- #1
jimbobian
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Homework Statement
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.
Homework Equations
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