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Improper Integral
10-13-2010, 10:31 AM
Post: #1
Improper Integral
(1) $\displaystyle{\int_0^{\infty}\frac{dx}{1+x^2\sin^2 x}}$
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10-13-2010, 04:34 PM
Post: #2
RE: Improper Integral
(1) Let $N_x = \lfloor x/\pi\rfloor$, then $\displaystyle{\int_0^x \frac{dt}{1+t^2\sin^2t} \ge \sum_{n=1}^{N_x} \int_{(n-1)\pi}^{n\pi} \frac{dt}{1+t^2\sin^2t}\ge \sum_{n=1}^{N_x} \int_{(n-1)\pi}^{n\pi} \frac{dt}{1+(n\pi)^2\sin^2t}}$
$\quad\quad \displaystyle{\ge \sum_{n=1}^{N_x} \int_0^{\pi /2} \frac{2dt}{1+(n\pi)^2\sin^2t} \ge 2\sum_{n=1}^{N_x} \int_0^{\pi /2} \frac{dt}{1+(n\pi)^2 t^2}=\sum_{n=1}^{N_x} \frac{2}{n\pi}\tan^{-1} \frac{n\pi^2}{2}\ge \frac{2}{\pi}\tan^{-1} \frac{n\pi^2}{2} \sum_{n=1}^{N_x} \frac{1}{n}}$
$\quad\quad$ Therefore $\displaystyle{\int_0^{\infty}\frac{dx}{1+x^2\sin^2x}}$ divergent
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