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Question. Prove the following polylogarithmic integral has the stated value:

$$I:=\int_{0}^{1}\frac{\operatorname{Li}_2{(1-x)}\log^2{(1-x)}}{x}\mathrm{d}x=-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}.$$

I was able to arrive at the proposed value for the integral through a combination of educated guessing and after-the-fact numerical checking (the value of the integral was approximately $I\approx0.457621...$), but I'm at a loss for how to derive it rigorously. Thoughts?

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  • $\begingroup$ Expanding and multiplying power series' then integrating term by term. $\endgroup$ Jul 19, 2014 at 23:05
  • $\begingroup$ It might help to note that $\log(1-x) = -\operatorname{Li}_1(x)$. $\endgroup$
    – Brad
    Jul 19, 2014 at 23:20
  • $\begingroup$ The integral can be evaluated in terms of the harmonic sum $ \displaystyle\sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{2}} $, which in turn can be evaluated using contour integration (or perhaps in some other manner of which I'm not aware). $\endgroup$ Jul 19, 2014 at 23:50
  • $\begingroup$ Have you tried integration by parts with regards to $\text{Li}_2(1-x)$ ? $\endgroup$
    – Lucian
    Jul 19, 2014 at 23:50
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    $\begingroup$ integralsandseries.prophpbb.com/topic413.html#p2719 Just make the substitution $u=1-x$ in (1). There is a proof of (1) later in the post. $\endgroup$ Jul 20, 2014 at 0:12

2 Answers 2

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Since no answers have been posted, I'll expand on my comment above.

There is a general formula that states $$\sum_{n=1}^{\infty}\frac{H_{n}^{(r)}}{n^{q}}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}\int_{0}^{1}\frac{\text{Li}_{q}(x) \log^{r-1}(x) }{1-x}dx $$ where $$ H_{n}^{(r)} = \sum_{k=1}^{n} \frac{1}{k^{r}} .$$

A proof can be found here.

Making the substitution $u = 1-x$, $$ \sum_{n=1}^{\infty}\frac{H_{n}^{(r)}}{n^{q}}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}\int_{0}^{1}\frac{\text{Li}_{q}(1-u) \log^{r-1}(1-u)}{u}dx .$$

Therefore, $$ \int_{0}^{1} \frac{\text{Li}_{2}(1-x)\log^{2}(1-x)}{x} \ dx = 2 \zeta (3) \zeta (2) - 2 \sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{2}} .$$

To simplify the evaluation of that Euler sum slightly, I'm first going to evaluate $ \displaystyle \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}}$ and then use the identity $$\sum_{n=1}^{\infty} \frac{H_{n}^{(r)}}{n^{q}} + \sum_{n=1}^{\infty} \frac{H_{n}^{(q)}}{n^{r}} = \zeta(r) \zeta(q) + \zeta(r+q) . \tag{1} $$

Consider $$f(z) = \frac{ \pi \cot (\pi z) \ \psi_{1}(-z)}{z^{3}} $$ where $\psi_{1}(z)$ is the trigamma function.

The function $f(z)$ has poles of order $3$ at the positive integers, a pole of order 6 at the origin, and simple poles at the negative integers.

On the sides of a square with vertices at $\pm \left( N+ \frac{1}{2} \right) \pm i \left( N+ \frac{1}{2} \right)$, $\cot (\pi z)$ is uniformly bounded.

And when $z$ is large in magnitude and not on the positive real axis, $\psi_{1}(-z)$ is approximately $ \displaystyle - \frac{1}{z}$.

So as $N \to \infty$ through the integers, $ \displaystyle \int f(z) \ dz$ will vanish on all four sides of the square.

Therefore, $$ \sum_{n=-\infty}^{\infty} \text{Res} [f(z), n] = 0.$$

The Laurent expansion of $\psi_{1}(-z)$ at the positive integers (including $0$) is $$ \psi_{1}(-z) = \frac{1}{(z-n)^{2}} + \sum_{m=0}^{\infty} (m+1) \left( (-1)^{m} H_{n}^{(m+2)} + \zeta(m+2) \right) (z-n)^{m} . $$

And the Laurent expansion of $\pi \cot \pi z$ at the integers is

$$ \pi \cot (\pi z) = \frac{1}{z-n} - 2 \sum_{m=1}^{\infty} \zeta(2m) (z-n)^{2m-1} .$$

So at the positive integers, $$f(z) = \frac{1}{z^{3}} \left(\frac{1}{(z-n)^{3}} + \frac{H_{n}^{(2)} - \zeta(2)}{(z-n)} + \mathcal{O}(1) \right) $$

which implies

$$ \begin{align} \text{Res} [f(z), n] &= \text{Res} \left[\frac{1}{z^{3}(z-n)^{3}}, n \right] + \text{Res} \left[\frac{H_{n}^{(2)}-\zeta(2)}{z^{3}(z-n)}, n \right] \\ &= \frac{6}{n^{3}} + \frac{H_{n}^{(2)}}{n^{3}} -\frac{\zeta(2)}{n^{3}} . \end{align} $$

