Please help me to evaluate this integral in a closed form: $$I=\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$$ Using integration by parts I found that it could be expressed through integrals of elementary functions: $$I_1=\int_0^\infty\log\left(1-e^{-\pi x}\right)\log\left(1+x^2\right)dx$$ $$I_2=\int_0^\infty x\log\left(1-e^{-\pi x}\right)\arctan x\,dx$$


1. Solution

Here is another solution: Notice that

\begin{align*} &\int_{0}^{\infty} \operatorname{Li}_2(e^{-\pi x}) \arctan x \, dx \\ &\qquad = \overbrace{\left[ -\tfrac{1}{\pi} \operatorname{Li}_3(e^{-\pi x}) \arctan x \right]_{0}^{\infty}}^{=0} + \int_{0}^{\infty} \frac{\operatorname{Li}_3(e^{-\pi x})}{\pi(1 + x^2)} \, dx \\ &\qquad = \sum_{n=1}^{\infty} \frac{1}{n^3} \int_{0}^{\infty} \frac{e^{-n\pi x}}{\pi(1+x^2)} \, dx \\ &\qquad = \sum_{n=1}^{\infty} \frac{1}{n^2} \int_{0}^{\infty} \frac{e^{-x}}{(\pi n)^2 + x^2} \, dx \qquad (n\pi x \mapsto x) \\ &\qquad = \frac{\pi^2}{2} \int_{0}^{\infty} \left( \frac{1}{x^4} + \frac{1}{3x^2} - \frac{\coth x}{x^3} \right) e^{-x} \, dx \end{align*}

In order to evaluate the last integral, we introduce the function $I(s)$ defined by

$$ I(s) = \frac{\pi^2}{2} \int_{0}^{\infty} \left( \frac{1}{x^4} + \frac{1}{3x^2} - \frac{\coth x}{x^3} \right) x^{s} e^{-x} \, dx. $$

Then $I(s)$ is analytic for $\Re(s) > -1$ and our integral can be written as $I(0)$. Now assume for a moment that $\Re(s) > 3$. Then

\begin{align*} I(s) &= \frac{\pi^2}{2} \int_{0}^{\infty} \left( \frac{1}{x^4} + \frac{1}{x^3} + \frac{1}{3x^2} - \frac{2}{x^3(1 - e^{-2x})} \right) x^{s} e^{-x} \, dx \\ &= \frac{\pi^2}{2} \left( \Gamma(s-3) + \Gamma(s-2) + \frac{1}{3}\Gamma(s-1) - 2 \sum_{n=0}^{\infty} \frac{\Gamma(s-2)}{(2n+1)^{s-2}} \right) \\ &= \frac{\pi^2}{2} \left( \Gamma(s-3) + \Gamma(s-2) + \frac{1}{3}\Gamma(s-1) - 2 \Gamma(s-2)(1 - 2^{2-s})\zeta(s-2) \right). \end{align*}

By the principle of analytic continuation, this relation continues to hold on $\Re(s) > -1$. So we only need to take limit as $s \to 0$. To this end, we treat two parts separately:

  1. It is easy to check that $\Gamma(s-3) + \Gamma(s-2) + \frac{1}{3}\Gamma(s-1) \to \frac{1}{9}$ as $s \to 0$.

  2. Using the functional equation of $\zeta(s)$ and the reflection formula for $\Gamma(s)$, we have $$ 2 \Gamma(s-2)(1 - 2^{2-s})\zeta(s-2) = \frac{(1 - 2^{s-2})\pi^{s-2}}{\cos(\pi s/2)} \zeta(3-s). $$ Taking $s \to 0$, this converges to $\frac{3}{4\pi^2}\zeta(3)$.

Combining these two observations, we have

$$ I(0) = \frac{\pi^2}{18} - \frac{3}{8}\zeta(3). $$

2. Generalization

We can generalize the problem by considering the family of integrals

$$ I_n = \int_{0}^{\infty} \operatorname{Li}_{n}(e^{-\pi x}) \arctan x \, dx. $$

When $n$ is even, we can use the same technique as in our solution to obtain

Claim. For $m = 0, 1, 2, \cdots$ we have $$ I_{2m} = \frac{(-1)^m \pi^{2m}}{2} \sum_{k=0}^{2m+1} \frac{2^k B_k H_{2m+1-k}}{k!(2m+1-k)!} - \frac{1}{2}\eta(2m+1), \tag{1} $$ where $(B_k)$ are the Bernoulli numbers, $(H_k)$ are the harmonic numbers, and $$\eta(s) = (1-2^{1-s})\zeta(s)$$ is the Dirichlet eta function.

