Convergence of $\int_1^\infty\int_1^\infty(x+y)^{-a}dxdy$

Question

My question is as follows:

Find $a>0$ so that $I=\int_1^\infty\int_1^\infty(x+y)^{-a}dxdy$ converges.

My attempt: Assume that $I$ converges.

$I=\int_1^\infty x^{-a}\int_1^\infty(1+y/x)^{-a}dydx =\int_1^\infty x^{-a+1}\int_{1/x}^\infty(1+u)^{-a}dudx$, where $u:=y/x$.

If $a \le 1$ then $\int_{1/x}^\infty(1+u)^{-a}du\to \infty$ and $x^{-a+1} \ge 0$ which implies $I$ diverges. Let us assume that $a >1$. Then $$I \le \int_1^\infty x^{-a+1}dx \le \int_{0}^\infty(1+u)^{-a}du\int_1^\infty x^{-a+1}dx.$$ Hence $I$ is finite if and only if $a>1$ and $a-1 > 1$, which is equivalent to $a>2$.

Is my proof correct?

Answer 1

By exploiting the positivity of the integrand function and applying Fubini's theorem we have: $$ I(a)=\iint_{(1,+\infty)}(x+y)^{-a}\,d\mu =\int_{1}^{+\infty}\frac{dy}{(a-1)(1+y)^{a-1}}=\frac{1}{2^{a-2}(a-1)(a-2)} $$ for any $\color{red}{a>2}$.