<s>[INST] A difficult improper integral 
What is the value of the $$\int_{1}^{\infty}\frac{dx}{ x^2(e^x+1)}?$$

I got  $2$ as a result.  Wolfram has given $0.11111$ as a result. I am confused.
Is it correct? Or incorrect ? Please tell me...
Thanks in advance
 [/INST] As an alternative expression, we have
$$ \int_{1}^{\infty} \frac{dx}{x^2(e^x + 1)}
= C + \int_{0}^{1} \left( \frac{1}{e^x + 1} \right)'' \log x \, dx $$
where $C$ is given by
\begin{align*}
C
&= \frac{1}{4} + \frac{1}{1+e} + \frac{\gamma}{4} + \frac{\log 2}{3} - 3\log A \\
&\approx 0.14803096668067398473\cdots.
\end{align*}
and $\gamma$ is the Euler-Mascheroni constant and $A$ is the Glaisher-Kinkelin constant. So if we plug the Taylor expansion of $\frac{1}{e^x+1} = \frac{1}{2}(1-\tanh(x/2))$, we obtain
$$ \int_{1}^{\infty} \frac{dx}{x^2(e^x + 1)}
= C - \sum_{n=2}^{\infty} (-1)^n \frac{2n-1}{n-1} \cdot \frac{2^{2n}-1}{(2\pi)^{2n}} \zeta(2n). $$
Finally, expanding $\zeta(2n) = \sum_{k=1}^{\infty} \frac{1}{k^{2n}} $ and interchanging the order of summation, we find that
\begin{align*}
\sum_{n=2}^{\infty} (-1)^n \frac{2n-1}{n-1} \cdot \frac{2^{2n}-1}{(2\pi)^{2n}} \zeta(2n)
&= \frac{3-e}{4(1+e)} + \sum_{k=1}^{\infty} \frac{1}{(\pi k)^2} \log\left(1 + \frac{1}{(\pi k)^2} \right) \\
&\hspace{6em} - \sum_{k=1}^{\infty} \frac{1}{(2\pi k)^2} \log\left(1 + \frac{1}{(2\pi k)^2} \right)
\end{align*}
which might be better suited for studying this quantity.

Addendum. Here is a numerical computation:

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