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How to obtain the rational function integration recursive formula

  1. Jan 21, 2014 #1
    I've been dealing with several integrals involving rational functions. I have encountered myself arriving to an integral that requires the application of the following recursive formula:

    [itex] \int\frac{1}{(u^2+α^ 2)^m} \, du= \frac{u}{2 α^ 2 (m-1)(u^2+α^ 2)^{m-1}}+\frac{2m-3}{2 α^2 (m-1)}\int\frac{1}{(u^2+α^ 2)^{m-1}} \, du [/itex]

    as stated in Apostol's Calculus.

    However, I am curious about the demonstration of this formula. Apostol states in his book that it is obtained by integrating by parts, but I don't see how... does anybody have any ideas?


    Calculus, Volume 1, One-variable calculus, with an introduction to linear algebra, (1967) Wiley, ISBN 0-536-00005-0, ISBN 978-0-471-00005-1
  2. jcsd
  3. Jan 21, 2014 #2


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    hi mr_sparxx! :smile:

    put Im = ##\int\frac{1}{(u^2+α^2)^m} \, du##

    then Im-1 = ##\int\frac{u^2+α^2}{(u^2+α^2)^m} \, du##

    = ##\int\frac{u^2}{(u^2+α^2)^m} \, du## + α2Im

    = (something)##\left[\frac{u}{(u^2+α^2)^{m-1}}\right]## + (something)Im-1 + α2Im

    carry on from there :wink:
  4. Jan 21, 2014 #3
    Awesome! :)
    Thanks tiny-Tim (specially for leaving some of the fun for me) ;P
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