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I am trying to integrate $$\int_1^\infty x(1+x^2)^p dx$$

I get down to $$\frac1{2(p+1)}\left(\lim \limits_{t \to \infty}(1+t^2)^{p+1}\right)-\frac1{2(p+1)}\lim \limits_{t \to \infty}2^{p+1}$$

The solution says that the next step is $$\frac1{2(p+1)}(0)-\frac{2^{p+1}}{2(p+1)}$$

That requires $$\lim \limits_{t \to \infty}(1+t^2)^{p+1} = 0$$ but how can this be true?

$p$ is a constant, but it can be negative or positive.

Blue
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Khris Davis
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1 Answers1

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The integral converges iff $p <-1$ so, when $p <-1$ what is claimed in the solution is correct. To see that the integral converges iff $p<-1$ compare it with the following improper integral $$\int_1^{\infty} (x) (x^{2p})dx.$$

Mikasa
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Kavi Rama Murthy
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  • I can see that by plugging in values less than -1 into the integral, but is there a way to see that intuitively? – Khris Davis Mar 17 '20 at 06:14
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    @KhrisDavis: What did you mean by saying "intuitively"? – Mikasa Mar 17 '20 at 06:48
  • Why is it that the integral converges iff p < -1? Could it be that if p < -37483 (an arbitrary number), the integral converges? Is the -1 an arbitrary number? If not, how do we know that it is -1. – Khris Davis Mar 19 '20 at 05:18
  • @KhrisDavis Write down the value of $\int_1^{M} x^{2p+1} dx$ explicitly using an anti-derivative. Then see when this integral has a finite limit as $M \to \infty$. – Kavi Rama Murthy Mar 19 '20 at 05:23