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Value of a definite integral

  1. Feb 28, 2009 #1

    I am looking for a value of an integral
    [tex]\int^{\infty}_0 {r^{3-\epsilon} \over (r^2+N^2)^2}dr[/tex]

    I have tried looking up a book by Gradshteyn and Ryzhik, however, its structure is quite complicated. Should I rewrite the integrand in some other non-obvious way to find it? Would you recommend using some other resource?

    The answer is known (it involves Gamma functions), as the integral is a part of a paper about the "ABC theory" (toy QFT) by Kraus and Griffiths. However, I would like to 1) discover the optimal way to check complicated definite integrals in future and 2) check the value of this particular integral.

    Thank you.
  2. jcsd
  3. Feb 28, 2009 #2
  4. Feb 28, 2009 #3
    I have tried it, this integrator does not do the definite integration. Input of the integrand above gives some "hypergeometric" functions as an output. I still hope that something better exists, but thank you anyway.

    Maple or Mathematica would be a possibility, but I usually do not have them at hand.
  5. Mar 8, 2009 #4
    I have failed to find the value in the integral tables, but I post the solution here, in case someone needs it.

    \int_0^\infty {r^{3-\epsilon}dr\over \left[r^2+\Lambda^2\right]^2}
    = \int_0^\infty {rdr(r^2)^{1-\epsilon /2}dr\over \left[r^2+\Lambda^2\right]^2}
    = \left[ \substack{r^2+\Lambda^2=\Lambda^2 / y \\ 2rdr = -\Lambda^2 dy / y^2} \right][/tex]
    [tex]= \int_1^0 \left( -{\Lambda^2dy \over 2y^2 } \right)
    \left[ \Lambda^2 \left( {1-y\over y} \right) \right]^{1-\epsilon /2}
    \left( y \over \Lambda^2 \right)^2
    ={1 \over 2\Lambda^{\epsilon}} \int_0^1 \left( {1-y \over y} \right)^{1-\epsilon /2}dy.[/tex]

    Then we have to know what Euler Beta and Gamma functions are.

    [tex]B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}dt={\Gamma (a)\Gamma (b)\over \Gamma (a+b)}.[/tex]

    And finally

    [tex]R = \dfrac{1}{2\Lambda^{\epsilon}} B({\epsilon \over 2}, 2-{\epsilon \over 2})
    = {\Gamma(\epsilon /2)\Gamma(2-\epsilon /2) \over 2\Lambda^{\epsilon}\Gamma(2)}
    = {\Gamma(\epsilon /2)\Gamma(2-\epsilon /2) \over 2\Lambda^{\epsilon}}.[/tex]
  6. Mar 8, 2009 #5
    Using Mathematica I get
    [tex] R = \frac{\pi(2-\epsilon)}{4\Lambda^{\epsilon}\sin(\epsilon\pi/2)}[/tex]
    which appears to be equivalent.
  7. Mar 8, 2009 #6
    Yes by [tex]\frac{\pi}{\sin(\pi x)}=\Gamma(x)\Gamma(1-x)[/tex] one gets that result.
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