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Validity of Reduction Formula

by jimbobian
Tags: formula, reduction, validity
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jimbobian
#1
May11-12, 10:31 AM
P: 52
1. The problem statement, all variables and given/known data

If In denotes [tex]\int_0^∞ \! \frac{1}{(1+x^2)^n} \, \mathrm{d} x[/tex]
Prove that [itex]2nI_{n+1} = (2n-1)I_n[/itex], and state the values of n for which this reduction formula is valid.

2. Relevant equations



3. The attempt at a solution

[tex]I_n=\int_0^∞ \! \frac{1}{(1+x^2)^n} \, \mathrm{d} x[/tex]
[tex]=\int_0^∞ \! (1+x^2)^{-n} \, \mathrm{d} x[/tex]
By parts:
[tex]=\left[ x(1+x^2)^{-n} \right]_0^∞ + 2n\int_0^∞ \! \frac{x^2}{(1+x^2)^{n+1}} \, \mathrm{d} x[/tex]
[tex]=0 + 2n\int_0^∞ \! \frac{(1+x^2)-1}{(1+x^2)^{n+1}} \, \mathrm{d} x[/tex]
[tex]=2n\int_0^∞ \! \frac{(1+x^2)}{(1+x^2)^{n+1}} \, \mathrm{d} x - 2n\int_0^∞ \! \frac{1}{(1+x^2)^{n+1}} \, \mathrm{d} x[/tex]
[tex]=2nI_{n}-2nI_{n+1}[/tex]
[tex]2nI_{n+1}=(2n-1)I_{n}[/tex]
as required.

It's the next bit where I'm stuck - the range of values for which n is valid. Obviously when part of the integral has been evaluated (following parts), this requires that n>0 otherwise the expression doesn't converge. I can't see anywhere else in the method where there is a restriction for n to be a specific value so I went with n>0 as my answer, but my book says n>1/2, can anyway shed some light on this for me.

Thanks
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andrien
#2
May11-12, 10:56 AM
P: 1,020
the expression inside the integral is always positive and so does the integral.at n=1/2 the integral does not converge and observe that at n=1/2 ,I(n+1) is zero and for 0<n<1/2, I(n+1) is negative.
Dickfore
#3
May11-12, 11:29 AM
P: 3,014
you should use:
[tex]
\frac{1}{(1 + x^2)^n} = \frac{1 + x^2 - x^2}{(1 + x^2)^n} = \frac{1}{(1 + x^2)^{n - 1}} - \frac{x^2}{(1 + x^2)^n}
[/tex]
For the integral of the second term, use integration by parts:
[tex]
-\int_{0}^{\infty}{x \, \frac{x}{(1 + x^2)^n} \, dx}
[/tex]
[tex]
u = x \Rightarrow du = dx
[/tex]
[tex]
dv = \frac{x}{(1 + x^2)^n} \, dx \Rightarrow v = \int{ \frac{x}{(1 + x^2)^n} \, dx} \stackrel{t = 1 + x^2}{=} \frac{1}{2} \, \int{t^{-n} \, dt} = \frac{t^{1- n}}{2(1 - n)} = -\frac{1}{2 (n - 1) (1 + x^2)^{n - 1}}
[/tex]
Combine everything, identify the relevant integrals with [itex]I_n[/itex], and [itex]I_{n - 1}[/itex], and see what you get.

Dickfore
#4
May11-12, 11:31 AM
P: 3,014
Validity of Reduction Formula

Oh, I see you already did the steps. As for the range of validity, answer these questions:

1) What is the value of the integral [itex]\int{t^{-n} \, dt}[/itex] for [itex]n = 1[/itex]?

2) When does the integrated out part [itex]\frac{x}{2(n - 1)(1 + x^2)^{n - 1}}[/itex] converge when [itex]x \rightarrow \infty[/itex]?
Whovian
#5
May11-12, 11:34 AM
P: 643
Quote Quote by Dickfore View Post
Oh, I see you already did the steps. As for the range of validity, answer these questions:

1) What is the value of the integral [itex]\int{t^{-n} \, dt}[/itex] for [itex]n = 1[/itex]?

2) When does the integrated out part [itex]\frac{x}{2(n - 1)(1 + x^2)^{n - 1}[/itex] converge when [itex]x \rightarrow \infty[/itex]?
Please preview your posts, LaTeX errors happen all the time.
jimbobian
#6
May11-12, 01:23 PM
P: 52
Quote Quote by Dickfore View Post
Oh, I see you already did the steps. As for the range of validity, answer these questions:

1) What is the value of the integral [itex]\int{t^{-n} \, dt}[/itex] for [itex]n = 1[/itex]?

2) When does the integrated out part [itex]\frac{x}{2(n - 1)(1 + x^2)^{n - 1}}[/itex] converge when [itex]x \rightarrow \infty[/itex]?
Thanks for your response.

1) ln(t), so this would suggest it doesn't converge for n=1?

2) Firstly I'm not sure where you've got this fraction from, I can't find it in any working of yours or mine? It would converge for n>1 for sure. For n=1 it won't converge, but for n<1 I have no idea?
Dickfore
#7
May11-12, 08:08 PM
P: 3,014
what is [itex]u v[/itex] in the integration by parts?


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