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 $B_n(0,0)$.

Thanks in advance.

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

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

2 Answers 2


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 $$I(a)=\iint_E \frac{dx\,dy}{(x^2+y^2)^2},\qquad E=\{(x,y):x^2+y^2\leq 1, x> 0, 0<y<x^a\}.$$ enter image description here

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 $0<b<1.$ 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


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


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 $a>3.$


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