Convergence of improper double integral.

Question

Please help me to determine $\alpha$ and $p$, such that the integral $$ I = \iint_G \frac{1}{(x^{\alpha}+y^3)^p} \ dx dy $$ converges, where $G = {x>0, y >0, x+y <1}$ and $\alpha >0, p>0$.

I am comfortable with proper double integrals. I am also comfortable with improper double integrals when $f$ is continuous in $G$ everywhere except the one point. However, I do not understand how to handle this case - it looks like $f$ is not continuous at 2 segments. I am also not sure why and when I can change the order of integration in this case. My textbook shows example when double integral diverges, while both iterated integral converges. Unfortunately I do no know where to start. I guess that I should compute the integral itself in terms of $\alpha$ and $p$ first, but again, I am not sure.

Thanks a lot for your help!

Answer 1

We can write the integral of interest $I$ as

$$\begin{align} I&=\int_0^1 \int_0^{1-y}\frac{1}{(x^\alpha+y^3)^p}\,dx\,dy\\\\ \end{align}$$

Next we enforce the substitution $x=u^{2/\alpha}\cos^{2/\alpha}(v)$ and $y=u^{2/3}\sin^{2/3}(v)$. The determinant of the Jacobian, $J$, is given by

$$ \begin{align} |J|&=\begin{vmatrix}\frac{\partial x}{\partial u}&&\frac{\partial x}{\partial v}\\\\\frac{\partial y}{\partial u}&&\frac{\partial y}{\partial v}\end{vmatrix}\\\\ &=\frac{4}{3\alpha}u^{2/\alpha+2/3-1}\cos^{2/\alpha-1}(v)\sin^{-1/3}(v) \end{align}$$

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Then, $I$ can be written

$$I=\frac{4}{3\alpha}\int_{D_{u,v}}u^{2/\alpha+2/3-1-2p}\cos^{2/\alpha-1}(v)\sin^{-1/3}(v)\,du\,dv$$

where the domain $D_{u,v}$ contains $(u,v)=(0,0)$. The integral over $v$ is well-behaved for all $\alpha$.

The integral over $u$ converges if and only if $\frac2\alpha+\frac23-1-2p>-1$ or $p<\frac1\alpha+\frac13$.

$I$ converges if and only if $p<\frac1\alpha+\frac13$.

Answer 2

Here's a different approach: For all $a,p >0,$ the integrand is bounded on the upper right triangle of the square $[0,1]^2,$ so it's enough to consider the integral over this square. This makes life a little easier.

Now let $x =y^{3/a}t.$ Then we have

$$\tag 1 \int_0^1\int_0^1 \frac{1}{(x^a+y^3)^p}\,dx\,dy = \int_0^1y^{3/a-3p}\int_0^{y^{-3/a}} \frac{1}{(t^a+1)^p}\,dt\,dy.$$

The inner integral on the right of $(1),$ as a function of $y,$ is bounded below by $\int_0^{1} 1/(t^a+1)^p\,dt.$ Thus a necessary condition for convergence of $(1)$ is $3/a-3p > -1,$ or $p<1/a +1/3.$ To see this is also a sufficient condition requires a little work in estimating this inner integral (as a function of $y$). I'll leave that to the reader for now.