With reference to my previous question : Show that $\int_0^\infty \frac{x\log(1+x^2)}{e^{2\pi x}+1}dx=\frac{19}{24} - \frac{23}{24}\log 2 - \frac12\log A$

The following integral should be manageable :

$$\int_0^\infty \frac{\log(1+x^2)}{e^{2\pi x}+1}dx = ?$$

But it seams that wolframalpha struggles to find a closed form solution.

Maybe the answer should be in terms Barnes G function or the derivative of the Hurwtiz zeta function.


My attempt so far, consider

$$F(a,b) = \int^\infty_0 \frac{\log(a+t^2)}{e^{2\pi (t+b)}-1}\,dt$$

We consider the derivative

$$\frac{\partial }{\partial a}F(a,b) = \int^\infty_0 \frac{dt}{(t^2+a)(e^{2\pi (t+b)}-1)}\,dt$$

Use the following expansion

$$\frac{2t}{e^{2\pi t}-1} =\frac{1}{\pi}-t+\frac{2t^2}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2} $$

Hence have

$$\frac{1}{e^{2\pi (t+b)}-1} =\frac{1}{2(t+b)\pi}-\frac{1}{2}+\frac{t+b}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+(t+b)^2} $$

$$\frac{\partial }{\partial a}F(a,b) = \int^\infty_0 \frac{1}{t^2+a}\left\{ \frac{1}{2(t+b)\pi}-\frac{1}{2}+\frac{t+b}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+(t+b)^2}\right\}dt$$

This reduces to $$\frac{\partial }{\partial a}F(a,b)= \frac{1}{2\pi}\int^\infty_0 \frac{1}{(t^2+a)(t+b)} dt-\frac{1}{2}\int^\infty_0 \frac{dt}{t^2+a}+\\\frac{1}{\pi}\sum_{k=1}^\infty\int^\infty_0\frac{t+b}{(t^2+a)(k^2+(t+b)^2)}dt$$

The first and second are simple it remains to evaluate


Then integrate with respect to $a$.

Or evaluate


Where $b = i+1/2$.

  • $\begingroup$ If you make a few substitutions and take the Taylor expansion of the log, you get a series in terms of the Riemann zeta function. $\endgroup$ – Simply Beautiful Art Jan 1 '17 at 13:59
  • $\begingroup$ i have only found a numerical value $\endgroup$ – Dr. Sonnhard Graubner Jan 1 '17 at 14:33
  • $\begingroup$ @SimpleArt, that series is not that simple. $\endgroup$ – Zaid Alyafeai Jan 2 '17 at 3:45
  • 4
    $\begingroup$ Have a look at the answer by pisco125 and you will be surprised: math.stackexchange.com/questions/1311929/… The best you can get here is something in terms of @Olivier Oloa's Poly Stieltjes constants...:/ $\endgroup$ – tired Jan 2 '17 at 9:44
  • $\begingroup$ @ZaidAlyafeai Obviously, if it were, I'd already have my answer checked in :-( $\endgroup$ – Simply Beautiful Art Jan 2 '17 at 12:35

