $e^{ix}$ common. Then, using the above equation and switching summation and integral, and some trivial simplifications, we have our desired equality. In between, one non trivial sum arises, i.e,

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

which is left as an exercise for the reader.

The answer is $= { Li }_{ 4 }\left(\dfrac { 1 }{ 2 } \right)-\dfrac { 19{ \pi }^{ 4 } }{ 2880 } +\dfrac { { \log }^{ 4 }(2) }{ 24 } +\dfrac { { \pi }^{ 2 }{ \log }^{ 2 }(2) }{ 12 }$

So we have $A=1,B=2,C=19,D=4,E=2880,F=4,G=2,\\ H=24,M=2,N=2,P=2,Q=12$