0
$\begingroup$

Let $f:(0,1]\to\mathbb R$ be continuous and positive, and: $$\lim_{t\to 0}f(t) = +\infty,\quad \int_0^1 f(t)dt = \lim_{\epsilon\to 0^+}\int_\epsilon^1 f(t)dt<\infty$$ Show that $F:B(0,1)\to\mathbb R$ defined by $F(x,y) = f(||(x,y)||)$ (where $B(0,1) = \{(x,y)\in\mathbb R\mid x^2+y^2\leq 1\})$ satisfies: $$\int_{B(0,1)} F(x,y)dxdy<\infty$$

I've gotten as far as making a variable change to polar coordinates and ending up with an integral of $f(r) * r $ from $0 $ to $ 1$ but I don't know how to prove that that converges.

Thanks in advance :)

  • 0
    Unfortunately the picture isn''t in English...Could you offer us a translation? – 2017-02-22
  • 0
    Let $f:(0,1] \rightarrow \mathbb{R}$ be continuous and positive, such that [equation], show that if $F:B(0,1) \rightarrow \mathbb{R}$ is defined by $F(x,y) = f(||(x,y)||)$, (where $B(0,1) = $...), then: [equation] – 2017-02-22
  • 0
    Thanks @Alex! You beat me to it! :) – 2017-02-22

1 Answers 1

2

We have that the integral is: $$\int_{B(0,1)} f(||(x,y)||)dxdy = \int_{B(0,1)} f(\sqrt{x^2+y^2})dxdy$$ Letting $x = r\cos\theta$ and $y = r\sin\theta$, we get that this is (recalling that the Jacobian determinant of this is $r$): $$\int_{r = 0}^1\int_{\theta = 0}^{2\pi} f(r)r d\theta dr$$ Now, the $\theta$ integral is easy, and just introduces a factor of $2\pi$, so we get: $$2\pi\int_0^1 rf(r)dr$$ Now, as $0

Note that in this last line it's very important that $f$ is positive, as otherwise we couldn't argue that $rf(r)\leq f(r)$ on the entire domain of integration.

  • 0
    Thanks @Mark!!! the last line of removing r just because its between 0 to 1 was what I couldn't see. Thanks a lot! :) – 2017-02-22