Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been struggling with the following problem:

$$\lim_{a\to\infty} \int_0^1 \frac {x^2e^x}{(2+ax)} dx .$$

My first reaction was to attempt to move the integral outside the limit, which would make the answer 0. However, I have read a little about Fatou's lemma and am not sure if I can do this. Any help would be greatly appreciated. Thank you

share|cite|improve this question

Note that $$0 \leq \dfrac{x^2 e^x}{2+ax} \leq \dfrac{e}{2+ax} \,\,\,\,\, \forall x \in [0,1]$$ Hence, we have $$0 \leq \int_0^1 \dfrac{x^2 e^x}{2+ax} dx \leq \int_0^1 \dfrac{edx}{2+ax} = e \left.\dfrac{\log(2+ax)}a \right \vert_{x=0}^1 = \underbrace{\dfrac{e}a \log \left(\dfrac{2+a}2\right)}_{\to 0 \text{ as } a \to \infty}$$

If you are lazy to evaluate any integrals, then you could also prove using by Lebesgue dominated convergence theorem, by noting the fact that $$f_a(x) = \dfrac{x^2 e^x}{2+ax} \leq \dfrac{x^2 e^x}2 = g(x)$$ and $\displaystyle \int_0^1 \dfrac{x^2e^x}2 dx < \infty$, since the integral is over a compact set and integrand is continuous over this compact set. Hence, we have $$\lim_{a \to \infty} \int_0^1 \dfrac{x^2 e^x}{2+ax} dx = \int_0^1 \lim_{a \to \infty} \dfrac{x^2 e^x}{2+ax} dx = \int_0^1 0dx = 0$$

share|cite|improve this answer
So since the answer is 0, does this mean I was allowed to pull the integral outside the limit to begin with? Or is this just a coincidence? Thank you. – Brian Mar 22 '13 at 4:13
In general, you are not allowed to get the limit inside the integral. For instance, consider this. $$\lim_{a \to \infty} \int_0^1 \dfrac{dx}{ax}$$ The answer to this $\infty$, whereas if you get the limit inside the integral, the answer is $0$ i.e. $$\lim_{a \to \infty} \int_0^1 \dfrac{dx}{ax} = \infty \neq 0 = \int_0^1 \lim_{a \to \infty} \dfrac{dx}{ax}$$ – user17762 Mar 22 '13 at 4:17

Your Answer


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.