Does the integral $$\int_0^\infty \frac{1}{1+(x\sin x)^2} \ \, \mathrm{d}x$$ converge?
I know that I need to look at:
$$\sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi} \frac{1}{1+(x\sin x)^2}\ \, \mathrm{d}x$$
but not sure how utilize this.
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Zain Patel
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jmiller
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Between $x=n\pi$ and $x=n\pi-1/n$, we have $|\sin x|<1/n$, so $f(x)>1/(1+\pi^2)$. How does that affect the total integral.
Empy2
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I guess we could look at $\int_0^\infty \frac{1}{1+x^2\left|\sin x\right|} \ dx$. Is that your intention? – jmiller Jul 05 '15 at 13:34
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1No, sum the narrow little pieces. – Empy2 Jul 05 '15 at 13:35
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I see. Thank you @Micael. – jmiller Jul 05 '15 at 13:37