2

I'm thinking about when can be stated that

$\lim\limits_{s\to s_0}\displaystyle\int f(x,s)dx = \int \lim\limits_{s\to s_0}f(x,s)dx$

Can you help me with this?

What are the hypothesis about $f$ to assure the statement?

Particularly I was trying to calculate

$$\iint_{\mathbb{R}^2}(x^2+y^2)e^{-(x^2+y^2)}dA$$

so I was thinking in doing this:

First use that

$$\iint_{\mathbb{R}^2}e^{-t(x^2+y^2)}dA = \frac{\pi}{t}$$

And then as $\frac{d}{dt}e^{-t(x^2+y^2)} = -(x^2+y^2)e^{-t(x^2+y^2)}$

Use $$\iint_{\mathbb{R}^2}\frac{d}{dt}e^{-t(x^2+y^2)}dA = \frac{-\pi}{t^2}$$

But I'm not sure if I can do that. That's why I'm asking about swap limit and integral.

MathGuest
  • 433
  • See dominated convergence theorem and monotone convergence theorem. Then investigate how to get a continuous limit given that you know sequential limits. – GEdgar Oct 19 '13 at 23:31
  • It's correct for that example, because the derivative is dominated by an integrable function (as long as you keep $t \geqslant \varepsilon > 0$). – Daniel Fischer Oct 19 '13 at 23:45
  • @DanielFischer I thought the usual criterion would be that the integral converges uniformly, not just that the function is integrable. – Betty Mock Oct 20 '13 at 00:47
  • @BettyMock The derivative $\frac{\partial f}{\partial t}(t,x,y)$ is dominated by the same integrable function for all $t\in [\varepsilon,\infty)$, I guess that means the integral converges uniformly. – Daniel Fischer Oct 20 '13 at 01:19
  • It looks like the dominated convergence theorem allows one to switch these limits under slightly less rigorous conditions. I see though that it applies to Lebesgue integrals. Then again with such a nice integrand it probably doesn't matter. – Betty Mock Oct 21 '13 at 02:11

0 Answers0