# Really tricky integral

## Homework Statement

$$I=\int_{0}^{\infty} \frac{x^3}{e^x-1}\ln(e^x-1)\,dx$$

## Homework Equations

Any identity involving $\zeta(s)$.

## The Attempt at a Solution

I noted that:
$$\ln(e^x - 1) = \ln(x) + \sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!}x^{n+1}$$
Therefore, the integral can be split into two parts:
$$I = \int_{0}^{\infty} \frac{x^3}{e^x-1}\ln(x)\,dx + \int_{0}^{\infty} \frac{x^3}{e^x-1}\sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!}x^{n+1}\,dx = I_{1} +I_{2}$$

Ignoring the first one for now,
$$I_{2} = \sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!} \int_{0}^{\infty} \frac{x^{n+4}}{e^x-1}\,dx = \sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!} \Gamma(n+5)\zeta(n+5) = \sum_{n=0}^{\infty} B_{n+1}(1) \frac{(n+4)(n+3)(n+2)}{n+1}\zeta(n+5) = 12\zeta(5) + \sum_{m=0}^{\infty} B_{2m}\frac{(2m+5)(2m+4)(2m+3)}{2m+2}\zeta(2m+6)$$
Now I suppose that you can use the relationship between $B_{2m}$ and $\zeta(2m)$ but I don't see how that would help. I want to believe that $I_{1}$ can be done via contour integration, but I can't come up with anything. Thanks for reading this mess.