Evaluating Accelerator's integral $\int_0^{\infty} \frac{e^{4x}-e^{2x}}{x(e^{2x}+1)^3}dx$ I met this integral in a post of Accelerator (Contour Integral Involving $e^z$, a Semicircle, and Triangle Inequality).
WA instantly evaluated it as $I=\int_0^{\infty} \frac{e^{4x}-e^{2x}}{x(e^{2x}+1)^3}dx =\frac{7\zeta(3)}{4\pi^2}$. Then, it is stuck in my mind.
I posted a simpler version of this integral here: Definite integral by Malmsten's formula
I evaluated this integral trickly but unfortunately not so rigorously in the following way:
Taking the second derivative of both sides of
$\frac{1}{e^{2x}+1}=\sum_{k=1}^{\infty} (-1)^{k-1}e^{-2kx}$
and simplying we get
$$\frac{e^{4x}-e^{2x}}{(e^{2x}+1)^3}=\sum_{k=1}^{\infty} (-1)^{k-1}k^{2}e^{-2kx}.$$
Since,
$$\sum_{k=1}^{\infty} (-1)^{k-1}k^{2}=\eta(-2)=(1-2^{3})\zeta(-2)=0,\tag{1}$$
we have
$$I=\sum_{k=1}^{\infty} (-1)^{k-1}k^{2}\left(\int_0^{\infty}\frac{e^{-2kx}-e^{-2x}}{x}\right).\tag{2}$$
Then, by Frullani's theorem (Proof of Frullani's theorem), we get
$$I=\sum_{k=1}^{\infty} (-1)^{k}k^{2}\ln k=\eta'(-2)=(1-2^3)\zeta'(-2)=(-7)\left(-\frac{\zeta(3)}{4\pi^2}\right)=\frac{7\zeta(3)}{4\pi^2}.$$
How can I make this way more rigorous? In $(2)$, interchange of the summation and integration is not obvious. $(1)$ is divergent so is $(2)$!
Can you suggest another solution? Thanks in advance.
 A: Substitute $t=e^{2x}$
$$I=\int_0^{\infty} \frac{e^{4x}-e^{2x}}{x(e^{2x}+1)^3}dx
=\int_1^\infty \frac{t-1}{\ln t (t+1)^3}\overset{t\to\frac1t}{dt}
= \frac12\int_0^\infty \frac{t-1}{\ln t \ (t+1)^3}dt
$$
Let $J(a)= \int_0^\infty \frac{t^a}{\ln t \ (t+1)^3}dt$
$$J’(a)=\int_0^\infty \frac{t^a}{(1+t)^3}dt=\frac{\pi a(1-a)}{2\sin \pi a}
$$
Then
\begin{align}
I=& \ \frac12(J(1)-J(0))\\
=& \ \frac\pi4\int_0^1 \frac{a(1-a)}{\sin \pi a}da
\overset{ibp}=\frac 12 \int_0^1 x\ln \tan \frac{\pi x}2 
\overset{t=\tan\frac{\pi x}2}{dx}\\
=& \ \frac2{\pi^2}\int_0^\infty \frac{\ln t\tan^{-1}t}{1+t^2}dt
=  \frac2{\pi^2}\int_0^\infty \int_0^1 \frac{t \ln t}{(1+t^2)(1+y^2 t^2)} \overset{t\to \frac1{yt}}{ dy} \ dt\\
 = &\ \frac1{\pi^2}\int_0^\infty \int_0^1 \frac{-t\ln y}{(1+t^2)(1+y^2 t^2)} dt\ dy
= \frac1{\pi^2}\int_0^1 \frac{\ln^2 y}{1-y^2}dy \\
= &\ \frac1{\pi^2}\cdot \frac{7\zeta(3)}4 = \frac{7 \zeta(3)}{4\pi^2}
\end{align}
A: Here, we use contour integration.  Proceeding we have
$$\begin{align}
I&=\int_0^\infty \frac1x\frac{e^{4x}-e^{2x}}{(1+e^{2x})^3}\,dx\\\\
&=\frac18\int_{-\infty}^\infty  \frac1x\frac{\sinh(x)}{\cosh^3(x)}\,dx \\\\
&=\frac{\pi i}{4} \sum_{n=0}^\infty \text{Res}\left(\frac1z\frac{\sinh(z)}{\cosh^3(z)}, z=i(2n+1)\pi/2\right)\\\\
&=\frac{\pi i}{4} \sum_{n=0}^\infty\left(-\frac{8i}{\pi^3(2n+1)^3}\right)\\\\
&=\frac{7\zeta(3)}{4\pi^3}
\end{align}$$
And we are done!

There is a question regarding whether contour integration works here.  In particular, the question is whether the integration over a semi-circular arc approaches $0$ as the radius of the semicircle approaches $0$.  We proceed now to show that contour integration is applicable.
In THIS ANSWER, I showed that $|\cot(\pi z)|$ is uniformly bounded on the circle $|z|=(N+1/2)$ for $N\in \mathbb{N^+}$.  Analogously, it is straightforward to show that $|\tanh(z)|$ is uniformly bounded on the circle $N\pi$, $N\in \mathbb{N^+}$.
Let $B$ represent an upper bound of $|\tanh(z)|$ on the circle $N\pi$, $C_N$ be the semi-circular arc with radius $N\pi$ in the upper-half plane, and $I_N=\int_{C_N}\frac{\sinh(z)}{z\cosh^3(z)}\,dz$.  Then we have the estimates on the semi-circular arc of radius $N\pi$
$$\begin{align}
|I_n|&=\left|\int_{C_N}\frac{\sinh(z)}{z\cosh^3(z)}\,dz\right|\\\\
&\le \int_{C_N}\frac{1}{N\pi}\left|\tanh(z)\right|\,\left|\text{sech}^2(z)\right|\,|dz|\\\\
&\le B \int_0^\pi \left|\text{sech}(N\pi e^{i\phi})\right|^2\,d\phi\\\\
&=4B \int_0^{\pi} \frac{|e^{-N\pi e^{i\phi}}|^2}{|1+e^{-2N\pi e^{i\phi}}|^2}\,d\phi\\\\
&\le \frac{8B}{\inf_{\phi\in[0,\pi], N\in\mathbb{N^+}} {|1+e^{-2N\pi e^{i\phi}}|^2}}\int_0^{\pi/2} e^{-2N\pi \cos(\phi)}\,d\phi\\\\
&\le \frac{8B}{\inf_{\phi\in[0,\pi], N\in\mathbb{N^+}} {|1+e^{-2N\pi e^{i\phi}}|^2}}\int_0^{\pi/2} e^{-2N\pi (1-2\phi/\pi)}\,d\phi\\\\
&\le \frac{8B}{\inf_{\phi\in[0,\pi], N\in\mathbb{N^+}} {|1+e^{-2N\pi e^{i\phi}}|^2}}\frac{1-e^{-2N\pi}}{4N}
\end{align}$$
which clearly approaches $0$ as $N\to \infty$.
So, by judiciously selecting the radius of the semi-circle to be $N\pi$, we construct a sequence of integrals $I_N$ such that $\lim_{N\to\infty}I_N=0$.
