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

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3 Answers 3

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.

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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 at 22:08
    
@lovelesswang:Yes you would get 0. Also, I just edit in another solution without using dominated convergence. –  Gina Jun 23 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 $p \to \infty$.

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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 $\lt \epsilon$.

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