Simple looking nasty problem

Proposition : $\displaystyle \int_{0}^{1} x^{n} \operatorname{Li}_{s}(x) \mathrm{d}x = (-1)^{s+1} \dfrac{H_{n+1}}{(n+1)^{s}} + \sum_{r=2}^{s} \left[(-1)^{s+r} \dfrac{\zeta(r)}{(n+1)^{s-r+1}} \right] \ ; \ s \geq 2 \ ; \ s,n \in \mathbb{Z^{+}}$

Proof : Let, $\displaystyle \text{J}(s) = \displaystyle \int_{0}^{1} x^{n} \operatorname{Li}_{s}(x)$
Using I.B.P.,

$\displaystyle \text{J}(s) = \left[\dfrac{x^{n+1}}{n+1} \operatorname{Li}_{s}(x) \right]_{0}^{1} - \dfrac{1}{n+1} \int_{0}^{1} x^{n} \operatorname{Li}_{s-1}(x) \mathrm{d}x$

$\displaystyle = \dfrac{\zeta(s)}{n+1} - \dfrac{1}{n+1} \text{J}(s-1)$

$\displaystyle \implies \text{J}(s) = \dfrac{\zeta(s)}{n+1} - \dfrac{1}{n+1} \text{J}(s-1)$

$\displaystyle \implies \text{J}(r) = \dfrac{\zeta(r)}{n+1} - \dfrac{1}{n+1} \text{J}(r-1)$

Multiplying both sides with $\displaystyle (-1)^r (n+1)^{r}$ , we have,

$\displaystyle (-1)^r (n+1)^{r} \text{J}(r) = (-1)^r (n+1)^{r-1} \zeta(r) + (-1)^{r-1} (n+1)^{r-1} \text{J}(r-1)$

$\displaystyle \implies (-1)^r (n+1)^{r} \text{J}(r) - (-1)^{r-1} (n+1)^{r-1} \text{J}(r-1) = (-1)^r (n+1)^{r-1} \zeta(r)$

Taking summation on both sides, we have,

$\displaystyle \sum_{r=2}^{s} [f(r) - f(r-1)] = \sum_{r=2}^{s} [(-1)^r (n+1)^{r-1} \zeta(r)] \quad (*)$

where $\displaystyle f(r) = (-1)^r (n+1)^{r} \text{J}(r)$ . Clearly, $(*)$ telescopes. Evaluating it, we have,

$\displaystyle f(s) = f(1) + \sum_{r=2}^{s} [(-1)^r (n+1)^{r-1} \zeta(r)]$

Note that,

$\displaystyle \text{J}(1) = -\int_{0}^{1} x^{n} \ln(1-x) \mathrm{d}x$

$\displaystyle = \sum_{r=1}^{\infty} \int_{0}^{1} \dfrac{x^{n+r}}{r} \mathrm{d}x$

$\displaystyle = \sum_{r=1}^{\infty} \dfrac{1}{r(r+n+1)}$

$\displaystyle = \dfrac{1}{n+1} \sum_{r=1}^{\infty} \left( \dfrac{1}{r} - \dfrac{1}{r+n+1} \right)$

$\displaystyle = \dfrac{1}{n+1} \sum_{r=1}^{\infty} \left( \left(\dfrac{1}{r} - \dfrac{1}{r+1} \right) + \left(\dfrac{1}{r+1} - \dfrac{1}{r+2} \right) + \ldots + \left(\dfrac{1}{r+n} - \dfrac{1}{r+n+1} \right) \right)$

$\displaystyle = \dfrac{H_{n+1}}{n+1}$

$\implies f(1) = - H_{n+1}$

Putting these values in the recurrence relation, we have,

$\displaystyle \text{J}(s) = (-1)^{s+1} \dfrac{H_{n+1}}{(n+1)^{s}} + (-1)^{s} \sum_{r=2}^{s} \left[(-1)^r \dfrac{\zeta(r)}{(n+1)^{s-r+1}} \right] \quad \square$

Now,

$\displaystyle \text{I} = \int_{0}^{1} x [\operatorname{Li}_{4}(x)]^2 \mathrm{d}x$

