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How to evaluate $$\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx\ ?$$

where $\displaystyle\operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}$ , $|x|\leq1$

I came across this integral while I was working on $\displaystyle \displaystyle\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx\ $ and here is how I established a relation between these two integrals:

$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-\underbrace{\int_1^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx}_{x\mapsto 1/x}$$

$$=\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-\int_0^1 \frac{\operatorname{Li}_3(1/x)}{1+x^2}\ dx$$ $$\left\{\color{red}{\text{add the integral to both sides}}\right\}$$

$$2\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx+\int_0^1 \frac{\operatorname{Li}_3(x)-\operatorname{Li}_3(1/x)}{1+x^2}\ dx$$

$$\{\color{red}{\text{use}\ \operatorname{Li}_3(x)-\operatorname{Li}_3(1/x)=2\zeta(2)\ln x-\frac16\ln^3x+i\frac{\pi}2\ln^2x}\}$$

$$=\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx+2\zeta(2)\underbrace{\int_0^1\frac{\ln x}{1+x^2}\ dx}_{-G}-\frac16\underbrace{\int_0^1\frac{\ln^3x}{1+x^2}\ dx}_{-6\beta(4)}+i\frac{\pi}2\underbrace{\int_0^1\frac{\ln^2x}{1+x^2}\ dx}_{2\beta(3)}$$

$$=\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-2\zeta(2)G+\beta(4)+i\pi \beta(3)$$

Then

$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=\frac12\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-\zeta(2)G+\frac12\beta(4)+i\frac{\pi}2 \beta(3)\tag{1}$$

where $\displaystyle\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\ $ is the the Dirichlet beta function.

So any idea how to evaluate any of these two integrals?

Thanks.

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    $\begingroup$ What are you asking? $\endgroup$
    – clathratus
    Commented Jun 13, 2019 at 2:17
  • $\begingroup$ Note the formula you got from the triglogarithmic identity involved you being cavalier about branch cuts; you are missing a factor of $\frac{16 i \pi^4}{512}$ $\endgroup$ Commented Jun 13, 2019 at 4:04
  • $\begingroup$ @clathratus I am not sure if you are being sarcastic or not but I'll take as not. If you want to know what I'm asking just ignore the body and read the title please. $\endgroup$ Commented Jun 13, 2019 at 4:28
  • $\begingroup$ @BrevanEllefsen wolfram gives the closed form. $\endgroup$ Commented Jun 13, 2019 at 4:30
  • $\begingroup$ @BrevanEllefsen I dont think I'm missing any. To make sure, just take x=1/2 and compare the two sides. $\endgroup$ Commented Jun 13, 2019 at 4:45

3 Answers 3

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Finally I got the answer:

Using the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.

$$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $n=2$ we get

$$\operatorname{Li}_{3}(x)=\frac12\int_0^1\frac{x\ln^2 u}{1-xu}\ du$$

we can write

$$\int_0^\infty\frac{\operatorname{Li}_{3}(x)}{1+x^2}\ dx=\frac12\int_0^1\ln^2u\left(\int_0^\infty\frac{x}{(1-ux)(1+x^2)}\ dx\right)\ du$$ $$=\frac12\int_0^1\ln^2u\left(-\frac12\left(\frac{\pi u}{1+u^2}+\frac{2\ln(-u)}{1+u^2}\right)\right)\ du,\quad \color{red}{\ln(-u)=\ln u+i\pi}$$

$$=-\frac{\pi}{4}\underbrace{\int_0^1\frac{u\ln^2u}{1+u^2}\ du}_{\frac3{16}\zeta(3)}-\frac12\underbrace{\int_0^1\frac{\ln^3u}{1+u^2}\ du}_{-6\beta(4)}-i\frac{\pi}2\underbrace{\int_0^1\frac{\ln^2u}{1+u^2}\ du}_{2\beta(3)}$$

Then

$$\int_0^\infty\frac{\operatorname{Li}_{3}(x)}{1+x^2}\ dx=-\frac{3\pi}{64}\zeta(3)+3\beta(4)-i\pi\beta(3)\tag{2}$$


Bonus:

By combining $(1)$ in the question body and $(2)$, the imaginary part $i\pi\beta(3)$ nicely cancels out and we get

