does $\int_0^\infty x/(1+x^2 \sin^2x) \mathrm dx$ converge or diverge?

Question

$$\int_0^\infty x/(1+x^2\sin^2x) \mathrm dx$$

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.

Answer 1

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 $$\frac{x}{1+x^2\sin^2 x} \ge \frac{1}{2x}.$$

Answer 2

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 $F(n) \approx n\pi.$

Answer 3

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.

Answer 4

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*}

Answer 5

Hint

$$\lim_{n \to \infty} \frac{n\pi}{1+ n^2 \pi^2 \sin^2(n \pi)} = \infty$$