So I have an improper integral:
$$\int_{1}^{\infty} \frac{1}{x^a(x-1)^b} dx$$
I have to find for which $a,b\in \mathbb{R}$ it converges.
So my initial step is:
$$\int_{1}^{\infty} \frac{1}{x^a(x-1)^b} \leq ?\int_{1}^{\infty} \frac{1}{x^ax^b}dx$$
But since $x>1$ this does not seem to be correct, should i use this:
$$\int_{1}^{\infty} \frac{1}{x^a(x-1)^b} \leq \int_{1}^{\infty} \frac{1}{(x-1)^a(x-1)^b}dx$$
It gives me the same result: $a+b>1$; is this the correct way or am I missing something? Any help would be appreciated.
Hint: Divide
$$\int_{1}^{\infty}\dfrac{1}{x^{a}\left(x-1\right)^{b}}{\rm d}x=\int_{1}^{2}\dfrac{1}{x^{a}\left(x-1\right)^{b}}{\rm d}x+\int_{2}^{\infty}\dfrac{1}{x^{a}\left(x-1\right)^{b}}$$
and study the convergence of each integral separately.