Let's do this in more general terms. In other words, let's solve

$\displaystyle \displaystyle I= \int_0^{\infty}\frac{x^{\alpha}}{\text{sinh}(x)}dx \quad\alpha>0$

To do this we merely note that

$\displaystyle \displaystyle \begin{aligned}\int_0^{\infty}\frac{x^{\alpha}}{\t ext{sinh}(x)}dx & = \int_0^{\infty}\frac{x^{\alpha}}{\frac{e^x-e^{-x}}{2}}dx\\ &= 2\int_0^{\infty}\frac{e^{-x}x^{\alpha}}{1-\left(e^{-x}\right)^2}dx\end{aligned}$

Note thought that since $\displaystyle e^{-2x}\leqslant 1$ we may apply the geometric series to arrive at

$\displaystyle \displaystyle \begin{aligned}I &= 2\int_0^{\infty}\frac{e^{-x}x^{\alpha}}{1-e^{-2x}}dx\\ &= 2\int_0^{\infty}e^{-x}x^{\alpha}\sum_{n=0}^{\infty}e^{-2nx}dx\\ &= 2\sum_{n=0}^{\infty}\int_0^{\infty}x^{\alpha}e^{-(2n+1)x}dx\end{aligned}$

But, making the substitution $\displaystyle (2n+1)x=z$ gives us

$\displaystyle \begin{aligned}I &= 2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{\alpha+1}}\in t_0^{\infty}z^{\alpha}e^{-\alpha}dz\\ &=2\Gamma(\alpha+1)\sum_{n=0}^{\infty}\frac{1}{(2n +1)^{\alpha+1}}\end{aligned}$

Now, to compute this last sum we note that

$\displaystyle \displaystyle \begin{aligned}\zeta(\alpha) &= \sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}\\ &=\sum_{n=1}^{\infty}\frac{1}{(2n)^{\alpha}}+\sum_ {n=0}^{\infty}\frac{1}{(2n+1)^{\alpha}}\\ &= \frac{\zeta(\alpha)}{2^{\alpha}}+\sum_{n=0}^{\inft y}\frac{1}{(2n+1)^{\alpha}}\end{aligned}$

And thus

$\displaystyle \displaystyle \sum_{n=0}^{\infty}\frac{1}{(2n+1)^{\alpha}} = \zeta(\alpha)\left(1-\frac{1}{2^\alpha}\right)$

Thus, applying this we may finally conclude that

$\displaystyle \displaystyle \begin{aligned}I &= 2\Gamma(\alpha+1)\sum_{n=0}^{\infty}\frac{1}{(2n+1 )^{\alpha+1}}\\ &= 2\Gamma(\alpha+1)\zeta(\alpha+1)\left(1-\frac{1}{2^{\alpha+1}}\right)\\ &= \Gamma(\alpha+1)\zeta(\alpha+1)\left(2-\frac{1}{2^{\alpha}}\right)\end{aligned}$