calculus - Integrate $int_{0}^infty frac{tanh(x)}{x^3}-frac{operatorname{sech}(x)}{x^2}dx$

I would like to integrate:

$$\int_{0}^\infty \frac{\tanh(x)}{x^3}-\frac{\operatorname{sech}(x)}{x^2}dx$$
I'm not sure where I found this integral, but I have a feeling I wrote it down because its solution. I want to say it's related to the Zeta Function, but I'm not sure. I've managed to rewrite it as:
$$\small \sum_{n=1}^\infty (-1)^{n+1}\left(2\ln\left(\frac{2n-1}{n-1}\right)+4n\ln\left(\frac{n-1}{2n-1}\right)+2n^2\ln\left(\frac{n}{n-1}\right)+4n\ln(2)-2n-2\ln(2)+1\right)$$

Above follows by writing the hyperbolic functions in terms of exponential functions and then using series. Then I used differentiating under the integral.

This makes me think otherwise about the Zeta Function/having a closed form for the original integral. I would appreciate any help in solving this.

Answer

Start by splitting the integral into two convergent parts:

$$I=\int_0^\infty \frac{\tanh x-x+x-x\operatorname{sech} x}{x^3}dx=\int_0^\infty \frac{\tanh x-x}{x^3}dx+\int_0^\infty \frac{1-\operatorname{sech} x}{x^2}dx$$
$$I_1=\int_0^\infty \frac{\tanh x-x}{x^3}dx\overset{IBP}=\frac12 \int_0^\infty \frac{\operatorname{sech}^2 x -1}{x^2}dx$$
$$\overset{IBP}=-\int_0^\infty \frac{\tanh x\operatorname{sech}^2 x}{x}dx=-\frac{7\zeta(3)}{\pi^2}$$

$$I_2=\int_0^\infty \frac{1-\operatorname{sech} x}{x^2}dx\overset{IBP}=\int_0^\infty \frac{\tanh x\operatorname{sech} x}{x}dx=\frac{4G}{\pi}$$

---

$$I_2=\int_0^\infty \frac{\tanh x \operatorname{sech} x}{x}dx\overset{x=\ln t}=2\int_1^\infty \frac{(t^2-1)}{(t^2+1)\ln t}dt\overset{t=\frac{1}{x}}=\int_0^\infty \frac{x^2-1}{(x^2+1)^2\ln x}dx$$
Above the two integrals were averaged after the reciprocal subtitution was done.

Now we will use Feynman's trick alongside beta function:

$$I(a)=\int_0^\infty \frac{x^a-1}{(x^2+1)^2 \ln x}dx\Rightarrow I'(a)=\int_0^\infty \frac{x^a}{(x^2+1)^2}dx=\frac12 \left(\frac{1-a}{2}\right)\frac{\pi}{\sin\left(\frac{\pi(a+1)}{2}\right)}$$
$$I(0)=0\Rightarrow I_2=\frac{\pi}{4}\int_0^2 \frac{1-a}{\sin\frac{\pi(a+1)}{2}}da =\frac{2}{\pi} \int_{0}^\frac{\pi}{2}\frac{t}{\sin t}dt=\frac{4 G}{\pi}$$
$I_1$ can be found here.