For $\displaystyle\int_0^{\infty}x^n\frac{\sinh ax}{\cosh bx}\text{d}x$, $\left|a\right|<b$
I think we can evaluate $\displaystyle\int_{0}^{\infty }{{{\text{e}}^{cx}}\frac{\sinh ax}{\cosh bx}\text{d}x}$ first. And use taylor expansion on parameter $c$? But I m not confident on that. Could anyone finish this? : )
We already have the solution mentioned here for the integral $$\int_0^{\infty} x^n \dfrac{\sinh(ax)}{\cosh(bx)}dx, \,\,\,\,\,\,\,\, \vert a \vert < b$$
Here is another solution, if you want to proceed by evaluating the integral $$I = \int_0^{\infty} e^{cx} \dfrac{\sinh(ax)}{\cosh(bx)}dx$$ Choose a small enough positive $c$ or choose $c<0$. As in the earlier proof, we have \begin{align} I & = \sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} \left(e^{(a-2kb-b+c)x} - e^{-(a+2kb+b-c)x}\right)dx\\ & = \sum_{k=0}^{\infty} (-1)^k \left(\dfrac1{2kb+b-a-c} - \dfrac1{2kb+b+a-c}\right)\\ & = \sum_{k=0}^{\infty} (-1)^k \left(\dfrac1{2kb+b-a}\left(\sum_{j=0}^{\infty} \left(\dfrac{c}{2kb+b-a} \right)^j\right) - \dfrac1{2kb+b+a} \left(\sum_{j=0}^{\infty} \left(\dfrac{c}{2kb+b+a} \right)^j\right)\right)\\ & = \sum_{j=0}^{\infty} \left(\sum_{k=0}^{\infty} (-1)^k \left(\left(\dfrac1{2kb+b-a} \right)^{j+1} - \left(\dfrac1{2kb+b+a} \right)^{j+1} \right) c^j\right) \end{align} But the integral is also equal to $$I = \int_0^{\infty} e^{cx} \dfrac{\sinh(ax)}{\cosh(bx)} dx = \int_0^{\infty} \sum_{j=0}^{\infty} \dfrac{c^j x^j}{j!} \dfrac{\sinh(ax)}{\cosh(bx)} dx = \sum_{j=0}^{\infty} \left(\int_0^{\infty} \dfrac{x^j}{j!} \dfrac{\sinh(ax)}{\cosh(bx)} dx \right)c^j$$ Now comparing the coefficient of $c^j$, we can now conclude that $$\int_0^{\infty} \dfrac{x^j}{j!} \dfrac{\sinh(ax)}{\cosh(bx)} dx = \sum_{k=0}^{\infty} (-1)^k \left(\left(\dfrac1{2kb+b-a} \right)^{j+1} - \left(\dfrac1{2kb+b+a} \right)^{j+1} \right)$$ Now as before, write $$\tan(\pi x/2) = \dfrac1{\pi} \lim_{N \to \infty}\left(\sum_{k=-N}^N \dfrac2{1-x+2k}\right)$$ Plugging in $x=a/b$ and taking derivative term by term you should be able to obtain your desired result.