# Integral $\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$

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.

• Amazing answer! Thanks Commented Dec 31, 2015 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.

• 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) Commented Aug 4, 2015 at 13:57
• Your proof isn't rigorous. Commented Jan 16, 2022 at 2:33
• @aaaaaaaaabbbbbbbbbcccccc Can you explain your claim? Commented Jan 17, 2022 at 12:30
• Why does your use of the zeta function (and implicit use of $\eta$) work, when you are working outside the domain of validity of the series representations? Commented Jul 22, 2022 at 17:31
• @FShrike Good point, I hadn't noticed it. I think it is some kind of analytic continuation but, at this moment, I am not able to prove it. Commented Jul 29, 2022 at 5:51

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

• How many matching decimal digits have you got? Commented Aug 2, 2015 at 19:51
• @Vladimir, up to 90 decimals. Commented Aug 2, 2015 at 20:29
• I checked more than $1500$ digits, they match the form you conjectured. Commented Aug 4, 2015 at 2:25

The following is another way to show that $$\sum_{n=1}^{\infty} \frac{(-1)^{k} \operatorname{Si}(k \pi )}{k^{3}}= - \frac{\pi^{3}}{18}.$$

(I'm surprised that replacing $$\operatorname{Si}(k \pi)$$ with its Maclaurin series and then switching the order of summation leads to the correct result because Fubini's theorem is not satisfied and $$\sum_{k=1}^{\infty} (-1)^{k}k^{2n-4} = -\eta(4-2n)$$ only when $$n < 2$$.)

The Fourier series of the piecewise continous function $$f(x) = \begin{cases} x &\text{if} - \pi < x< \pi\\ f(x+ 2\pi) \, &\text{otherwise} \end{cases}$$ is $$f(x) = 2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \sin(kx).$$

Term-by-term integration gives $$\frac{x^{2}}{2} - \frac{\pi^{2}}{6} = 2 \sum_{k=1}^{\infty} \frac{(-1)^{k} \cos(nx)}{k^{2}}, \quad - \pi \le x \le \pi.$$

And term-by-term integration a second time gives $$\frac{x^{3}}{6} - \frac{\pi^{2}x}{6} = 2 \sum_{k=1}^{\infty} \frac{(-1)^{k} \sin(kx)}{k^{3}}, \quad - \pi \le x \le \pi.$$

Therefore, for $$- \pi \le x \le \pi$$, we have \begin{align} \sum_{k=1}^{\infty} \frac{(-1)^{k}\operatorname{Si}(kx)}{k^{3}} &= \sum_{n=1}^{\infty} \frac{(-1)^{k}}{k^{3}}\int_{0}^{kx} \frac{\sin(t)}{t} \, \mathrm dt \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{k}}{k^{3}}\int_{0}^{x} \frac{\sin(ku)}{u} \, \mathrm du \\ &= \int_{0}^{x} \frac{1}{u} \sum_{k=1}^{\infty} \frac{(-1)^{k} \sin(ku)}{k^{3}} \\ &= \int_{0}^{x} \frac{1}{u} \left(\frac{u^{3}}{12}- \frac{\pi^{2}u}{12} \right) \, \mathrm du \\ &= \frac{1}{12} \int_{0}^{x} \left(u^{2} - \pi^{2} \right) \, \mathrm du \\ &=\frac{1}{12} \left( \frac{x^{3}}{3}- \pi^{2}x \right). \end{align}

Letting $$x= \pi$$, we get $$\sum_{k=1}^{\infty} \frac{(-1)^{k}\operatorname{Si}(k\pi)}{k^{3}} = \frac{1}{12} \left( \frac{\pi^{3}}{3}- \pi^{3} \right)= - \frac{\pi^{3}}{18}.$$

• +1. I completely agree with your beginning comment, Marco's answer is... spooky Commented Jul 22, 2022 at 17:33

Some Lemmas..

$$\displaystyle{L1:\quad \int\limits_0^\infty {{e^{ - \pi nx}} \cdot arc\tan x\;dx} = \frac{{ - 1}}{{n \cdot \pi }}\int\limits_0^\infty {{{\left( {{e^{ - \pi nx}}} \right)}{'}} \cdot arc\tan x\;dx} = \frac{{ - 1}}{{n \cdot \pi }}\left[ {{e^{ - \pi nx}}arc\tan x} \right]_0^\infty + \int\limits_0^\infty {\frac{{{e^{ - \pi nx}}}}{{1 + {x^2}}}dx} = \int\limits_0^\infty {\frac{{{e^{ - \pi nx}}}}{{1 + {x^2}}}dx} }$$

$$\displaystyle{L2:\quad \frac{1}{{1 + {x^2}}} = \int\limits_0^\infty {\sin y \cdot {e^{ - xy}}dy\,} }$$ (Laplace transform of the function \displaystyle{\sin y} )

$$\displaystyle{L3:\quad \int\limits_0^\infty {{e^{ - x\left( {n\pi + y} \right)}}} dx\, = \frac{{ - 1}}{{n\pi + y}}\left[ {{e^{ - x\left( {n\pi + y} \right)}}} \right]_0^\infty = \frac{1}{{n\pi + y}}}$$

$$\displaystyle{L4:\quad \int\limits_0^\infty {\frac{{\sin y}}{y}dy} = \frac{\pi }{2}}$$(considered well known)