$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=2\beta(4)-\zeta(2)G-\frac{3\pi}{128}\zeta(3)$$

where $\beta(4)$ $=\frac{1}{768}\psi^{(3)}(1/4)-\frac{\pi^4}{96}$

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  • $\begingroup$ The formula \begin{align}\Re\int_0^\infty\frac{\operatorname{Li}_{3}(x)}{1+x^2}\ dx&=\frac12\int_0^1\ln^2u\left(\int_0^\infty\frac{x}{(1-ux)(1+x^2)}\ dx\right)\ du\end{align} is weird. In the right side it's $\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx$ not its real part. There is no indication for the values of x involved in the formula about polylogarithm in the beginning of your post. Anyway your closed-form matches numeric evaluation. $\endgroup$
    – FDP
    Commented Nov 22, 2019 at 8:45
  • $\begingroup$ its edited now. thanks for pointing this out. $\endgroup$ Commented Nov 22, 2019 at 8:58
  • $\begingroup$ The formula for polylogarithm in the beginning of your post is valid for $|x|<1$ only. If you take $x=2$ the denominator is $1-2u$ and $1-2u=0$ when $u=1/2\in [0;1]$ $\endgroup$
    – FDP
    Commented Nov 22, 2019 at 15:42
  • $\begingroup$ Yes .. actually i got the right formula yesterday and I'll fix the problem today. $\endgroup$ Commented Nov 22, 2019 at 16:18
  • $\begingroup$ When you consider the integral $$\int_0^\infty\frac{x}{(1-ux)(1+x^2)}\ dx$$ as a CauchyPrincipalValue you don't get the imagniary part. It arises from the $\epsilon$ contour around that singularity. $\endgroup$
    – Diger
    Commented Nov 22, 2019 at 22:05
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For a different solution, use the first result from A simple idea to calculate a class of polylogarithmic integrals by using the Cauchy product of squared Polylogarithm function by Cornel Ioan Valean.

Essentially, the main new results in the presentation are:

Let $a\le1$ be a real number. The following equalities hold: \begin{equation*} i) \ \int_0^1 \frac{\log (x)\operatorname{Li}_2(x) }{1-a x} \textrm{d}x=\frac{(\operatorname{Li}_2(a))^2}{2 a}+3\frac{\operatorname{Li}_4(a)}{a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a}; \end{equation*} \begin{equation*} ii) \ \int_0^1 \frac{\log^2(x)\operatorname{Li}_3(x) }{1-a x} \textrm{d}x=20\frac{\operatorname{Li}_6(a)}{a}-12 \zeta(2)\frac{\operatorname{Li}_4(a)}{ a}+\frac{(\operatorname{Li}_3(a))^2}{a}. \end{equation*} For a fast proof, see the paper above (series expansion combined with the Cauchy product of squared Polylogarithms)

The use of these new results with integrals allows you to obtain your result elegantly, but also other results that are (very) difficult to obtain by other means, including results from the book, (Almost) Impossible Integrals, Sums, and Series.

BONUS: Using these results you may also establish that (or the versions with integration by parts applied).

$$i) \ \int_0^1 \frac{\arctan(x) \operatorname{Li}_2(x)}{x}\textrm{d}x$$ $$=\frac{1}{384}\left(720\zeta(4)+105\pi\zeta(3)+384\zeta(2)G-\psi^{(3)}\left(\frac{1}{4}\right)\right),$$ $$ii)\ \int_0^1 \frac{\arctan(x) \operatorname{Li}_2(-x)}{x}\textrm{d}x$$ $$=\frac{1}{768}\left(\psi^{(3)}\left(\frac{1}{4}\right)-384\zeta(2)G-126\pi\zeta(3)-720\zeta(4)\right).$$


EXPLANATIONS (OP's request): The following way in large steps shows the amazing possible creativity in such calculations.

We'll want to focus on the integral, $\displaystyle \int_0^1 \frac{\arctan(x)\operatorname{Li}_2(x)}{x}\textrm{d}x$ which is a translated form of the main integral.

Now, based on $i)$ where we plug in $a=i$ and then consider the real part, we obtain an integral which by a simple integration by parts reveals that

$$\int_0^1 \frac{\arctan(x)\operatorname{Li}_2(x)}{x}\textrm{d}x=\int_0^1 \frac{\arctan(x)\log(1-x) \log(x)}{x}\textrm{d}x+\frac{17}{48}\pi^2 G+\frac{\pi^4}{32}-\frac{1}{256}\psi^{(3)}\left(\frac{1}{4}\right).$$

Looks like we need to evaluate one more integral and we're done. Well, if you read the book (Almost) Impossible Integrals, Sums, and Series (did you?), particularly the solutions in the sections 3.24 & 3.25 you probably observed the powerful trick of splitting the nonnegative real line at $x=1$ with the hope of getting the same integral in the other side but with an opposite sign. Therefore, with such a careful approach (since we need to avoid the divergence issues), we obtain immediately that $$\int_0^1 \frac{\arctan(x)\log(1-x) \log(x)}{x}\textrm{d}x$$ $$=\frac{1}{2} \underbrace{\int_0^1 \frac{\arctan(x)\log^2(x)}{x}\textrm{d}x}_{\text{Trivial}}+\frac{\pi}{4}\underbrace{\int_0^1 \frac{\log(x)\log(1-x)}{x}\textrm{d}x}_{\text{Trivial}}$$ $$-\frac{1}{2}\Re\left \{\int_0^{\infty}\frac{\arctan(1/x) \log(1-x)\log(x)}{x}\textrm{d}x\right \},$$

and the last integral works simply nice with Cornel's strategy described in the second part of this post (it involves the use of Cauchy Principal Value) https://math.stackexchange.com/q/3488566.

