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I am hoping to find a closed form for the following

$$\tag{1} \sum_{k\geq 1}\frac{H_k}{k^3} x^k $$

Using the generating function

$$\sum_{k\geq 1}H^{(n)}_k x^k = \frac{\operatorname{Li}_n(x)}{1-x}$$

I could find this by simple integration

So I am stuck at evaluating

$$\tag{2}\int^x_0 \frac{\operatorname{Li}_2(1-t)\log(1-t)}{t}\, dt$$

For $x=\pm 1$ the problem can be solved , but what about the general case ?

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    $\begingroup$ @MhenniBenghorbal: Oh, you did. And I do not criticize your work lightly. Unlike you, I actually carry out operations by hand and check with a second source like Mathematica before I make a judgment. There is no doubt in my mind that the integral posted by the OP takes the value $-\pi^4/120$ at $x=1$. The value in your graph, on the other hand, is $-\pi^4/72$ at $x=1$, again allowing for your elastic definition of a dilogarithm. Your answer does not agree with the OP's integral at least at $x=1$, and I am sure everywhere else outside of $x=0$. I am sorry to need to be blunt about it. $\endgroup$
    – Ron Gordon
    Aug 21, 2013 at 2:59
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    $\begingroup$ @MhenniBenghorbal: how about allowing the OP to define what the various conventions mean? His integral has no reference to any "dilog," which looks exactly the same as what's inside the integral anyway. It is you that is confused, not I nor the OP. $\endgroup$
    – Ron Gordon
    Aug 21, 2013 at 3:31
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    $\begingroup$ @ZaidAlyafeai: This is equivalent to your integral at $x=1$. That is the point I have been trying to make. I hope you can now see your way through this miasma. $\endgroup$
    – Ron Gordon
    Aug 21, 2013 at 3:35
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    $\begingroup$ @RonGordon, don't worry I know what I am doing :) $\endgroup$ Aug 21, 2013 at 3:38
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    $\begingroup$ @mhennibenghorbal , thanks :) $\endgroup$ Aug 22, 2013 at 20:07

5 Answers 5

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$$\begin{align} &\int^x_0 \frac{\operatorname{Li}_2(1-t)\log(1-t)}{t}\,dt=\\ &\frac{\operatorname{Li}_2^2(1-x)}2-2\operatorname{Li}_4\left(1-\frac1x\right)-2\operatorname{Li}_4(1-x)+2\operatorname{Li}_4(x)-\operatorname{Li}_2\left(1-\frac1x\right)\log^2\left(\frac1x-1\right)+\\ &\operatorname{Li}_2(x)\left(\log^2\left(\frac1x-1\right)+\log(1-x)\log(x)\right)+2\operatorname{Li}_3(x)\log\left(\frac1x\right)+\\&2\operatorname{Li}_3\left(1-\frac1x\right)\log\left(\frac1x-1\right)+2\operatorname{Li}_3(1-x)\left(\log\left(\frac1x-1\right)+\log(x)\right)-\\ &\frac14\log^4\left(\frac1x\right)+\frac13\log^3(1-x)\left(2\log(x)-\log\left(\frac1x\right)\right)-\\ &\log(1-x)\left(\log^3\left(\frac1x\right)+\frac13\pi^2\log\left(\frac1x\right)+\frac{\pi^2}6\log(x)\right)+\\&\log^2(1-x)\left(-\log^2\left(\frac1x\right)+\frac12\log^2(x)+\log(x)\log\left(\frac1x\right)-\frac{\pi^2}6\right)-\frac{11\pi^4}{360},\end{align}$$ that can be checked by taking derivatives from both sides.

See the corresponding indefinite integral at WolframAlpha.

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The form of the integrand suggests the use of the reflection formula might come handy. We have: \begin{eqnarray} &&\int\limits_0^x Li_2(1-t) \frac{\log(1-t)}{t} dt = \int\limits_0^x \left(\frac{\pi^2}{6} - \log(t) \log(1-t) - Li_2(t) \right) \cdot \frac{\log(1-t)}{t} dt \\ &&=-\frac{\pi^2}{6} Li_2(x) + \frac{1}{2} Li_2(x)^2 - \int\limits_0^x \log(t) \frac{\log(1-t)^2}{t} dt \\ &&=-\frac{\pi^2}{6} Li_2(x) + \frac{1}{2} Li_2(x)^2 - \left(-2 S_{2,2}(x) + 2 \log(x) S_{1,2}(x)\right) \end{eqnarray} where $S_{p,q}(x)$ are the Nielsen generalized polylogarithms (see http://mathworld.wolfram.com/NielsenGeneralizedPolylogarithm.html for definition).

