
#1
May1112, 10:31 AM

P: 52

1. The problem statement, all variables and given/known data
If I_{n} denotes [tex]\int_0^∞ \! \frac{1}{(1+x^2)^n} \, \mathrm{d} x[/tex] Prove that [itex]2nI_{n+1} = (2n1)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}=(2n1)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 



#2
May1112, 10:56 AM

P: 977

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.




#3
May1112, 11:29 AM

P: 3,015

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. 



#4
May1112, 11:31 AM

P: 3,015

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]? 



#5
May1112, 11:34 AM

P: 642





#6
May1112, 01:23 PM

P: 52

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? 



#7
May1112, 08:08 PM

P: 3,015

what is [itex]u v[/itex] in the integration by parts?



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