Evaluate the integral of x'dx over (ex - 1) limits from zero to infinity.

To evaluate the integral $\int_{0}^{\infty} \frac{x \, dx}{e^x - 1}$, we will follow a structured approach:

1. Break Down the Problem

   We need to evaluate the improper integral $\int_{0}^{\infty} \frac{x \, dx}{e^x - 1}$. This integral is known as the Bose-Einstein integral and appears in statistical mechanics.

2. Relevant Concepts

   The integral can be related to the Riemann zeta function and the Gamma function. Specifically, the integral can be expressed in terms of the Riemann zeta function $\zeta(s)$ and the Gamma function $\Gamma(s)$.

3. Analysis and Detail

   The integral can be rewritten using the series expansion of $\frac{1}{e^x - 1}$: $\frac{1}{e^x - 1} = \sum_{n=1}^{\infty} e^{-nx}$ Thus, the integral becomes: $\int_{0}^{\infty} x \left( \sum_{n=1}^{\infty} e^{-nx} \right) \, dx = \sum_{n=1}^{\infty} \int_{0}^{\infty} x e^{-nx} \, dx$ The integral $\int_{0}^{\infty} x e^{-nx} \, dx$ can be solved using integration by parts. Let $u = x$ and $dv = e^{-nx} dx$, then $du = dx$ and $v = -\frac{1}{n} e^{-nx}$. Applying integration by parts: $\int x e^{-nx} \, dx = \left[ -\frac{x}{n} e^{-nx} \right]_{0}^{\infty} + \frac{1}{n} \int_{0}^{\infty} e^{-nx} \, dx$ The boundary term evaluates to zero, and the remaining integral is: $\frac{1}{n} \int_{0}^{\infty} e^{-nx} \, dx = \frac{1}{n^2}$ Therefore, the original integral becomes: $\sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2)$

4. Verify and Summarize

   The Riemann zeta function $\zeta(2)$ is known to be $\frac{\pi^2}{6}$. Therefore, the value of the integral is $\frac{\pi^2}{6}$.

Final Answer

The integral $\int_{0}^{\infty} \frac{x \, dx}{e^x - 1}$ evaluates to $\frac{\pi^2}{6}$.