Solving the improper integral $1/(x^a+y^b)$

Question

I want to discuss the convergence of this improper integral:

$$\int_{1}^{\infty }dy\int_{1}^{\infty }dx \frac{1}{x^\alpha +y^\beta} \text{ with } \alpha,\beta>0$$

I know by polar coordinates that $\alpha=\beta=2$ is divergent (because it's like $\int\int\frac{1}{\rho}$), but what about $\alpha=4,\beta=2$ or $\alpha=8,\beta=5$? What general way can I use to solve it?

Answer 1

Substituting $\phi(x,y) = (u, v) = (x^\alpha, y^\beta)$ gives $$ J_\phi = \det \frac{d(x,y)}{d(u,v)} = \frac{1}{\det\frac{d(u,v)}{d(x,y)}} = \frac{1}{\alpha\beta\cdot u^a \cdot s^b} $$ where $a = \frac{\alpha-1}{\alpha}= 1- \frac{1}{\alpha}, b = \frac{\beta-1}{\beta}= 1-\frac{1}{\beta}$.

Now:

$$\int_{x,y\geq1} \frac{dx \, dy}{x^\alpha+y^\beta}= \frac{1}{\alpha\beta}[\int_{u\geq s\geq 1} \frac{du \, ds}{(u+s)\cdot u^a \cdot s^b} + \int_{s > u \geq 1} \dots]$$

The first integrand ($u\geq s$) converges and diverges the same as $$\int_{u\geq max(1,s)} \frac {du \, ds}{2u \cdot u^a\cdot s^b} = \int_{1}^{\infty} \int_{1}^{u}\frac{ds \, du}{2u^{a+1}\cdot s^b} = \int_{1}^{\infty} \frac{\frac{1}{u^{b-1}}-1}{2u^{a+1}(1-b)}\,du$$

So for them to converge we need $a+1+b-1 > 1 \implies \frac{1}{\alpha} + \frac{1}{\beta} < 1$, and also $a+1 > 1\implies \frac{1}{\alpha}<1$, and if we'll take the second integrand instead we get $b+1 > 1 \implies \frac{1}{\beta} < 1$ so overall we just need $\frac{1}{\alpha} + \frac{1}{\beta} < 1 \, \cap \,max(\frac{1}{\alpha}, \frac{1}{\beta}) < 1$. (this is a re-writing of an answer I've given for $\int_{|x|+|y|\geq 1} \frac{1}{x^p+y^q}$)

Answer 2

Suppose that $\alpha,\beta>2$. Take $r$ such that $2<r<\min(\alpha,\beta)$. Then, if $x,y>1$ we have $$ \frac{1}{x^\alpha+y^\beta}\le\frac{1}{x^r+y^r}\le\frac{C}{(x^2+y^2)^{r/2}} $$ for some constant $C>0$. Since $r/2>1$, the integral is convergent.

The case $\alpha=2$ and $\beta>2$ is more delicate. You can perform the integration in $x$ explicitly, and get a one dimensional integral in $y$.