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I would like your help with deciding whether the following integral converges or not:

$\int_{0}^{\infty}\frac{dx}{1+(x\sin5x)^2}.$

I tried to compare it to other functions and to change the variables, but it didn't work for me.

Thanks a lot!

  • 0
    fyi, Mathematica 8.0.4 says it does not converge2012-11-27

3 Answers 3

3

Consider the intervals $I_k=[(k-\frac{1}{2})\frac{\pi}{5},(k+\frac{1}{2})\frac{\pi}{5}]$. The $I_k$ are the periods of $\sin^2(5x)$. Therefore, $ \begin{align} \int_{I_k}\frac{\mathrm{d}x}{1+x^2\sin^2(5x)} &\ge\left|\int_{I_k}\frac{\cos(5x)\;\mathrm{d}x}{1+(k+\frac{1}{2})^2\pi^2/25\;\sin^2(5x)}\right|\\ &=\frac{1}{(k+\frac{1}{2})\pi}\int_{-1}^1\frac{(k+\frac{1}{2})\pi/5\;\mathrm{d}t}{1+(k+\frac{1}{2})^2\pi^2/25\;t^2}\\ &=\frac{2}{(k+\frac{1}{2})\pi}\tan^{-1}\left(\frac{(k+\frac{1}{2})\pi}{5}\right) \end{align} $ Since $\tan^{-1}\left(\frac{(k+\frac{1}{2})\pi}{5}\right)>\frac{\pi}{4}$ for $k>1$, the integral on $I_k$ is greater than $\frac{1}{2k+1}$. Therefore, the integral diverges.

6

Basic idea: Near $x_k = {2\pi k \over 5}$ the integrand is comparable to ${1 \over 1 + 25x_k^2(x - x_k)^2}$. Thus integrates to a term of magnitude $O({1 \over x_k})$. Add up over all $k$ and the integral diverges.

  • 4
    Since this is probably a homework question I didn't want to give away too much.2011-11-29
5

Looks like it does not converge. You can argue as follows. Split the integral up into pieces:

$\int_0^\infty \frac{dx}{1+ (x\sin 5x)^2} \geq \sum_{k=0}^\infty \int_{k\pi/5}^{(k+1/2)\pi/5} \frac{dx}{1 + (x\sin 5x)^2}. $

For $k\pi/5 \leq x \leq (k+1/2)\pi/5$, note that

$ \frac{1}{1+(x\sin 5x)^2} \geq \frac{1}{1+25x^2(x-k\pi/5)^2} \geq \frac{1}{1+ (k+1/2)^2\pi^2(x-k\pi/5)^2}.$

It follows that

$\int_{k\pi/5}^{(k+1/2)\pi/5} \frac{dx}{1 + (x\sin 5x)^2} \geq \int_{k\pi/5}^{(k+1/2)\pi/5} \frac{dx}{1+ (k+1/2)^2\pi^2(x-k\pi/5)^2} = \frac{\arctan((k+1/2)\pi^2/10)}{(k+1/2)\pi}.$

Substituting this into the sum above we find that the sum diverges and hence the integral does too.

  • 0
    Just use the fact that $|\sin 5x| \leq 5|x-k\pi/5|$ for $k\pi/5 \leq x \leq (k+1/2)\pi/5$2011-12-03