Tricky Integral: Splitting and Solving using Zeta and Bernoulli Numbers

[Screwdriver]

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.

 

[mighty2000]

Thank you for sharing your attempt at solving this integral. Your approach using the identity involving the Bernoulli numbers and the Riemann zeta function is a good start. However, I would like to suggest an alternative method using the residue theorem.

First, let's rewrite the integral as follows:
$$
I=\int_{0}^{\infty} \frac{x^3}{e^x-1}\ln(e^x-1)\,dx = \int_{0}^{\infty} \frac{x^3}{e^x-1}\left(\ln(x)+\ln\left(\sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!}x^{n+1}\right)\right)\,dx
$$
Now, let's consider the function $f(z)=\frac{z^3}{e^z-1}\left(\ln(z)+\ln\left(\sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!}z^{n+1}\right)\right)$ and integrate it over a contour made up of the real axis and a large semicircle in the upper half plane, centered at the origin. The contour integral can be written as follows:
$$
\oint_{C} f(z)\,dz = \int_{0}^{\infty} \frac{x^3}{e^x-1}\ln(x)\,dx + \int_{0}^{\infty} \frac{x^3}{e^x-1}\ln\left(\sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!}x^{n+1}\right)\,dx + \int_{C_R} f(z)\,dz
$$
where $C_R$ is the semicircle of radius $R$ in the upper half plane. Using the residue theorem, we can evaluate the contour integral as follows:
$$
\oint_{C} f(z)\,dz = 2\pi i \sum_{k=1}^{\infty} \mathrm{Res}(f,z_k)
$$
where $z_k$ are the poles of $