does $\int_0^\infty x/(1+x^2 \sin^2x) \mathrm dx$ converge or diverge? $$\int_0^\infty x/(1+x^2\sin^2x) \mathrm dx$$
I'd be very happy if someone could help me out and tell me, whether the given integral converges or not (and why?). Thanks a lot.
 A: Hint: The integral diverges, there is trouble when $x$ is large. For detail, use the fact that if $x \ge 1$, then
$1+x^2\sin^2 x \le 2x^2$ and therefore 
$$\frac{x}{1+x^2\sin^2 x} \ge \frac{1}{2x}.$$ 
A: Welcome to math.stackexchange! The divergence of this integral is easily seen by André Nicolas' method. We might be interested in something more quantitative: How quickly does this integral diverge? One method is to exploit the periodicity of $\sin(x).$ Let us study the asymptotic behavior of $$F(n)=\int^{n\pi}_0 \frac{x}{1+x^2 \sin^2 x} dx.$$
Let us break the integral into pieces of length $\pi$ like this:
$$ \int^{n\pi}_0 \frac{x}{1+x^2 \sin^2 x} dx = \sum_{k=1}^n \int^{k\pi}_{(k-1)\pi} \frac{x}{1+x^2 \sin^2 x } dx .$$
Now let $u= x-(k-1)\pi:$
$$ = \sum_{k=1}^{n} \int^{\pi}_0 \frac{u + (k-1)\pi}{1+(u+(k-1)\pi)^2 \sin^2 u} du .$$
Now all that remains is to carefully estimate the size of this integral. Using $ 0 \leq u \leq \pi$ and $\displaystyle \int^{\pi}_0 \frac{1}{1+\alpha \sin^2 t} dt = \frac{ \pi}{\sqrt{1+\alpha}}$ we find that 
$$ \frac{(k-1)\pi^2}{\sqrt{1+k^2\pi^2}} \leq \int^{\pi}_0 \frac{u + (k-1)\pi}{1+(u+(k-1)\pi)^2 \sin^2 u} du \leq \frac{k \pi^2}{\sqrt{ 1+(k-1)^2 \pi^2} }. $$ Thus we can see $F(n) \approx n\pi.$
A: The answer for the convergence/divergence may be easily obtained by using the following trivial inequality:
$$\int_{0}^{\infty} \frac{x}{1+x^{2}\sin^{2}{x}} \geq \int_{0}^{\infty} \frac{x}{1+x^{2}}= \bigl(\frac{1}{2}\log(x^2+1)\bigr)_{0}^{\infty}  \to \infty\ $$
The proof is complete.
A: Since $-1 \leq \sin{x} \leq 1$, you have $\sin^{2}{x} \leq 1$, and from this you can conclude that 
\begin{align*}
1+x^{2}\sin^{2}{x} &\leq 1+2x^{2} \\\ \Longrightarrow  \frac{1}{1+x^{2}\sin^{2}{x}} &\geq \frac{1}{1+2x^{2}} \\\ \Longrightarrow \frac{x}{1+x^{2}\sin^{2}{x}}&\geq \frac{x}{1+2x^{2}} \\\
 \Longrightarrow \int_{0}^{\infty} \frac{x}{1+x^{2}\sin^{2}{x}} \ dx &\geq \int_{0}^{\infty} \frac{x}{1+2x^{2}} \ dx
\end{align*} 
Now 
\begin{align*}
\int_{0}^{\infty} \frac{x}{1+2x^{2}} \ dx &= \frac{1}{4} \cdot \int_{0}^{\infty} \frac{4x}{1+2x^{2}} \ dx \\\ &= \frac{1}{4} \bigl(\log{x}\bigr)_{1}^{\infty} \to \infty
\end{align*}
A: Hint
$$\lim_{n \to \infty} \frac{n\pi}{1+ n^2 \pi^2 \sin^2(n \pi)} = \infty$$
