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Recurrence relation for an integral equation

  1. Aug 1, 2011 #1
    Hi guys, would appreciate any help with this problem. It's from a graduate school entrance examination which I am practicing.

    problem statement

    When n is a natural number, the indefinite integral [itex]I_n[/itex] is defined as

    [itex]I_n=\int\frac{1}{(x^2+a^2)^n}dx[/itex]

    Here, a is a constant real number, and not equal to zero. Answer the following questions.

    1. Express [itex]I_{n+1}[/itex] as a recurrence equation with [itex]I_n[/itex].
    2. Derive [itex]I_1[/itex] and [itex]I_2[/itex]

    where I'm at
    I can solve part 2 relatively easily by using the trigonometric substitution [itex]x=a \tan\theta[/itex] which gives the result (feel free to ask for working if you don't believe me):

    [itex]I_n=\frac{1}{a^{2n-1}}\int(\cos\theta)^{2n-2}d\theta[/itex]

    or:

    [itex]I_{n+1}=\frac{1}{a^{2n+1}}\int(\cos\theta)^{2n} d\theta[/itex]

    these are fairly easy to solve for fixed values of n using basic integration and backwards substitutions such as [itex]\tan\theta=(x/a)[/itex] etc, giving:

    [itex]I_1=\arctan(x/a)/a[/itex]

    [itex]I_2=\arctan(x/a)/(2a^3)+\frac{x}{2a^2(x^2+a^2)} [/itex]

    but I don't know how to generalize to the final recurrence equation.

    attempted solution
    I tried to solve the original equation using integration by parts by writing:

    [itex]I_{n+1}=\int\frac{1}{(x^2+a^2)^n}\frac{1}{(x^2+a^2)}dx[/itex]

    then setting:
    [itex]u=\frac{1}{(x^2+a^2)}[/itex], [itex]v'=\frac{1}{(x^2+a^2)^n}[/itex]


    [itex]u'=\frac{-2x}{(x^2+a^2)^2}[/itex], [itex]v=I_n[/itex]

    this gives:
    [itex]I_{n+1}=\frac{I_n}{(x^2+a^2)}+\int\frac{I_n 2x}{(x^2+a^2)^2}dx[/itex]

    but this doesn't agree with the calculations from above with fixed n, so it must be wrong. (I presume [itex]v=I_n[/itex] is invalid for some reason?).

    Unfortunately I don't really have any ideas on where to go from here...?
     
  2. jcsd
  3. Aug 1, 2011 #2

    SammyS

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    Try the following: (I haven't checked it out myself, so it may be bogus.)

    [itex]\displaystyle I_n=\int\frac{1}{(x^2+a^2)^n}dx[/itex]
    [itex]\displaystyle =\int\frac{x^2+a^2}{(x^2+a^2)^{(n+1)}}dx[/itex]

    [itex]\displaystyle =a^2I_{n+1}+\int (x)\,\frac{x}{(x^2+a^2)^{(n+1)}}dx[/itex]

    Now try integration by parts: u = x, [itex]\displaystyle dv=\frac{x}{(x^2+a^2)^{(n+1)}}dx[/itex]​
     
    Last edited: Aug 1, 2011
  4. Aug 1, 2011 #3
    That is very clever, it worked great thank you!
     
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