So I have to find for which $a\in \mathbb{R}$ te following improper integral converges:
$$\int_{0}^{\infty}\frac{dx}{(x^2-2x-3)^{2a}}$$.
Okay so the integral is:
$$\int_{0}^{\infty}\frac{dx}{(x^2-2x-3)^{2a}}=\int_{0}^{\infty}\frac{dx}{((x-3)(x+1))^{2a}}$$
So m first thought is to separate the integral like this:
$$\int_{0}^{\infty}\frac{dx}{((x-3)(x+1))^{2a}}=\int_{0}^{3}\frac{dx}{((x-3)(x+1))^{2a}}+\int_{3}^{4}\frac{dx}{((x-3)(x+1))^{2a}}+\int_{4}^{\infty}\frac{dx}{((x-3)(x+1))^{2a}}$$
And look for converge of each one by itself.
So is it correct to do it like this:
First for $a>1$, then $0<a<1$ and lastly $a<0$.
Lets say: $0<a<1$ $$\int_{0}^{3}\frac{dx}{((x-3)(x+1))^{2a}} \leq \int_{0}^{3}\frac{dx}{((x)(x+x))^{2a}}=\int_{0}^{3}\frac{dx}{((2x^2)^{2a}}\leq \frac{1}{2^{2a}}\int_{0}^{3}\frac{dx}{x^{4a}}$$ So this converges for $a>\frac{1}{4}$. But how can I look at this integrals convergence for $a>0$ and $a<0$. And is my approach correct?
Thank you for any help.