So I have this integral: $$\int_{0}^{\infty} \frac{1}{x^a(x+a)^a}\,dx.$$ So I know that $$\int_{a}^{\infty} \frac{1}{x^a}\,dx$$ converges for $a>1$.
But I need to find all real $a$ for which this integral will converge.
Any help would be appreciated.
Hint. Let $a>0$. The integral over $[1,+\infty)$ is convergent iff $2a>1$ because as $x\to +\infty$ $$\frac{1}{x^a(x+a)^a}\sim\frac{1}{x^{2a}}.$$ On the other hand the integral over $[0,1]$ is convergent iff $a<1$ because as $x\to 0^+$ $$\frac{1}{x^a(x+a)^a}\sim\frac{1}{x^{a}a^a}.$$ What happens when $a\leq 0$?