MHB How Can We Solve the Integral Involving the Dilogarithm Function?

[alyafey22]

This thread will be dedicated to try finding a general solution for the integral

$$\int^x_0 \frac{\operatorname{Li}_2(t)^2}{t}\, dt \,\,\,\,\,\,\, 0\leq x \leq 1$$
​

We define the following

$$\operatorname{Li}_2(x)^2 =\left( \int^x_0 \frac{\log(1-t)}{t}\, dt \right)^2$$​

This is NOT a tutorial , every member is encouraged to give thoughts on how to solve the question.

 

[alyafey22]

Let us try the special case $$x=1$$

$$
\begin{align}

\int^1_0 \frac{\operatorname{Li}_2(x)^2}{x}\, dx &= \int^1_0 \sum_{k\geq 1} \sum_{n\geq 1}\frac{1}{n^2 k^2}\, x^{k+n-1}\, dx \\
& = \sum_{k\geq 1}\sum_{n\geq 1 }\frac{1}{n^2 k^2 (n+k)}\\
& = \sum_{k\geq 1}\frac{1}{k^2} \sum_{n \geq 1}\left( \frac{1}{k \, n^2}-\frac{1}{k^2 \, n}+\frac{1}{k^2(n+k)} \right)\\
& = \sum_{k\geq 1}\frac{1}{k^3} \sum_{n\geq 1}\frac{1}{n^2}-\sum_{k\geq 1}\frac{1}{k^4} \sum_{n \geq 1} \frac{1}{n}-\frac{1}{(n+k)}\\
& = \zeta(3)\zeta(2)- \sum_{k \geq 1}\frac{H_k}{k^4}\\
&= 2\zeta(3)\zeta(2)-3\zeta(5)
\end{align}

$$

 

[alyafey22]

That made me think of finding an analytic solution of the special case $$x=\frac{1}{2}$$ . My chance might be slim but I will try it later.

 

[polygamma]

The fact that Wolfram Alpha can't express an antiderivative in terms of elementary functions and polylogs (or in terms of any functions for that matter) is a bit discouraging.

 

[alyafey22]

The fact that Wolfram Alpha can't express an antiderivative in terms of elementary functions and polylogs (or in terms of any functions for that matter) is a bit discouraging.

I think the solution if it exists will involve infinite sums of partial sums of poly logarithms. I think we should define a new function.

 

[alyafey22]

The solution of the integral seems to involve the following double sum

$$\sum_{k\geq 1} \sum_{n\geq 1} \frac{x^{n+k}}{(nk)^2 (n+k)}$$

Generally it will be interesting looking at the sum

$$\sum_{a_1\geq 1} \sum_{a_2\geq 1} \cdots \sum_{a_n\geq 1} \frac{x^{a_1+a_2+\cdots +a_n}}{(a_1a_2 \cdots a_n)^k(a_1+a_2+\cdots +a_n)}$$