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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!

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${1\over 1+(x\sin 5x)^2}\ge{1\over 1+x^2}$ –  David Mitra Nov 29 '11 at 21:18
8  
@DavidMitra : How does that help? Squeezing is the first thing anyone will think of, but the inequality goes in the wrong direction for that. –  Michael Hardy Nov 29 '11 at 21:21
    
erm, sorry... off day for me... –  David Mitra Nov 29 '11 at 21:23
    
fyi, Mathematica 8.0.4 says it does not converge –  Nasser Nov 27 '12 at 4:58

3 Answers 3

up vote 5 down vote accepted

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.

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Can you please explain the first inequality in the 4Th row? Thanks! –  Jozef Dec 2 '11 at 13:04
    
Just use the fact that $|\sin 5x| \leq 5|x-k\pi/5|$ for $k\pi/5 \leq x \leq (k+1/2)\pi/5$ –  Jeff Dec 3 '11 at 18:55

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.

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More rigor please! –  Potato Nov 29 '11 at 21:45
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Since this is probably a homework question I didn't want to give away too much. –  Zarrax Nov 29 '11 at 21:46

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.

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