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Here is a proof for your sum not the integral,

From here we have

$$\frac12\int_0^y\frac{\ln x\ln^2(1-x)}{x}dx=\operatorname{Li}_4(y)-\ln y\operatorname{Li}_3(y)+\ln y\sum_{n=1}^\infty\frac{H_n}{n^2}y^n-\sum_{n=1}^\infty\frac{H_n} {n^3}y^n$$

Substitute

$$\sum_{n=1}^\infty\frac{H_{n}}{n^2}y^{n}=\operatorname{Li}_3(y)-\operatorname{Li}_3(1-y)+\ln(1-y)\operatorname{Li}_2(1-y)+\frac12\ln y\ln^2(1-y)+\zeta(3)$$

and

$$\frac12\int_0^y\frac{\ln x\ln^2(1-x)}{x}dx$$ $$=\frac14\ln^2y\ln^2(1-y)-\frac16\ln^3y\ln(1-y)+\frac1{24}\ln^4(1-y)+\frac16\ln^3\left(\frac{y}{1-y}\right)\ln(1-y)\\-\frac12\ln^2\left(\frac{y}{1-y}\right)\operatorname{Li}_2\left(\frac{y}{y-1}\right)+\ln \left(\frac{y}{1-y}\right)\operatorname{Li}_3\left(\frac{y}{y-1}\right)-\operatorname{Li}_4\left(\frac{y}{y-1}\right)-\frac12\ln^2y\operatorname{Li}_2(y)\\+\ln y\operatorname{Li}_3(y)-\operatorname{Li}_4(y)+\frac12\ln^2(1-y)\operatorname{Li}_2(1-y)-\ln (1-y)\operatorname{Li}_3(1-y)+\operatorname{Li}_4(1-y)-\zeta(4)$$

we obtain that

$$\sum_{n=1}^\infty\frac{H_n}{n^3}y^n$$ $$=\zeta(4)-\frac1{24}\ln^4(1-y)+\frac16\ln^3y\ln(1-y)-\frac16\ln^3\left(\frac{y}{1-y}\right)\ln(1-y)+\frac14\ln^2y\ln^2(1-y)$$

$$-\frac12\ln^2(1-y)\operatorname{Li}_2(1-y)+\frac12\ln^2y\operatorname{Li}_2(y)+\ln (1-y)\operatorname{Li}_3(1-y)-\ln y\operatorname{Li}_3(y)$$

$$-\ln y\operatorname{Li}_3(1-y)+\ln y\ln(1-y)\operatorname{Li}_2(1-y)+\zeta(3)\ln y+2\operatorname{Li}_4(y)-\operatorname{Li}_4(1-y)$$

$$+\frac12\ln^2\left(\frac{y}{1-y}\right)\operatorname{Li}_2\left(\frac{y}{y-1}\right)-\ln \left(\frac{y}{1-y}\right)\operatorname{Li}_3\left(\frac{y}{y-1}\right)+\operatorname{Li}_4\left(\frac{y}{y-1}\right)$$


If we use Landen's identity

$$\operatorname{Li}_2(y)+\operatorname{Li}_2\left(\frac{y}{y-1}\right)=-\frac12\ln^2(1-y)$$

and

$$\operatorname{Li}_3(1-y)+\operatorname{Li}_3(y)+\operatorname{Li}_3\left(\frac{y}{y-1}\right)=\zeta(3)+\frac16\ln^3(1-y)-\frac12\ln^2y\ln(1-y)+\zeta(2)\ln y$$

the sum simplifies to

\begin{align} \sum_{n=1}^\infty\frac{H_n}{n^3}y^n&=\operatorname{Li}_4\left(\frac{y}{y-1}\right)-\frac12\operatorname{Li}_2^2\left(\frac{y}{y-1}\right)+2\operatorname{Li}_4(y)-\operatorname{Li}_4(1-y)-\ln(1-y)\operatorname{Li}_3(y)\\ &\quad +\frac12\ln^2(1-y)\operatorname{Li}_2(y)+\frac12\operatorname{Li}_2^2(y)+\frac16\ln^4(1-y)-\frac16\ln y\ln^3(1-y)\\ &\quad+\frac12\zeta(2)\ln^2(1-y)+\zeta(3)\ln(1-y)+\zeta(4) \end{align}


