# "I am Back" says Integration Part 5

**Calculus**Level 5

\[\large \int_{0}^{\infty}{\dfrac{x \ln x}{e^x -1} \, dx}\]

The integral above can be expressed as \[ \dfrac{\pi^a}{b} \left( \gamma + \ln(c\pi) - d\ln(A) - (\gamma -1)^{f} \right),\]

where \(a,b,c,d,f\) are positive integers.

Find \(a+b+c+d+f\).

**Notations**:

- \(\gamma\) denotes the Euler-Mascheroni constant, \( \displaystyle \gamma = \lim_{n\to\infty} \left( - \ln n + \sum_{k=1}^n \dfrac1k \right) \approx 0.5772 \).
- A denotes the Glaisher-Kinkelin constant, \( \displaystyle A = \lim_{n\to\infty} \dfrac{1^1 \cdot 2^2 \cdots n^n}{e^{-n^2/4} \cdot n^{n^2/2 + n/2 + 1/12}} \approx 1.2824\).