$\int_{\mathbb{R}^2} e^{-(2|x|^a+3|y|^b)}dxdy$

Question

As you see from the title, I'm asked to compute for which $a,b$ the following integral converges: $$\int_{\mathbb{R}^2} e^{-(2|x|^a+3|y|^b)}dxdy$$ By parity argument we can say that: $$\int_{\mathbb{R}^2} e^{-(2|x|^a+3|y|^b)}dxdy=4\int_{\mathbb{R}_+\times \mathbb{R}_+} e^{-(2|x|^a+3|y|^b)}dxdy$$ and I thought that we might move to common exponent by putting $u=x$ and $v^{\frac{b}{a}}=y$: $$4\int_{\mathbb{R}_+\times \mathbb{R}_+} e^{-2|x|^a+3|y|^b}dxdy=\frac{4a}{b}\int_{\mathbb{R}_+\times \mathbb{R}_+}v^{\frac{a}{b}-1}e^{-(u^a+v^a)}dvdu$$ but I don't really know how to proceed. Can you help please?

Answer 1

You may start from $$ \int_{0}^{+\infty} e^{-w^\alpha}\,dw =\frac{1}{\alpha}\int_{0}^{+\infty}z^{1/\alpha-1}e^{-z}\,dz = \Gamma\left(1+\frac{1}{\alpha}\right)$$ which holds for any $\alpha>0$, then apply your symmetry argument and Fubini's theorem.