At the negative integers,

$$\text{Res} [f(z), -n] = - \frac{\psi_{1}(n)}{n^{3}} = \frac{H_{n-1}^{(2)} - \zeta(2)}{n^{3}} = \frac{H_{n}^{(2)}}{n^{3}} - \frac{1}{n^{5}} - \frac{\zeta(2)}{n^{3}} .$$

And at the origin,

$$ f(z) = \frac{1}{z^{6}} - \frac{\zeta(2)}{z^{4}} + \frac{2 \zeta(3)}{z^{3}} + \left(\zeta(4) - 2 \zeta^{2}(2) \right) \frac{1}{z^{2}} + \Big(4 \zeta(5) - 4 \zeta(3) \zeta(2) \Big) \frac{1}{z} + \mathcal{O}(1)$$

while implies

$$ \text{Res}[f(z),0] = 4 \zeta(5) - 4 \zeta(3) \zeta(2) .$$

Summing up all the residues,

$$6 \zeta(5) + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} - \zeta(3) \zeta(2) + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} - \zeta(5) - \zeta(3) \zeta(2) + 4 \zeta(5) - 4 \zeta(3) \zeta(2) = 0$$

which implies

$$ \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} = 3 \zeta(3) \zeta(2) - \frac{9}{2} \zeta(5) .$$

Then using $(1)$,

$$\sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{2}} = \zeta(3) \zeta(2) + \zeta(5) - 3 \zeta(3) \zeta(2) + \frac{9}{2} \zeta(5) = - 2 \zeta(3) \zeta(2) + \frac{11}{2} \zeta(5). $$

So finally we have

$$ \begin{align} \int_{0}^{1} \frac{\text{Li}_{2}(1-x)\log^{2}(1-x)}{x} \ dx &= 2 \zeta (3) \zeta (2) - 2 \Big(- 2 \zeta(3) \zeta(2) + \frac{11}{2} \zeta(5) \Big) \\ &= 6 \zeta(3) \zeta(2) - 11 \zeta(5) . \end{align}$$

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  • $\begingroup$ Interesting evaluation. $\endgroup$ Aug 5, 2014 at 19:33
  • $\begingroup$ @OlivierOloa Thanks. $\endgroup$ Aug 5, 2014 at 19:44
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\begin{align} I&=\int_0^1\frac{\operatorname{Li}_2(1-x)\ln^2(1-x)}{x}\ dx=\int_0^1\frac{\operatorname{Li}_2(x)\ln^2x}{1-x}\ dx\\ &=\sum_{n=1}^\infty H_n^{(2)}\int_0^1x^n\ln^2x\ dx=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}\\ &=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}-2\zeta(5)\\ &=2\left(3\zeta(2)\zeta(5)-\frac92\zeta(5)\right)-2\zeta(5)\\ &=6\zeta(2)\zeta(5)-11\zeta(5) \end{align}


Proof:

\begin{align} S&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\sum_{n=1}^\infty\frac1{n^3}\left(\zeta(2)-\sum_{k=1}^\infty\frac1{(n+k)^2}\right)=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{n^3(n+k)^2}\\ &=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\sum_{n=1}^\infty\left(\frac{3}{k^4}\left(\frac1{n}-\frac1{n+k}\right)-\frac2{k^3n^2}-\frac1{k^3(n+k)^2}+\frac1{k^2n^3}\right)\\ &=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\left(\frac{3H_k}{k^4}-\frac{2\zeta(2)}{k^3}-\frac{\zeta(2)-H_k^{(2)}}{k^3}+\frac{\zeta(3)}{k^2}\right)\\ &=\zeta(2)\zeta(3)-3\sum_{k=1}^\infty\frac{H_k}{k^4}+2\zeta(2)\zeta(3)+\zeta(2)\zeta(3)-S-\zeta(2)\zeta(3)\\ 2S&=3\zeta(2)\zeta(3)-3\sum_{k=1}^\infty\frac{H_k}{k^4}\\ &=3\zeta(2)\zeta(3)-3\left(3\zeta(5)-\zeta(2)\zeta(3)\right)\\ &=6\zeta(2)\zeta(3)-9\zeta(5)\\ S&=3\zeta(2)\zeta(3)-\frac92\zeta(5) \end{align}


Addendum

Different proof:

By Cauchy product we have

$$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=\sum_{n=1}^\infty\left(\frac{6H_n}{n^4}+\frac{3H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}-\frac{10}{n^5}\right)x^n$$

set $x=1$ to get

$$\zeta(2)\zeta(3)=6\sum_{n=1}^\infty\frac{H_n}{n^4}+3\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-10\zeta(5)$$

Now lets use the well-known identity

$$\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^q}+\sum_{n=1}^\infty\frac{H_n^{(q)}}{n^p}=\zeta(p)\zeta(q)+\zeta(p+q)$$

set $p=2$ and $q=3$

$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)+\zeta(5)-\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}\tag{*}$$

So

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\frac92\zeta(5)-3\sum_{n=1}^\infty\frac{H_n}{n^4}$$

substitute $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ we get

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(5)-\frac92\zeta(2)\zeta(3)$$

and as a bonus, substitute the last result in (*) we get

$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)$$

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