$\displaystyle = \sum_{r=1}^{\infty} \dfrac{1}{r^4} \int_{0}^{1} x^{r+1} \operatorname{Li}_{4}(x) \mathrm{d}x$

Using the proposition,

$\displaystyle = \sum_{r=1}^{\infty} \dfrac{\zeta(2)}{r^4(r+2)^3} - \sum_{r=1}^{\infty} \dfrac{\zeta(3)}{r^4(r+2)^2} + \sum_{r=1}^{\infty} \dfrac{\zeta(4)}{z^4(r+2)} - \sum_{r=1}^{\infty} \dfrac{H_{r+2}}{r^4(r+2)^4}$

Now, using partial fractions and the elementary result $\displaystyle \sum_{r=1}^\infty \frac{H_r}{r^p}= \left(1+\frac{p}{2} \right)\zeta(p+1)-\frac{1}{2}\sum_{k=1}^{p-2}\zeta(k+1)\zeta(q-k)$ , it simplifies to,

$\displaystyle \text{I} = \dfrac{361}{256} - \dfrac{59}{384}\pi^2 - \dfrac{3}{32}\zeta(3) + \dfrac{19}{2880}\pi^4 - \dfrac{1}{24}\pi^2\zeta(3) -\dfrac38\zeta(5) + \dfrac{1}{2160}\pi^6 + \dfrac14\zeta(3)^2 - \dfrac{1}{180}\pi^4\zeta(3) + \dfrac{1}{16200}\pi^8$

$\displaystyle \implies \text{Ans.} = \boxed{22620}$

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In general,

$\displaystyle \int_{0}^{1} x^n \operatorname{Li}_{s} (x) \ \operatorname{Li}_{q} (x) \ \mathrm{d}x \ = \ \sum_{r=1}^\infty \dfrac{1}{r^s} \left(\dfrac{(-1)^{q+1}}{(n+r+1)^q}H_{n+r+1} + (-1)^{q}\sum_{k=2}^q (-1)^k\dfrac{\zeta(k)}{(n+r+1)^{q-k+1}} \right)$

which can be evaluated similarly by using partial fractions and the elementary Euler Sum.

According to this paper , we have $\begin{array}{rcl} \displaystyle \int_0^1 x \mathrm{Li}_4(x)^2\,dx & =& \displaystyle \tfrac{361}{256} - \tfrac{59}{64}\zeta(2) - \tfrac{3}{32}\zeta(3) + \tfrac{19}{32}\zeta(4) - \tfrac{1}{4}\zeta(2)\zeta(3) \\ & & {} \displaystyle - \tfrac38\zeta(5) + \tfrac{7}{16}\zeta(6) + \tfrac14\zeta(3)^2 - \tfrac12\zeta(3)\zeta(4) + \tfrac{7}{12}\zeta(8) \\ & = & \tfrac{361}{256} - \tfrac{59}{384}\pi^2 - \tfrac{3}{32}\zeta(3) + \tfrac{19}{2880}\pi^4 - \tfrac{1}{24}\pi^2\zeta(3) \\ & & {} -\tfrac38\zeta(5) + \tfrac{1}{2160}\pi^6 + \tfrac14\zeta(3)^2 - \tfrac{1}{180}\pi^4\zeta(3) + \tfrac{1}{16200}\pi^8 \end{array}$ Mathematica confirms this answer numerically to $50$ decimal places.

Thus we have $A=361$ , $B=256$ , $C=59$ , $D=2$ , $E=384$ , $F=3$ , $G=32$ , $H=3$ , $I=19$ , $J=4$ , $K=2880$ , $L=2$ , $M=24$ , $N=3$ , $O=3$ , $P=8$ , $Q=5$ , $R=1$ , $S=6$ , $T=2160$ , $U=2$ , $V=3$ , $W=4$ , $X=4$ , $Y=180$ , $Z=3$ , $\alpha=1$ , $\beta=8$ , $\gamma = 16200$ , making the answer $\boxed{22620}$ .