Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My professor posed a question to us last week about the limit of a function as $k \to \infty$. He asked us to prove that $\int \lim_{k \to \infty} f(x)e^{-\frac{x^2}{k}} dx = \int f(x)dx$. This seems fairly basic, since it can be directly shown that the exponential part of the integrand, $e^{-\frac{x^2}{k}} \to 1$ as $k \to \infty$. Is there something I am missing? Are there extra steps needed to show that the integral of the limit is equal to the limit of the integral? Is this even necessary? I feel like this question is too simple relative to the rest of the material in class, but I'm not sure what I am missing.

share|improve this question
    
well, $f=g$ implies $\int f=\int g$... –  leo Nov 20 '12 at 21:29
    
Are you sure he didn't ask $\lim_{k\rightarrow \infty}\int f(x)e^{-x^2/k}\,dx$? In that case, you need to justify switching the integral and the limit. –  asmeurer Dec 3 '12 at 5:12

1 Answer 1

Over an interval or compact set, $f(x) e^{-x^2/k} \to f(x)$ uniformly. Then you can get $\int \lim = \lim \int$ from Riemann integration.

In a Lebesgue theory class, looks like you're integrating with uniform measure on some measurable set. So perhaps you need dominated convergence.

Indeed, $f(x) e^{-x^2/k} \to f(x)$ pointwise and $|f(x) e^{-x^2/k}| \leq f(x)$. Therefore

$$ \lim_{k \to \infty} \int f(x) e^{-x^2/k} \, dx \to \int f(x) \, dx$$

So you just need to be sure everything is integrable.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.