Saturday, 5 December 2015

real analysis - Find limrtoinftyintpi0ercos2(theta)dtheta



lim




Here is what I did:



Since e^{-r\cos^2(\theta)} is continuous on [0,\pi] for any fixed r, we can use MVT:
There is c \in (0,\pi) such that



\displaystyle \int_{0}^{\pi} e^{-r\cos^2(\theta)} d\theta = e^{-r\cos^2(c)}\pi



Also e^x>0 \forall x \in \mathbb{R}, so:



\displaystyle 0 \leq \int_{0}^{\pi} e^{-r\cos^2(\theta)} d\theta = e^{-r\cos^2(c)}\pi




Using squeeze theorem, \displaystyle \lim_{r \to \infty} 0 = 0 = \lim_{r \to \infty} e^{-r\cos^2(\theta)}, then we have \displaystyle \lim_{r \to \infty} \int_{0}^{\pi} e^{-r\cos^2(\theta)} d\theta=0



Is this correct?
Thanks.



Also: I cannot pass the limit inside the integral, right? But why not, exactly?


Answer



For any r\ge 0 and for any \theta the following inequality holds e^{-r\cos^2\theta}\le 1. Choose a sequence of positive real numbers (r_n) that runs to the infinity. Then
e^{-r_n\cos^2\theta}\longrightarrow \begin{cases} 0 \text{ if } \theta\neq\frac{\pi}{2}\\ 1 \text{ if } \theta = \frac{\pi}{2} \end{cases}
pointwise as n goes to infinity (but not uniformly). Moreover such a limit does not depend on the sequence we have chosen before. Now apply the dominated Lebesgue’s theorem to any sequence to pass the limit under the integral (indeed, the pointwise convergence is sufficient, you don’t need the uniform convergence). Hence:



\lim_{r_n\to\infty}\int_0^\pi e^{-r_n\cos^2\theta}=\int_0^\pi \lim_{r_n\to\infty} e^{-r_n\cos^2\theta}=0


No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...