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

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.

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This technique would work here, too. –  sos440 Jun 1 '12 at 11:29
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5 Answers

up vote 7 down vote accepted

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

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Thank you for your post and I know how to do it now! –  Hao Wang Jun 1 '12 at 11:44
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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 $F(n) \approx n\pi.$

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Thank you for your answer, but it's too fancy for the class I'm taking now. It just covers some basic convergence tests like ratio, comparison and dirichlet. Andre's post will solve it. However, thanks again for I might be able to use it for upper classes. –  Hao Wang Jun 1 '12 at 11:43
    
Interesting. So if integrating from $0$ to $n\pi$ gives approximately $n\pi$, then in some way the function itself is on average approximately $1$? –  Théophile Jun 2 '12 at 15:58
    
@Théophile Precisely! Your intuitive idea of the average value of a function is correct. In fact, the "average" value of $f$ over $[a,b]$ is often defined by $ 1/(b-a) \int^b_a f(x) dx.$ So even though the function frequently spikes up ( at $x=k\pi$, $f(x) = k\pi$), the spikes only last for a very short time, and most of the time the values are between 0 and 2. Plotting the graph would illustrate this nicely. –  Ragib Zaman Jun 2 '12 at 16:04
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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*}

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Hint

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

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A function may very well have unbounded sub-sequences and be integrable. –  AD. Jun 1 '12 at 15:15
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The answer for the convergence/divergence may be easily obtained by using the following trivial inequality: $$\int_{0}^{\infty} \frac{x}{1+x^{2}}= \bigl(\frac{1}{2}\log(x^2+1)\bigr)_{0}^{\infty} \leq\int_{0}^{\infty} \frac{x}{1+x^{2}\sin^{2}{x}} \to \infty\ $$

The proof is complete.

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