To get your integral, integrate by parts

$$\int_0^y\frac{\ln(1-x)\operatorname{Li}_2(1-x)}{x}dx=\operatorname{Li}_2(y)\operatorname{Li}_2(1-y)+\int_0^y\frac{\ln x\operatorname{Li}_2(x)}{1-x}dx$$

$$=\operatorname{Li}_2(y)\operatorname{Li}_2(1-y)+\sum_{n=1}^\infty \left(H_n^{(2)}-\frac1{n^2}\right)\int_0^y x^{n-1}\ln x\ dx$$

$$=\operatorname{Li}_2(y)\operatorname{Li}_2(1-y)+\sum_{n=1}^\infty \left(H_n^{(2)}-\frac1{n^2}\right)\left(\ln y\frac{y^n}{n}-\frac{y^n}{n^2}\right)$$

$$=\operatorname{Li}_2(y)\operatorname{Li}_2(1-y)-\ln y\operatorname{Li}_3(y)+\operatorname{Li}_4(y)+\ln y\sum_{n=1}^\infty\frac{H_n^{(2)}}{n}y^n-\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}y^n$$

by Cauchy product we have

$$\frac12\operatorname{Li}_2^2(y)=2\sum_{n=1}^\infty\frac{H_n}{n^3}y^n+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}y^n-3\operatorname{Li}_4(y)$$

which gives

$$\int_0^y\frac{\ln(1-x)\operatorname{Li}_2(1-x)}{x}dx$$ $$=\operatorname{Li}_2(y)\operatorname{Li}_2(1-y)-\frac12\operatorname{Li}_2^2(y)-\ln y\operatorname{Li}_3(y)-2\operatorname{Li}_4(y)+\ln y\sum_{n=1}^\infty\frac{H_n^{(2)}}{n}y^n+2\sum_{n=1}^\infty\frac{H_n}{n^3}y^n$$

where

$$\sum_{n=1}^\infty\frac{H_{n}^{(2)}}{n}y^{n}=\operatorname{Li}_3(y)+2\operatorname{Li}_3(1-y)-\ln(1-y)\operatorname{Li}_2(1-y)-\zeta(2)\ln(1-y)-2\zeta(3)$$