$$\displaystyle{L5:\quad \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}} = - \frac{3}{4} \cdot \zeta \left( 3 \right)}$$ (obvious)

$$\displaystyle{L6:\quad L{i_3}\left( { - {e^{ix}}} \right) - L{i_3}\left( { - {e^{ - ix}}} \right) = \frac{i}{6}\left( {{x^3} - {\pi ^2}x} \right)}$$ because from here http://functions.wolfram.com/ZetaFuncti ... owAll.html we know that for $$\displaystyle{z \notin \left( {0,1} \right)}$$ it applies $$\displaystyle{L{i_3}\left( z \right) - L{i_3}\left( {\frac{1}{z}} \right) = - \frac{1}{6}\left( {{{\log }^3}\left( { - z} \right) + {\pi ^2}\log \left( { - z} \right)} \right)}$$ so $$\displaystyle{L{i_3}\left( { - {e^{ix}}} \right) - L{i_3}\left( { - {e^{ - ix}}} \right) = - \frac{1}{6}{\log ^3}\left( {{e^{ix}}} \right) - \frac{{{\pi ^2}}}{6}\log \left( {{e^{ix}}} \right) = \frac{i}{6}\left( {{x^3} - {\pi ^2}x} \right)}$$

On our topic..

$$\displaystyle{I = \int\limits_0^\infty {L{i_2}\left( {{e^{ - \pi x}}} \right) \cdot arc\tan x\;dx} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \int\limits_0^\infty {{e^{ - \pi nx}} \cdot arc\tan x\;dx} \mathop = \limits^{L1} \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} \int\limits_0^\infty {\frac{{{e^{ - \pi nx}}}}{{1 + {x^2}}}dx} \mathop = \limits^{L2} }$$

$$\displaystyle{ = \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} \int\limits_0^\infty {{e^{ - \pi nx}}\int\limits_0^\infty {\sin y \cdot {e^{ - xy}}dy\,} dx} = \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} \int\limits_0^\infty {\sin y\int\limits_0^\infty { \cdot {e^{ - x\left( {n\pi + y} \right)}}} dx\,dy} \mathop = \limits^{L3} }$$ $$\displaystyle{\frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} \int\limits_0^\infty {\frac{{\sin y}}{{n\pi + y}}dy} = }$$

$$\displaystyle{ = \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}} \int\limits_0^\infty {\frac{{\sin \left( {n\pi + y} \right)}}{{n\pi + y}}dy} \mathop = \limits^{n\pi + y \to y} \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}} \int\limits_{n \cdot \pi }^\infty {\frac{{\sin y}}{y}dy} = }$$ $$\displaystyle{\frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}} \left( {\int\limits_0^\infty {\frac{{\sin y}}{y}dy} - \int\limits_0^{n \cdot \pi } {\frac{{\sin y}}{y}dy} } \right)\mathop = \limits^{L4} }$$

$$\displaystyle{ = \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}} \left( {\frac{\pi }{2} - \int\limits_0^{n \cdot \pi } {\frac{{\sin y}}{y}dy} } \right) = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}} - \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}} \int\limits_0^{n \cdot \pi } {\frac{{\sin y}}{y}dy} }$$ $$\displaystyle{\mathop = \limits^{L5} \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^3}}}} \int\limits_0^{n \cdot \pi } {\frac{{\sin y}}{y}dy} - }$$

$$\displaystyle{ - \frac{3}{8} \cdot \zeta \left( 3 \right)\mathop = \limits^{y = nx} \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^3}}}} \int\limits_0^\pi {\frac{{\sin nx}}{x}dx} - \frac{3}{8} \cdot \zeta \left( 3 \right) = }$$ $$\displaystyle{\frac{1}{{2 \cdot \pi \cdot i}}\int\limits_0^\pi {\frac{1}{x}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {{e^{inx}} - {e^{ - inx}}} \right)}}{{{n^3}}}} } - \frac{3}{8} \cdot \zeta \left( 3 \right) = }$$

$$\displaystyle{ = - \frac{1}{{2 \cdot \pi \cdot i}}\int\limits_0^\pi {\frac{1}{x}\sum\limits_{n = 1}^\infty {\left( {\frac{{{{\left( { - {e^{ix}}} \right)}^n}}}{{{n^3}}} - \frac{{{{\left( { - {e^{ - ix}}} \right)}^n}}}{{{n^3}}}} \right)} \,dx} - \frac{3}{8} \cdot \zeta \left( 3 \right) = }$$ $$\displaystyle{ - \frac{1}{{2 \cdot \pi \cdot i}}\int\limits_0^\pi {\frac{1}{x}\left( {L{i_3}\left( { - {e^{ix}}} \right) - L{i_3}\left( { - {e^{ - ix}}} \right)} \right)dx} - \frac{3}{8} \cdot \zeta \left( 3 \right)\mathop = \limits^{L6} }$$

$$\displaystyle{ = - \frac{1}{{12 \cdot \pi }}\int\limits_0^\pi {\left( {{x^2} - {\pi ^2}} \right)dx} - \frac{3}{8} \cdot \zeta \left( 3 \right) = - \frac{1}{{12 \cdot \pi }}\left( { - \frac{{2{\pi ^3}}}{3}} \right) - \frac{3}{8} \cdot \zeta \left( 3 \right) = \frac{{{\pi ^2}}}{{18}} - \frac{3}{8} \cdot \zeta \left( 3 \right)}$$