Remark 1. Following @Marco Cantarini's calculation, we have

$$ I_n = \sum_{k=0}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!(2k+1)} \eta(n-2k) - \frac{1}{2}\eta(n+1). \tag{2} $$

This reduces to a finite summation when $n = 2m$ is even:

$$ I_{2m} = (-1)^m \pi^{2m} \sum_{k=0}^{2m} \frac{(1-2^{k-1})B_k}{(2m+1-k)!(2m+1-k)(k!)} - \frac{1}{2}\eta(2m+1). \tag{3} $$

So we have two different representations for $I_{2m}$.

Remark 2. Using (1), we find that for $|z| < 1$,

$$ \sum_{m=0}^{\infty} I_{2m}z^{2m} = \frac{1}{2}\left( \frac{\operatorname{Si}(\pi z)}{\sin (\pi z)} - \int_{0}^{\infty} \frac{\cosh(z t)}{e^t + 1} \, dt \right). \tag{4} $$

Thus in principle, we can find the values of $I_{2m}$ by differentiating (4) $2m$ times.

  • $\begingroup$ Amazing answer! Thanks $\endgroup$ – Nik Z. Dec 31 '15 at 21:02

We have $$\int_{0}^{\infty}\textrm{Li}_{2}\left(e^{-\pi x}\right)\arctan\left(x\right)dx=\sum_{k\geq1}\frac{1}{k^{2}}\int_{0}^{\infty}e^{-\pi kx}\arctan\left(x\right)dx=\frac{1}{\pi}\sum_{k\geq1}\frac{1}{k^{3}}\int_{0}^{\infty}\frac{e^{-\pi kx}}{1+x^{2}}dx $$ and this is the Laplace transform of $\frac{1}{1+x^{2}} $ at $s=\pi k $. This can be calculated (see for example here for $s=1 $) $$\frac{1}{\pi}\sum_{k\geq1}\frac{1}{k^{3}}\left(\textrm{Ci}\left(\pi k\right)\sin\left(\pi k\right)+\frac{\pi\cos\left(\pi k\right)}{2}-\textrm{Si}\left(\pi k\right)\cos\left(\pi k\right)\right)=$$ $$=-\frac{3}{8}\zeta\left(3\right)-\frac{1}{\pi}\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\textrm{Si}\left(\pi k\right) $$ where $\textrm{Ci}\left(x\right) $ and $\textrm{Si}\left(x\right) $ are the cosine and the sine integral. Now note, using the power series of $\textrm{Si}\left(x\right) $ $$\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\textrm{Si}\left(\pi k\right)=\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\sum_{n\geq1}\left(-1\right)^{n-1}\frac{\left(\pi k\right)^{2n-1}}{\left(2n-1\right)\left(2n-1\right)!}=$$ $$=\sum_{n\geq1}\left(-1\right)^{n-1}\frac{\pi^{2n-1}\left(2^{2n-3}-1\right)\zeta\left(4-2n\right)}{\left(2n-1\right)\left(2n-1\right)!} =\pi\left(2^{-1}-1\right)\zeta\left(2\right)-\frac{\pi^{3}\zeta\left(0\right)}{18}=$$ $$=-\frac{\pi^{3}}{18} $$ because $\zeta\left(-2n\right)=0,\,\forall n\geq1 $. So we have $$\int_{0}^{\infty}\textrm{Li}_{2}\left(e^{-\pi x}\right)\arctan\left(x\right)dx=-\frac{3}{8}\zeta\left(3\right)+\frac{\pi^{2}}{18} $$ as conjectured by dbanet.

  • $\begingroup$ Maybe you should add that u use $\zeta(0)=-1/2$ and give a reference how one arrives at this value . Otherwise: Quite nice! (+1) $\endgroup$ – tired Aug 4 '15 at 13:57

Some heuristics suggest $\displaystyle\int\limits_0^\infty\operatorname{Li}_2\left(e^{-\pi x}\right)\arctan(x)\operatorname{d}x=\frac{\pi^2}{18}-\frac{3\zeta\left(3\right)}{8}.$

  • $\begingroup$ How many matching decimal digits have you got? $\endgroup$ – Vladimir Reshetnikov Aug 2 '15 at 19:51
  • $\begingroup$ @Vladimir, up to 90 decimals. $\endgroup$ – dbanet Aug 2 '15 at 20:29
  • 6
    $\begingroup$ I checked more than $1500$ digits, they match the form you conjectured. $\endgroup$ – Vladimir Reshetnikov Aug 4 '15 at 2:25

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