Improper integral with two problem points

Question

I need to prove wheter the integral $\int_0^1\frac{x^{\alpha-1}}{e^x-1}dx$ converges or diverges. I managed to prove the it diverges for $\alpha = 1$, and for $0<\alpha<1$ it's simple to prove by comparison with the case $\alpha = 1$. But this obviously doesn't work for $\alpha>1$. Any sugestions?

Answer 1

HINT

Note that

$$\frac{x^{\alpha-1}}{e^x-1}=\frac{x}{e^x-1}\frac{x^{\alpha-1}}{x}\sim x^{\alpha-2}$$

Answer 2

Near $0$, we know that $$e^x-1\sim x $$

thus $$\frac {x^{\alpha-1}}{e^x-1}\sim x^{\alpha-2}$$

the integral converges iff $$2-\alpha <1$$ or $$\alpha>1.$$