Calculations for $\int_0^\infty \frac{2\sinh^2\left(\frac{x}{e}\right)}{e^x+1}dx$, or $\sum_{n=1}^\infty\frac{(4^n-1)\zeta(2n+1)}{e^{2n}}$

Question

From (all calculations, except the manipulations from Wolfram Alpha) $$\int_0^\infty\frac{x^{2n}}{e^x+1}dx=4^{-n}\left(4^n-1\right)\zeta(2n+1)\Gamma(2n+1)$$ that holds for $\Re n>-1/2$, then I wrote doing comparisons to get convergent series in RHS, the following that holds for each integer $n\geq 1$ $$\frac{4^n}{e^{2n}(2n!)}\int_0^\infty\frac{x^{2n}}{e^x+1}dx=\frac{(4^n-1)\zeta(2n+1)}{e^{2n}}.$$ Thus by Tonelli or absolute convergence, I wrote $$\int_0^\infty \frac{2\sinh^2\left(\frac{x}{e}\right)}{e^x+1}dx=\sum_{n=1}^\infty\frac{(4^n-1)\zeta(2n+1)}{e^{2n}},$$ because $$\sum_{n=1}^\infty\frac{1}{(2n)!}\left(\frac{2x}{e}\right)^{2n}=2\sinh^2\left(\frac{x}{e}\right).$$

Question. Can you provide me some evaluation (as a closed-form in term of special functions) of the integral orof the series to get an identity more elaborated than that I wrote? I say if you can deduce new calculations for $$\int_0^\infty \frac{2\sinh^2\left(\frac{x}{e}\right)}{e^x+1}dx=\text{something},$$ or well $$\sum_{n=1}^\infty\frac{(4^n-1)\zeta(2n+1)}{e^{2n}}=\text{something}.$$

I type here four codes:

integrate x^(2n))/(e^x+1)dx, from x=0 to infnite; sum (2/e x)^(2n)1/(2n)!, from n=1 to infnite; sum (4^n-1)zeta(2n+1)/e^(2n), from n=1 to 100 and integrate 2 sinh^2(x/e)/(e^x+1)dx, from x=0 to x=100.

Thanks in advance. If you can add in your answer a decimal approximation also it is good.

Answer 1

Hint. One may use the generating function of the zeta values $$ \sum_{n=1}^\infty\zeta(2n+1)\cdot z^{2n}=-\gamma-\frac12\psi(1-z)-\frac12\psi(1+z), \qquad |z|<1, $$ where $\psi(\cdot)$ is the digamma function.

Then one gets $$ \sum_{n=1}^\infty\frac{(4^n-1)\zeta(2n+1)}{e^{2n}}=\frac12\psi(1+1/e)+\frac12\psi(1-1/e)-\frac12\psi(1+2/e)-\frac12\psi(1-2/e) $$ which is equal to your integral.