How to calculate $\frac{d}{dα}\int_0^\infty \frac{x^{2-\alpha}}{e^x-1}\,dx$ and its convergence or divergence I will exactly show you the output of my calculations and the reason why I am struggling with this improper integral. I expanded a function acc. to a McLaurin serie for values of α<<1. I obtained a term that looks like this
$$ \frac{d}{dα}\int_0^\infty \frac{x^{2-α}}{e^x-1}\;\mathrm{d}x$$
As one can see, it is a derivative (not in the variable x but in α)  of an improper integral. What is the right calculation procedure?

*

*Should I check the convergence of the integral in order to first apply the derivative of integral argument and then integrate?
$$ \frac{d}{dα}\int_0^\infty \frac{x^{2-α}}{e^x-1}\;\mathrm{d}x = \int_0^\infty \frac{d}{dα}\frac{x^{2-α}}{e^x-1}\;\mathrm{d}x$$
The point is, does the integral converge? I cannot get a reliable result on this.


*Or should I first solve the integral and then derivate?
$$ \frac{d}{dα}\int_0^\infty \frac{x^{2-α}}{e^x-1}\;\mathrm{d}x = \frac{d}{dα} F(x,α)$$ where F(x,α) is the solution of the improper integral.
On this regards, how do I solve this improper integral?
Can you please kindly show me a procedure of calculation in detail ? It would be really appreciated. Resumed, my queries concerns:
convergence of the integral and calculation of the integral.
Have a nice day and thank you so much for your professional support!
 A: So you want to determine the convergence of $$-\int_0^{\infty}\frac{ x^{2-\alpha} \log x}{e^x-1} dx $$
Near $x=0$, $e^x-1 \sim x$ so the integrand is asymptotic to $$x^{1-\alpha} \log x \to 0$$
As $x\to \infty$, it’s asymptotic to$$x^{2-\alpha} \log x \cdot e^{-x} $$ and convergence is guaranteed because of the $e^{-x}$. So, the integral converges.
A: Too long for a comment. You can do both.
However this integral is a special one. This integral is related to Riemann Zeta function $\zeta(s)$ by the formula
$$ \zeta(s) \Gamma(s) = \int_0^{\infty} \frac{x^{s-1}}{e^x - 1}\, dx$$ for $s > 1$ where $\Gamma(s) = \int_0 ^ {\infty} e^{-x} x^{s-1} \, dx$ is the gamma function.
Hint:
This is just a hint and try to formalize the argument;
Try to show that series $\zeta(s) = \sum_{k=1} ^{\infty } \frac{1}{k^s}$ converges when $s > 1$ and write
$$ \frac{x^{s-1}}{e^x -1} = \frac{e^{-x} x^{s-1}}{1 -e^{-x}} =  \sum_{k = 1}^{\infty}e^{-kx} x^{s-1}$$  and try to write the integral in terms of $\zeta(s)$ and $\Gamma(s)$ by making proper substitutions since the convergence of both functions are wellknown.
