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Insane integral tricks

  1. Nov 3, 2008 #1
    Ingenious integral tricks

    In QFT today I learnt some insane tricks in calculating impossible integrals....I figure it'd be a good idea to see if others have similar tricks so we can all learn from each other.

    here goes:
    to integrate (bounds are assumed to be from negative inf to inf, k^2 means the vector dot product)
    edit: missing factors of 2pi

    [tex]\int \frac{d^d k}{(2\pi)^d}\frac{1}{(k^2 + m^2)^n}[/tex]
    [tex]\frac{1}{(k^2 + m^2)^n}=\frac{1}{\Gamma(n)}\int_0^\infty t^n e^{-t(k^2 + m^2)} dt[/tex]

    and get a gaussian in the integral. At the end of the day,
    [tex]\int \frac{d^d k}{(2\pi)^d}\frac{1}{(k^2 + m^2)^n}=\frac{\Gamma(n-d/2)(m^2)^{d/2-n}}{\Gamma(n)(4\pi)^{d/2}}[/tex]


    Quite crazy eh? I would've never thought of it myself... what about your favorite crazy tricks?
    Last edited: Nov 4, 2008
  2. jcsd
  3. Nov 3, 2008 #2
    Another I have to share (I read it in some analysis book...)

    take the following integral
    [tex]\binom{m+n+1}{m,n}\int_0^1 u^{m} (1-u)^n du[/tex]
    consider the integral as a sum, we see that given m+n+1 particles on (0,1), it sums the probability of finding m particles in (0, u), 1 particle around the point u and n particles in (u, 1), so it must be one. This gives the beta integral for integers m, n!!!
    [tex]\int_0^1 u^{m} (1-u)^n du=\frac{m!n!}{(m+n+1)!}[/tex]
  4. Nov 4, 2008 #3
    Re: Ingenious integral tricks

    I wonder if most of these crazy tricks can be found without guessing. This substitution suggests that the initial problem could be simply solved by Laplace transforms?!
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