Convergence of improper double integral

Calculus Level 5

Inital conditions of the problem:

Let be $D := \{ (x,y)\in \mathbb{R}^2 : x^2 + y^2 < 1 \}$

Consider the improper integral parametrized by $a$ :

$I_a := \iint_{D} \frac{dxdy}{(\sqrt{1-x^2-y^2})^{5a}}$

What you've to answer:

For what real values of $a$ does $I_a$ converge?

a > \frac{5}{2}

a > \frac{2}{5}

a < \frac{5}{2}

a \leq \frac{2}{5}

a \geq \frac{2}{5}

a < \frac{2}{5}

a \geq \frac{5}{2}

a \leq \frac{5}{2}

1 solution

Otto Bretscher
Dec 24, 2015

Changing variables to polar coordinates, we find that $I=\lim_{a\to 1^{-}}\int_{0}^{2\pi}\int_{0}^{a}(1-r^2)^{-k}rdrd\theta=\frac{1}{2-2k}$

for $k<1$ . For $k=1$ we have $I=-\pi\lim_{a\to 1^{-}} \ln(1-a^2)$ , which diverges to infinity, and for $k>1$ the integral will diverge as well, by dominance. Thus it is required that $k=\frac{5a}{2}<1$ , or $a<\boxed{\frac{2}{5}}$ .

Moderator note:

Good usage of the conversion to polar coordinates, as suggested by $x^2 + y^2$ and the integration about the circle in the question.

Convergence of improper double integral

1 solution

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