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$$\int_0^{ + \infty } {\left( {\frac{x}{{1 + {x^6}{{\sin^2 x}}}}} \right)dx}$$

I'd like your help with see why does?? I tried to compare it to other functions and to change the variables, but it didn't work for me.

Thanks a lot!

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    what do you mean by `why does ??`. fyi, Mathematica says this integral does not converge.2012-11-27

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We can divide up the ray $[0,\infty)$ into subintervals $[(n-\frac{1}{2})\pi,(n+\frac{1}{2})\pi]$, $n=1,2,\ldots$ (there a little piece $[0,\pi/2]$ left over, which we may ignore). On each subinterval, we compute $$ \begin{eqnarray*} \int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi}\frac{x}{1+x^6\sin^2 x}\,dx&\leq& (n+\frac{1}{2})\pi\int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi}\frac{1}{1+x^6\sin^2 x}\,dx\\ &\leq&(n+\frac{1}{2})\pi\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+(n-\frac{1}{2})^6\pi^2\sin^2 u}\,du\\ &\leq&(n+\frac{1}{2})\pi\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+4(n-\frac{1}{2})^6u^2}\,du\\ &\leq&(n+\frac{1}{2})\pi\int_{-\infty}^{\infty}\frac{1}{1+4(n-\frac{1}{2})^6u^2}\,du\\ &=&\frac{\pi^2(n+\frac{1}{2})}{2(n-\frac{1}{2})^3} \end{eqnarray*} $$

These integrals are decaying like $\frac{1}{n^2}$, so the sum over all $n$ converges.

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    +1. There might be little error though. How did you go from the second line to third line? You could directly integrate $$\int_{-\pi/2}^{\pi/2} \dfrac{du}{1+a^2 \sin^2(u)} = \dfrac{\pi}{\sqrt{1+a^2}}$$2012-11-27
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    I have a feeling there is an error somewhere also, because I checked with both Maple and Mathematica and both say this integral does not converge (Maple gives no answer, but Mathematica says it does not converge). Unless both have a bug :)2012-11-27
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    @Marvis: on $[-\frac\pi2,\frac\pi2]$, $4u^2\le\pi^2\sin^2(u)$.2012-11-27
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    @NasserM.Abbasi: it looks good to me. I think Mathematica 8 can't see this.2012-11-27
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    @robjohn Thanks. It makes sense now.2012-11-27
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    @Marvis: however, your identity would shorten the proof and give a constant of $\pi$ instead of $\frac{\pi^2}{2}$.2012-11-27