show that $\int_0^{\infty}\ln(2x)\frac{ax-u(1-e^{-2ax})}{\sinh^2(ax)}x^{2u-1}dx=\frac{1}{2u}(\frac{\pi}{a})^{2u}|B_{2u}|:u \in N$ $$\int_0^{\infty}\ln(2x)\frac{ax-u(1-e^{-2ax})}{\sinh^2(ax)}x^{2u-1}dx=\frac{1}{2u}(\frac{\pi}{a})^{2u}|B_{2u}|:u \in N$$
using real or complexe analysis
where  $B_{2u}$ is bernoulli number
 A: For first, let we compute:
$$\begin{eqnarray*} I_n &=& \int_{0}^{+\infty}\frac{x^{2n}}{\sinh^2 x}\,dx = \int_{1}^{+\infty}\frac{(\log z)^{2n}}{z\left(\frac{z-\frac{1}{z}}{2}\right)^2}\,dz=\int_{0}^{1}\frac{(\log z)^{2n}}{z\left(\frac{z-\frac{1}{z}}{2}\right)^2}\,dz\\&=&\int_{0}^{1}\frac{4z\,(\log z)^{2n}}{(1-z^2)^2}\,dz=\frac{2}{4^n}\int_{0}^{1}\frac{\left(\log u\right)^{2n}}{(1-u)^2}\,du = \frac{2}{4^n}\sum_{k\geq 0}(k+1)\int_{0}^{1}(\log u)^{2n}u^k\,du\\&=&\frac{2}{4^n}\sum_{k\geq 0}(k+1)\frac{(2n)!}{(k+1)^{2n+1}}=\frac{2\cdot(2n)!}{4^n}\cdot\zeta(2n)\tag{1}\end{eqnarray*}$$
as well as:
$$\begin{eqnarray*} J_n = \int_{0}^{+\infty}\frac{1-e^{-2x}}{\sinh^2 x}\,x^{2n-1}\,dx &=& \int_{1}^{+\infty}\frac{4(\log z)^{2n-1}}{z(1-z^2)}\,dz\\ &=& \frac{4(2n-1)!}{4^n}\,\zeta(2n)\tag{2} \end{eqnarray*}$$
hence our integral can be computed from $(1)$ and $(2)$ by differentiation under the integral sign, since $\frac{d}{du} x^{2u-1} = 2\log x\cdot x^{2u-1}$ and the original integral without the logarithmic factor can be computed from $(1)$ and $(2)$. At last, we just need the well-known identity connecting the Bernoulli numbers with the values of the zeta function at even integers.