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$0<x<1$, \begin{align} J(x)&=\int^x_0 \frac{\operatorname{Li}_2(1-t)\log(1-t)}{t}\, dt\\ \operatorname{Li}_2(1-x)&=-\int_0^{1-x}\frac{\ln(1-t)}{t}dt\\ &{\overset{y=1-t}=}-\int_x^1 \frac{\ln t}{1-t}\,dt\\ &=\Big[\ln(1-t)\ln t\Big]_x^1-\int_x^1 \frac{\ln(1-t)}{t}dt\\ &=-\ln(1-x)\ln x+\zeta(2)-\operatorname{Li}_2(x)\\ \int_0^x \frac{\ln t}{1-t}\,dt&=-\Big[\ln(1-t)\ln t\Big]_0^x+\int_0^x \frac{\ln(1-t)}{t}dt\\ &=-\ln(1-x)\ln x-\operatorname{Li}_2(x)\\ J(x)&=-\Big[\operatorname{Li}_2(1-t)\operatorname{Li}_2(t)\Big]_0^x+\int_0^x \frac{\operatorname{Li}_2(t)\ln t}{1-t}dt\\ &=-\operatorname{Li}_2(1-x)\operatorname{Li}_2(x)+\int_0^x \frac{\operatorname{Li}_2(t)\ln t}{1-t}dt\\ &=\ln(1-x)\ln x\operatorname{Li}_2(x)-\zeta(2)\operatorname{Li}_2(x)+\operatorname{Li}^2_2(x)+\int_0^x \frac{\operatorname{Li}_2(t)\ln t}{1-t}\,dt\\ \int_0^x \frac{\operatorname{Li}_2(t)\ln t}{1-t}\,dt&=\left[\left(\int_0^t \frac{\ln u}{1-u}\,du\right)\operatorname{Li}_2(t)\right]_0^x+\int_0^x \left(\int_0^t \frac{\ln u}{1-u}\,du\right)\frac{\ln(1-t)}{t}\,dt\\ &=-\ln(1-x)\ln x\operatorname{Li}_2(x)-\operatorname{Li}^2_2(x)-\int_0^x \frac{\Big(\ln(1-t)\ln t+\operatorname{Li}_2(t)\Big)\ln(1-t)}{t}dt\\ &=-\ln(1-x)\ln x\operatorname{Li}_2(x)-\frac{1}{2}\operatorname{Li}^2_2(x)-\int_0^x \frac{\ln^2(1-t)\ln t}{t}dt\\ J(x)&=\frac{\operatorname{Li}^2_2(x)}{2}-\zeta(2)\operatorname{Li}_2(x)-\int_0^x \frac{\ln^2(1-t)\ln t}{t}dt\\ &=\frac{\operatorname{Li}^2_2(x)}{2}-\zeta(2)\operatorname{Li}_2(x)-\Big[\frac{\ln^2(1-t)\ln^2 t}{2}\Big]_0^x -\int_0^x \frac{\ln(1-t)\ln^2 t}{1-t}dt\\ &=\frac{\operatorname{Li}^2_2(x)}{2}-\zeta(2)\operatorname{Li}_2(x)-\frac{\ln^2(1-x)\ln^2 x}{2}-\int_0^x \frac{\ln(1-t)\ln^2 t}{1-t}dt\\ K(x)&=-\int_0^x \frac{\ln(1-t)\ln^2 t}{1-t}\,dt\\ &=\int_0^x \left(\sum_{n=1}^\infty \text{H}_n t^n\right)\ln^2 t dt\\ &=\sum_{n=1}^\infty \text{H}_n \left(\int_0^x t^n\ln^2 t \,dt\right)\\ &=\sum_{n=1}^\infty \left(\text{H}_{n+1}-\frac{1}{n+1}\right) \left(\int_0^x t^n\ln^2 t dt\right)\\ &=\left(\ln^2 x\sum_{n=1}^\infty \frac{\text{H}_{n+1}x^{n+1}}{n+1}-2\ln x\sum_{n=1}^\infty \frac{\text{H}_{n+1}x^{n+1}}{(n+1)^2}+2\sum_{n=1}^\infty \frac{\text{H}_{n+1}x^{n+1}}{(n+1)^3}\right)-\\ &\left(\ln^2 x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^2}-2\ln x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^3}+2\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^4}\right)\\ &=\ln^2 x\sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n}-2\ln x\sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n^2}+2\sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n^3}-\operatorname{Li}_2(x)\ln^2 x+\\ &2\operatorname{Li}_3(x)\ln x-2\operatorname{Li}_4(x)\\ \sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n}&=\operatorname{Li}_2(x)+\frac{1}{2}\ln^2(1-x)\\ K(x)&=-2\ln x\sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n^2}+2\sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n^3}+\frac{1}{2}\ln^2(1-x)\ln^2 x+\\ &2\operatorname{Li}_3(x)\ln x-2\operatorname{Li}_4(x)\\ \end{align} $\boxed{\displaystyle J(x)=\frac{1}{2}\operatorname{Li}^2_2(x)-\zeta(2)\operatorname{Li}_2(x)+2\operatorname{Li}_3(x)\ln x-2\operatorname{Li}_4(x)-2\ln x\sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n^2}+2\sum_{n=1}^\infty \frac{\text{H}_{n}x^{n}}{n^3}}$

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Start with using the reflection formula of the dilogarithm function,

$$\text{Li}_2(1-x)=\zeta(2)-\ln(x)\ln(1-x)-\text{Li}_2(x)$$

$$\Longrightarrow I=\zeta(2)\underbrace{\int_0^1\frac{\ln(1-x)}{x}\ dx}_{-\zeta(2)}-\underbrace{\int_0^1\frac{\ln(x)\ln^2(1-x)}{x}\ dx}_{\text{Beta function:}\ -\frac12\zeta(4)}-\underbrace{\int_0^1\frac{\ln(1-x)\text{Li}_2(x)}{x}\ dx}_{-\frac12\text{Li}_2^2(x)|_0^1=-\frac54\zeta(4)}=-\frac34\zeta(4)$$

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