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

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.

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.

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.