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    $\begingroup$ The explanation is really amazing and helpful. thank you. $\endgroup$ Commented Dec 30, 2019 at 20:07
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I assume we're using the principal branch of the trilogarithm, and we're integrating on the lower side of the branch cut (which is what Mathematica assumes).

Let's integrate the function $$f(z) = \frac{\operatorname{Li}_{s}(z)}{1+z^{2}}, \quad s>0,$$

around a contour consisting of the real axis from $z= -R$ to $z=1-\epsilon$, a small clockwise-oriented semicircle of radius $\epsilon$ about the point $z=1$, the lower side of the branch cut from $z=1+\epsilon $ to $z= R$, and a counterclockwise-oriented semicircle in the lower half-plane of radius $R$.

As $ R \to \infty$, the integral vanishes on the semicircle since $\frac{\operatorname{Li}_{s}(z)}{1+z^{2}} $ is asymptotic to $-\frac{\ln^{s}(-z)}{\Gamma(s+1)z^{2}} $ as $|z| \to \infty$. (See here.)

And the integral vanishes on the small semicircle as $\epsilon \to 0$ since $s>0$. (When $s = 0$, the singularity at $z=1$ becomes a simple pole.)

Therefore, we have $$ \begin{align} \int_{-\infty}^{\infty} \frac{\operatorname{Li}_{s}(x)}{1+x^{2}} \, \mathrm dx &= \int_{0}^{\infty} \frac{\operatorname{Li}_{s}(-x)}{1+x^{2}} \, \mathrm dx + \int_{0}^{\infty} \frac{\operatorname{Li}_{s}(x)}{1+x^{2}} \, \mathrm dx \\ &= \color{red}{-} 2 \pi i \operatorname*{Res}_{z=-i} f(z) \\ &= \pi \operatorname{Li}_{s}(-i) \\ &\overset{\clubsuit}{=} -\pi \left(2^{-s} \eta(s)+i \beta(s) \right). \end{align}$$

In a previous answer I used contour integration to show that $$\int_{0}^{\infty} \frac{\operatorname{Li}_{s}(-x)}{1+x^{2}} \, \mathrm dx = - s \beta(s+1) - \pi 2^{-s-1} \eta(s), \quad s >0. $$

Using that result, we get $$ \int_{0}^{\infty} \frac{\operatorname{Li}_{s}(x)}{1+x^{2}} \, \mathrm dx = -\pi \, 2^{-s-1} \eta(s) + s \beta(s+1) - i \pi \beta(s). $$

For $s=3$ we have

$$ \begin{align} \int_{0}^{\infty} \frac{\operatorname{Li}_{3}(x)}{1+x^{2}} \, \mathrm dx &= - \frac{\pi}{16} \eta(3) + 3 \beta(4) - i \pi \beta(3) \\ &= - \frac{\pi}{16} \left(1-2^{-2} \right) \zeta(3) + 3 \beta(4) - i \pi \left(\frac{\pi^{3}}{32} \right) \\ &= - \frac{3 \pi}{64} \zeta(3) +3 \beta(4) - \frac{i \pi^{4}}{32}. \end{align}$$


$\clubsuit$ https://en.wikipedia.org/wiki/Polylogarithm#Relationship_to_other_functions


Wolfram Alpha seems to have a bit of difficulty approximating the value of the integral when $s$ is between $0$ and $1$. I think this has to do with the behavior of $\operatorname{Li}_{s}(z)$ near $z=1$ when $s <1$.

When $s>1$, $\lim_{z \to 1} \operatorname{Li}_{s}(z)$ exists and is finite. But when $s <1$, $\operatorname{Li}_{s}(z)$ behaves like $\Gamma(1-s)(- \ln z)^{s-1}$ near $z=1$. See the discussion here.

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    $\begingroup$ Thank you Random (+1). I managed to find more generalized integrals in this preprint researchgate.net/publication/… $\endgroup$ Commented Mar 26 at 1:47
  • $\begingroup$ +1$\phantom{aaaaa}$ $\endgroup$ Commented Apr 27 at 19:55

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