Convergent integral

  • Thread starter talolard
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  • #1
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Homework Statement



Does [tex] \int_{0}^{\infty}\frac{dx}{1+\left(xsinx\right)^{2}} [/tex] converge?

I don't know if this is a legitimate solution. Any insight? Thanks
Tal

The Attempt at a Solution


No.
[tex]
f(x)=\frac{1}{1+\left(xsinx\right)^{2}}\geq g(x)=\begin{cases}
\frac{1}{1+\left(xsin(x)\right)^{2}} & 2\pi k\leq x\leq2\pi k+\pi\\
0 & \else\end{cases} [/tex]

Then [tex] \int_{0}^{\infty}\frac{dx}{1+\left(xsinx\right)^{2}}\geq\int_{0}^{\infty}g(x)=\sum_{k=0}\int_{2\pi k}^{2\pi k+\pi}\frac{1}{1+\left(xsin(x)\right)^{2}}\geq\sum_{k=0}\int_{2\pi k}^{2\pi k+\pi}\frac{1}{1+\left(\frac{\pi}{2}\right)^{2}}=\sum_{k=0}^{\infty}\frac{\pi}{1+\left(\frac{\pi}{2}\right)^{2}} [/tex]
 

Answers and Replies

  • #2
Gib Z
Homework Helper
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[tex]\sum_{k=0}\int_{2\pi k}^{2\pi k+\pi}\frac{1}{1+\left(xsin(x)\right)^{2}}\geq\sum_{k=0}\int_{2\pi k}^{2\pi k+\pi}\frac{1}{1+\left(\frac{\pi}{2}\right)^{2}} [/tex]

I don't see why that step should be true, since [itex] \pi/2 [/itex] doesn't bound x sin x. This attempted solution was incorrect unfortunately although you are correct in that it does not converge.

To start the problem, write down [tex] \int^{\infty}_0 \frac{1}{1+x^2 \sin^2 x} dx = \sum_{k=0}^{\infty} \int^{(k+1)\pi}_{k\pi} \frac{1}{1+x^2 \sin^2 x} dx [/tex].

Then let [itex] x = u + k \pi [/itex] and try to bound the integral.
 

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