General formula for $\int^1_0 x^\alpha \log(1-x)\operatorname{Li}_2 (x)\, \mathrm dx$

Question

Consider a following definite integral

$$I(\alpha) = \int^1_0 x^\alpha \log(1-x)\operatorname{Li}_2 (x)\, \mathrm dx$$

Mathematica is able to provide a result for many $\alpha$ integers. See table here:

$$\begin{array}{c} I(0)=-2 \zeta (3)+3-\frac{\pi ^2}{6} \\ I(1)=-\zeta (3)+\frac{25}{16}-\frac{\pi ^2}{8} \\ I(2)=-\frac{2 \zeta (3)}{3}+\frac{227}{216}-\frac{11 \pi ^2}{108} \\ I(3)=-\frac{\zeta (3)}{2}+\frac{5443}{6912}-\frac{25 \pi ^2}{288} \\ I(4)=-\frac{2 \zeta (3)}{5}+\frac{677309}{1080000}-\frac{137 \pi ^2}{1800} \\ I(5)=-\frac{\zeta (3)}{3}+\frac{673333}{1296000}-\frac{49 \pi ^2}{720} \\ I(6)=-\frac{2 \zeta (3)}{7}+\frac{229495199}{518616000}-\frac{121 \pi ^2}{1960} \\ I(7)=-\frac{\zeta (3)}{4}+\frac{1824315697}{4741632000}-\frac{761 \pi ^2}{13440} \\ I(8)=-\frac{2 \zeta (3)}{9}+\frac{16317321883}{48009024000}-\frac{7129 \pi ^2}{136080} \\ I(9)=-\frac{\zeta (3)}{5}+\frac{1946452513}{6401203200}-\frac{7381 \pi ^2}{151200} \\ I(10)=-\frac{2 \zeta (3)}{11}+\frac{64401732356723}{234300040128000}-\frac{83711 \pi ^2}{1829520} \\ \end{array}$$

Is there a closed form expression for $f,g,h$ in $I(\alpha) = f(\alpha) + g(\alpha)\pi^2 + h(\alpha)\zeta(3)$?

Answer 1

This is immediately revealed by combining the Cauchy product $$ \log(1-x)\operatorname{Li}_2(x)=3\sum_{n=1}^{\infty} \frac{x^n}{n^3}-2\sum_{n=1}^{\infty} x^n\frac{H_n}{n^2}- \sum_{n=1}^{\infty} x^n \frac{H_n^{(2)}}{n} ,\ |x|\le1 \land x\neq1$$ and well-known harmonic series in the mathematical literature. You might find useful results in (Almost) Impossible Integrals, Sums, and Series and in the paper Euler sum involving tail by Moti Levy and C.I. Valean, here.

More precisely, we need slightly adjusted forms (or we let them the way they are and rearrange the series result obtained after integration) of $$\sum_{k=1}^{\infty} \frac{H_k}{(k+1)(k+n+1)}=\frac{H_n^2+H_n^{(2)}}{2n}$$ and $$\sum_{k=1}^{\infty} \frac{H_k^{(2)}}{(k+1)(k+n+1)}= \zeta(2)\frac{H_n}{n}-\frac{1}{n}\sum_{i=1}^n\frac{H_i}{i^2},$$ as they appear in the generalization from Sect. 4.16, page 289 in (Almost) Impossible Integrals, Sums, and Series.

Answer 2

You did a great job in finding $g(\alpha)$ and $h(\alpha)$. Yet in my opinion there is no way to find a general expression for $f(\alpha)$.

My doubts are explained through a simple analysis of the first six pure coefficients:

$$3 ~~~~~ \frac{25}{16} ~~~~~ \frac{227}{216} ~~~~~ \frac{5443}{6912} ~~~~~ \frac{677309}{1080000} ~~~~~ \frac{673333}{1296000}$$

Writing them in terms of primes we have ($p$ denotes a pure prime):

$$p ~~~~~ \frac{5^2}{2^4} ~~~~~ \frac{p}{2^3 3^3} ~~~~~ \frac{p}{2^8 3^3} ~~~~~\frac{p}{2^6 3^3 5^4} ~~~~~ \frac{419 \times 1607}{2^7 3^4 5^3}$$

• It's a rather messy series, where second and sixth numerators are not primes, whilst others are.
• Analysing the denominators, we do not recognise a pattern in terms of the exponents concerning the first three primes.
• The sixth numerator is relatively close to the previous one, rather irregular for a pattern.

So this is my opinion: there is no general expression for $f(\alpha)$.

However I could be wrong, and I could have missed something very subtle someone else might notice! Maybe my very rude and simple analysis can lead someone to something else (better, new, more argumentative), so who knows!

That's the beauty of maths and numbers.

Answer 3

Long Comment On Coaxing an Answer From Mathematica 12: (in answer to comment from O.P.)

Mathematica can't symbolically evaluate $$I(\alpha)=\int^1_0 x^\alpha \log(1-x)\operatorname{Li}_2 (x)\, \mathrm dx$$ directly.

Therefore trying integrating by parts with $\int x^{\alpha } \log (1-x) \, dx=\frac{x^{\alpha +1} \log (1-x)+B_x(\alpha +2,0)}{\alpha +1}$ and $\frac{d\, \text{Li}_2(x)}{dx}=-\frac{\log (1-x)}{x}$, both of which Mathematica is happy with.

Mathematica can find the limits $x\to 1$ and $x\to 0$ to evaluate

$$\left[\frac{\text{Li}_2(x) \left(x^{\alpha +1} \log (1-x)+B_x(\alpha +2,0)\right)}{\alpha +1}\right]^1_0$$

e.g. using

Limit[((Beta[x, 2 + \[Alpha], 0] + x^(1 + \[Alpha]) Log[1 - x])/(1 + \[Alpha])) PolyLog[2, x], x -> 1]    

Mathematica can evaluate the integral,

$$-\int_0^1 \frac{x^{\alpha } \log ^2(1-x)}{\alpha +1} \, dx = -\frac{-6 \psi ^{(1)}(\alpha +2)+6 \left(H_{\alpha +1}\right){}^2+\pi ^2}{(\alpha +1) (6 \alpha +6)}$$

but then fails to evaluate:

$$\int_0^1 \frac{(-\log (1-x)) B_x(\alpha +2,0)}{(\alpha +1) x} \, dx.$$

However this is easily solved by substituting in the infinite series for the Incomplete Beta Function in this case

$$B_x(\alpha +2,0)=\sum _{n=0}^{\infty } \frac{x^{\alpha +n+2}}{\alpha +n+2}$$

and then by changing the order of the resulting integral and summation we have

$$\sum _{n=0}^{\infty } \int_0^1 \frac{x^{\alpha +n+1}(-\log (1-x)) }{(\alpha +1) (\alpha +n+2)} \, dx=\sum _{n=0}^{\infty } \frac{H_{n+\alpha +2}}{(\alpha +1) (\alpha +n+2)^2}$$

Then its just a matter of rearranging, changing the summations to start from $n=1$ and simplifying where necessary.