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July 29th, 2014, 07:10 AM   #1
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Integral

Hi Everybody!

I am trying to prove analytically that the integral
$\int_0^\infty \frac{r^2}{r^3+0.8} \frac{0.2\sin(11r^3+1.8 )}{11r^3+1.8}dr $
will be nonnegative. Can somebody give me an idea how to do it please. Is it even possible?
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July 29th, 2014, 07:38 AM   #2
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Try writing the integral as an infinite sum of integrals such that the integrand is alternately positive and negative for the whole of the range of the integral. I think that you should be able to prove that each negative integral has a smaller magnitude that the preceding positive integral. This would give your solution.
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July 29th, 2014, 08:27 AM   #3
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Quote:
Originally Posted by v8archie View Post
Try writing the integral as an infinite sum of integrals such that the integrand is alternately positive and negative for the whole of the range of the integral. I think that you should be able to prove that each negative integral has a smaller magnitude that the preceding positive integral. This would give your solution.
I were thinking about that, but what to do with the first two areas?
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July 29th, 2014, 06:30 PM   #4
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They are very, very close in area, but I think the first is a fraction larger. You might want to prove it though!

The first area.
The second area.
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July 31st, 2014, 07:53 AM   #5
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Its almost obvious that magnitude will decrease and that the distance between zeroes will become smaller and smaller. Thus, assuming that shape will remain the same, it follows that
$\left|\int_{r_i}^{r_{i+1}}Integrand \ \ \ dr \right|\ge \left| \int_{r_{i+1}}^{r_{i+2}}Integrand \ \ \ dr\right|$,
where $r_i$ are zeroes of the integrand. But what to do with the first two areas I have no idea.......I mean how to show analitically that the first area will be greater or equal to the second one?
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July 31st, 2014, 08:03 AM   #6
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For the first four regions ($r \le 1$), you might have to settle for a numerical solution. But for $r \gt 1$, you ought to be able to get an analytic solution.
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July 31st, 2014, 08:05 AM   #7
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Thank you.
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July 31st, 2014, 09:40 AM   #8
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Warning: that root isn't at 1, but $r^3 + 1.8 = 4\pi \implies r \gt 1$.
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