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

$$\lim_{p\rightarrow\infty}\int_0^1e^{-px}(\cos x)^2\text{d}x$$ I tried to prove the integrand is uniformly convergent so that the limit and integral can be exchanged. But I failed. Any ideas?

share|cite|improve this question
up vote 9 down vote accepted

There is no need for uniform convergence; using dominated convergence will do the job just fine. In fact, this is not uniform convergence at all (remember that the uniform convergence of a continuous function is continuous, but this is not the case, as the function it converges to has a discontinuity at $x=0$).

Also, if you cannot use dominated convergence, even the squeeze theorem will do. We have $$\begin{align} 0 \leq \int\limits_{0}^{1} e^{-px}(\cos(x))^{2}\;\mathrm{d}x &\leq \int\limits_{0}^{1} e^{-px}\;\mathrm{d}x \\ &=\left. \frac{-e^{-px}}{p} \right|_{x=0}^{1} \\[5pt] &=\frac{1}{p} - \frac{e^{-p}}{p} \end{align}$$ Now $\lim\limits_{p\rightarrow\infty}\frac{e^{-p}}{p}=0$ using L'Hopital, so this does indeed squeeze in.

share|cite|improve this answer
So just apply dominated convergence thm and change the order of limit and integral? Do I get 0 for the final answer? – lovelesswang Jun 23 '14 at 22:08
@lovelesswang:Yes you would get 0. Also, I just edit in another solution without using dominated convergence. – Gina Jun 23 '14 at 22:10

This is actually a simple integration by parts question and you do not necessarily need to invoke any fancy theorems. The integral can be evaluated by elementary means. Use $\cos^2 x = [1 + \cos(2x)]/2$. Then integrate $e^{-px} \cos(2x)$ by parts (twice). I do not think you should have any trouble integrating $e^{-px}$. Lastly, take the limit as $p \to \infty$.

share|cite|improve this answer

Let $\epsilon\gt 0$. Then there is a $P$ such that if $p\gt P$, and $x\gt \epsilon/2$, then $e^{-px}\lt \epsilon/2$. So for $p\gt P$, our integral is $\lt (1)(\epsilon/2)+(\epsilon/2)(1-\epsilon/2)$, which is $\lt \epsilon$.

share|cite|improve this answer

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.