I Was Vey Amaze At The Solution 2

Use the substitution $x = 1 - e^{-u}$ to rewrite the integral: $\int_0^\infty \frac{u}{e^u-1}\,du = \int_0^\infty \frac{u}{1 - e^{-u}}\,(e^{-u}du) = \int_0^1 \frac{-\ln(1-x)}{x}\,dx$

Then using the usual Maclaurin series, for $|x|<1$ , $\frac{-\ln(1-x)}{x} = \frac{1}{x}\cdot \sum_{n=1}^\infty \frac{1}{n}x^n = \sum_{n=1}^\infty \frac{1}{n}x^{n-1}$ and integrating, $\int_0^1 \frac{-\ln(1-x)}{x}\,dx = \sum_{n=1}^\infty \int_0^1 \frac{1}{n} x^{n-1} = \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} \approx \boxed{1.645}$

$\int_{0}^{\infty} \dfrac{u}{e^u-1} du \\ = \int_{0}^{\infty} \dfrac{u e^{-u}}{1-e^{-u}} du \\ =\int_{0}^{\infty} u \sum_{r=1}^{\infty} e^{-ru} du$

Now evaluate $\int_{0}^{\infty} u e^{-ru} du=\dfrac{1}{r^2}$ by using Integration By-Parts.

So, the given integral becomes $\sum_{r=1}^{\infty} \int_{0}^{\infty} u e^{-ru} du = \sum_{r=1}^{\infty} \dfrac{1}{r^2} =\dfrac{\pi^2}{6}$

I hope it makes sense

By the fact that:

$\begin{aligned} \displaystyle \zeta(z) \times \Gamma(z) = \int_{0}^{\infty} \cfrac{u^{z - 1}}{e^u - 1} \ \text{du} \implies \zeta(2) \times \Gamma(2) & = \int_{0}^{\infty} \cfrac{u^{2 - 1}}{e^u - 1} \ \text{du} \\ \implies \cfrac{\pi^2}{6} \times (2 - 1)! & = \int_{0}^{\infty} \cfrac{u}{e^u - 1} \ \text{du} \\ \implies \boxed{\cfrac{\pi^2}{6} \approx 1.6449340668482262 } & = \int_{0}^{\infty} \cfrac{u}{e^u - 1} \ \text{du} \end{aligned}$