Show that the integral is divergent $\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}$

Question

Show that the integral is divergent $$\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}$$

It has no point of discontinuity in range of integration. Also, I have found $\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}>\frac{\pi}{2}$, but that seems to be of no consequence. I don't how to move ahead with this.

Answer 1

We have that \begin{align*}\int_0^\infty\frac{dx}{1+x^2\sin^2x} &=\sum_{n=0}^{\infty} \int_0^{\pi}\frac{dx}{1+(x+n\pi)^2\sin^2(x+n\pi)} \\&\geq \sum_{n=0}^{\infty} \int_0^{\pi/2}\frac{2dx}{1+\pi^2(1+n)^2\sin^2(x)}\\ &=\sum_{n=0}^{\infty}\frac{\pi}{\sqrt{1+\pi^2(n+1)^2}}\\ &\geq \frac{\pi}{\sqrt{1+\pi^2}}\sum_{n=0}^{\infty}\frac{1}{n+1}=+\infty \end{align*} where we used the fact that $$\int\frac{\ dx}{1+a^2\sin^2(x)}=\frac{\arctan(\sqrt{1+a^2}\tan(x))}{\sqrt{1+a^2}}+C.$$ P.S. We can also use the inequality $\sin^2(x)\leq x^2$ and evaluate $$\int_0^{\pi/2}\frac{2dx}{1+\pi^2(1+n)^2x^2}=\frac{2\arctan\left((n+1)\pi^2/2\right)}{\pi(n+1)}.$$

Answer 2

As an alternative to Robert Z' fine answer, you may consider that for any $n\in\mathbb{N}^+$ the function $x^2 \sin^2 x$ is bounded by $2\pi^2$ on the interval $I_n=\left[\pi n-\frac{1}{n},\pi n+\frac{1}{n}\right]$. Since the integrand function is non-negative and these intervals are disjoint,

$$ \int_{0}^{+\infty}\frac{dx}{1+x^2 \sin^2 x}\geq \int_{\bigcup I_n}\frac{dx}{1+2\pi^2} \geq \frac{1}{1+2\pi^2}\sum_{n\geq 1}\left|I_n\right|=+\infty $$ since $|I_n|=\frac{2}{n}$ and the harmonic series is divergent.