By integration by parts, such integral equals $$ I=\frac{1}{\pi}\int_{0}^{+\infty}\frac{x\log\left(1+e^{-2\pi x}\right)}{1+x^2}\,dx $$ and since $\mathcal{L}^{-1}\left(\frac{x}{1+x^2}\right)=\cos(s)$ and $\mathcal{L}\left(\frac{(-1)^{k+1} e^{-2\pi k x}}{k}\right) = \frac{(-1)^{k+1}}{k(s+2\pi k)}$, we also have: $$ I = 2\int_{0}^{+\infty}\cos(s)\sum_{k\geq 1}\frac{(-1)^{k+1}}{2\pi k(s+2\pi k)}\,ds $$ so $I$ is a series of weighted sine and cosine integrals depending on a Lerch transcendent.
That is not surprising since $$ \int_{0}^{+\infty}\log(1+x^2)e^{-2k\pi x}\,dx = -\frac{\text{Ci}(2\pi k)}{\pi k}=\frac{1}{\pi k}\int_{2\pi k}^{+\infty}\frac{\cos x}{x}\,dx=\frac{1}{\pi k}\int_{0}^{+\infty}\frac{\cos x}{x+2\pi k}\,dx $$ So: $$\begin{eqnarray*} I &=& \frac{1}{\pi}\int_{0}^{+\infty}\cos(2\pi u)\sum_{k\geq 1}\frac{(-1)^{k+1}}{k (u+k)}\,du\\&=&\int_{0}^{+\infty}\frac{\cos(2\pi u)}{2\pi u}\left[\log(4)+\psi\left(\tfrac{1+u}{2}\right)-\psi\left(\tfrac{2+u}{2}\right)\right]\,du\tag{1}\end{eqnarray*}$$ Additionally, from $$ \zeta(s)=\frac{1}{\Gamma(s)}\left(1-\frac{2}{2^s}\right)^{-1}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx \tag{2}$$ holding for every $s$ such that $\text{Re}(s)>0$, it is not difficult to derive $$ \int_{0}^{+\infty}\frac{\log x}{e^{2\pi x}+1}\,dx = -\frac{\log(2)\log(8\pi^2)}{4\pi} \tag{3} $$ through a substitution and differentiation under the integral sign. We also have: $$\begin{eqnarray*} I &=& \int_{0}^{+\infty}\frac{\log(x+i)}{e^{2\pi x}+1}\,dx + \int_{0}^{+\infty}\frac{\log(x-i)}{e^{2\pi x}+1}\,dx\\ &=& \left(\int_{i}^{+\infty+i}+\int_{-i}^{+\infty-i}\right)\frac{\log x}{e^{2\pi x}+1}\,dx \tag{4}\end{eqnarray*}$$ hence our problem is equivalent to finding a closed form for the principal value of the integral $$ J = \text{PV}\int_{0}^{i}\text{Re}\left(\frac{\log x}{e^{2\pi x}+1}\right)\,dx \tag{5}$$ by choosing a suitable contour avoiding the singularity at $x=\frac{i}{2}$.
The singularity is associated with the term corresponding to $h=0$ in $$ \frac{1}{e^{2\pi x}+1}=\frac{1}{2}-\sum_{h\geq 0}\frac{4x}{(2h+1)^2 \pi+4\pi x^2}\tag{6}$$ and for every $h>0$ we have $$ \text{Re}\int_{0}^{i}\log(x)\frac{4x}{(2h+1)^2 \pi+4\pi x^2}\,dx = \frac{1}{4\pi}\,\text{Li}_2\left(\frac{4}{(2h+1)^2}\right)\tag{7}$$ while the singular term associated with $h=0$ gives a contribute of $\frac{1}{2}\text{arctanh}\frac{1}{2}$. It follows that the problem of finding a closed form for $I$ is equivalent to finding a closed form for: $$ S = \sum_{h\geq 1}\text{Li}_2\left(\frac{1}{(h+1/2)^2}\right) =\sum_{n\geq 1}\frac{\zeta\left(2n,\tfrac{3}{2}\right)}{n^2}.\tag{8}$$ Now we may exploit: $$ \sum_{n\geq 1}\zeta\left(2n,\tfrac{3}{2}\right)x^{2n} = \frac{4x^2}{4x^2-1}+\frac{\pi x\tan(\pi x)}{2}\tag{9} $$ and check that $S$ just depends on $$\text{PV}\int_{0}^{1}\frac{x\log x}{4x^2-1}\,dx,\qquad \color{blue}{\text{PV}\int_{0}^{1}\log(x)\tan(\pi x)\,dx}.\tag{10}$$ According to Mathematica, the first integral is a combination of $\pi^2,\log^2(2)$ and $\text{Li}_2\left(\tfrac{1}{4}\right)$.
That is not difficult to prove, and reduces the whole problem to finding a closed form for the blue integral. The computation of the blue integral boils down to the computation of $$ \int_{0}^{1}\text{arctanh}(x)\cot\left(\tfrac{\pi x}{2}\right)\,dx =-\frac{2}{\pi}\int_{0}^{1}\frac{\log\sin\tfrac{\pi x}{2}}{1-x^2}\,dx=-\frac{2}{\pi}\int_{0}^{1}\frac{\log\cos\tfrac{\pi x}{2}}{x(2-x)}\,dx\tag{11}$$ where $$ \log\cos\tfrac{\pi x}{2}=\sum_{k\geq 1}(-1)^k\frac{1-\cos(\pi k x)}{k} $$ has a nice Fourier series, but $\frac{1}{x(2-x)}$ does not seem to. However, a decent numerical approximation can be recovered from the fact that over the interval $\left(0,1\right)$ we have: $$ \text{arctanh}(x)\cot\left(\tfrac{\pi x}{2}\right)\approx\frac{2}{\pi}\sqrt{1-x^2}. \tag{12}$$

  • $\begingroup$ Nice work (+1). I'm curious if you are able to find a closed form (i have to confess i'm sceptical)...good luck! $\endgroup$ – tired Jan 3 '17 at 10:25
  • $\begingroup$ @tired: after my last update, I am starting to being skeptical too :D $\endgroup$ – Jack D'Aurizio Jan 3 '17 at 18:06

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