Does $\int_{0}^{\infty}\frac{dx}{1+(x\sin5x)^2}$ converge? 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! 
 A: 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. 
A: 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.
A: 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.
A: Consider the intervals where
$$|x\sin(5x)|<1,$$ where the function value exceeds $\dfrac12$.
For large $x$, the bounds of the intervals are close to $\dfrac{k\pi}5$ and asymptotically
$$\left|x\left(5\left(x-\frac{k\pi}5\right)\right)\right|<1.$$
Thus after solving the equation, you see that the length of the intervals is asymptotic to
$$a\sqrt{k^2+b}-a\sqrt{k^2-b}\sim \frac {ab}k$$ for some $a$ and $b$, leading to an Harmonic series.
