Is $\int_E \frac{1}{(x^2+y^2)^2}dxdy$ convergent?

I have to tell whether this integral is convergent: $$\int_E \frac{1}{(x^2+y^2)^2}dxdy$$ where $$E=\{0\leq y \leq x^a\} \cap \{x^2+y^2\leq 1\}$$.

I'm asked for which $$a \geq 0$$ the integral converges. How am I supposed to act when I find this kind of integrals? I mean, these domains determined by intersections of a curve with $$[0,1]^2$$ or with $$B_n(0,0)$$.

EDIT: I suppose I should do it for $$x \geq 0$$ even if not specified.

• $a \in \mathbb R$ ? Jun 2, 2020 at 13:33
• $a \geq 0$, sorry. Jun 2, 2020 at 13:35

If $$x<0$$ we have that $$x^a$$ is not necessarily defined, so I am going to assume that the actual problem is to discuss the convergence of $$I(a)=\iint_E \frac{dx\,dy}{(x^2+y^2)^2},\qquad E=\{(x,y):x^2+y^2\leq 1, x> 0, 0

With these assumptions we have $$I(a) = \int_{0}^{1}\frac{L(\rho)}{\rho^4}\,d\rho$$ hence the problem boils down to estimating $$L(\rho)$$ for $$\rho\to 0^+$$. If $$a\leq 1$$ we have $$L(\rho)\geq c\rho$$ and the integral is clearly divergent. It follows that we may assume that $$x^a$$ is a convex function on $$[0,1]$$. This easily leads to $$L(\rho)\sim \rho^a\quad\text{as }\rho\to 0^+$$ and to the fact that the integral is convergent for $$\color{red}{a>3}$$.

The curves $$y=x^a, y=\sqrt {1-x^2}$$ intersect at some $$(b,b^a),$$ where $$0 Thus $$E= \{(x,y): 0\le x \le b,0\le y\le x^a\}.$$ We don't need the exact value of $$b$$ to do the problem.

The integral is thus

$$\int_0^b\int_0^{x^a}\frac{1}{(x^2+y^2)^2}\,dy\,dx.$$

In the inner integral, let $$y=xt.$$ The integral becomes

$$\tag 1 \int_0^b\frac{1}{x^3}\int_0^{x^{a-1}}\frac{1}{(1+t^2)^2}\,dt\,dx.$$

If $$a\le 1,$$ then $$x^{a-1} > 1.$$ So $$(1)$$ is at least

$$\int_0^b\frac{1}{x^3}\int_0^{1}\frac{1}{(1+t^2)^2}\,dt\,dx.$$

The inner integral is a constant $$C,$$ so $$(1)$$ is at least $$C\int_0^b\dfrac{1}{x^3}\,dx = \infty.$$

If $$a>1,$$ then $$x^{a-1} <1.$$ $$(1)$$ is then bounded above by

$$\int_0^b\frac{1}{x^3}\int_0^{x^{a-1}}1\,dt\,dx = \int_0^b \frac{1}{x^3}\cdot x^{a-1}\,dx = \int_0^b x^{a-4}\,dx.$$

This converges iff $$a-4>-1,$$ or $$a>3.$$

It follows that the given integral in the problem converges iff $$a>3.$$