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Re: Integrating Feynman integrals in mathematica

In article <ciu5iv$81d$1 at>, pabird at (Xman) 

> I am trying to find an explicit form of the following 4-dimensional
> fourier transforms. Can anyone help? ( x and k are 4 dimensional
> vectors) They are from physics.
> 1)
> f(x) =Intregral[    e^(i x.k) / (k.k -m^2)  ]dk^4 

Shouldn't this be k.k + m^2 (to agree with your equation for f below)?
> 2)
> g(x)=Intregral[    e^(i x.k) / (k.k -m^2)^2  ]dk^4 
> I know that the first is of the form:
> f(x) = 1/|x.x| + log|x.x| * P((m^2/4) |x.x|) + Q((m^2/4) |x.x|)
> (when m=0 this becomes 1/|x.x|)
> Where P and Q stand for infinite polynomial series and that I think
> P(y) = Sum( y^n /(n!(n+1)!) ,y=0..infinity )

Apart from the error in syntax and summation index, Mathematica can 
compute this sum in closed form. Try

 Sum[y^n/(n! (n+1)!),{n,0,Infinity}]

> and that in the second one
> g(x)  =  log|x.x| * R((m^2/4) |x.x|) + S((m^2/4) |x.x|)
> (when m=0 this becomes log|x.x|)
> where R(y) = Sum( y^n /(n!n!) ,y=0..infinity )

and this one too.
> But the functions Q and S are more difficult to find.
> Plus does anyone know if the series P and R (=P') or Q and S can be
> written in terms of simple functions?
> It may help to know that f and g satisfy the following 4 dimensional
> wave equations:
> ( d/dx . d/dx - m^2) f(x) = delta(x)          (=0 for x=/=0)
> ( d/dx . d/dx - m^2)^2 g(x) = delta(x)          (=0 for x=/=0)
> I am particularly interested in g(x).

As pointed out by Steve Luttrell, if you can compute f(x), you know g(x) 
by parametric differentiation. Let me indicate how you can get 
Mathematica to compute the integral over 4 dimensional space of the 

  h[x_][k_] := Exp[I x.k]/(k.k + m^2)

(related to yours) in closed form. As you have observed, the integral 
only depends on the length of the 4 dimensional vector x. Call this r. 
Let p denote the magnitude of the vector k. If you express k in 
hyperspherical co-ordinates, say

 k = p {Sin[a]Sin[b]Sin[c],Sin[a]Sin[b]Cos[c],Sin[a]Cos[b],Cos[a]} 

where p is in [0,Infinity), c is in [0,2Pi], and a and b are in [0,Pi], 
then the volume element reads

 dk^4 = p^3 dp Sin[a]^2 da Sin[b] db dc

Now, because the angular part of the k integral is over all possible 
orientations of the vector k, you can choose an _arbitrary_ orientation 
for x (this is why the integral only depends on r). The choice resulting 
in the simplest integrand is

 x = {0,0,0,r}

so that the integrand becomes

 Exp[I p r Cos[a]]/(p^2 + m^2)

where a is the angle between x and k. The integrals over b and c are now 

  Integrate[Sin[b], {b,0,Pi},{c,0,2Pi}]

leading to an overall multiplicative factor of 4 Pi.

You are left with the following computation: first compute the integral 
over a,

   Assuming[r > 0 && m > 0 && p > 0, 
      4 Pi Integrate[Exp[I p r Cos[a]]/(p^2 + m^2) Sin[a]^2, {a, 0, Pi}]]

followed by the integral over p

   Assuming[r > 0 && m > 0, Integrate[% p^3, {p, 0, Infinity}]]

which evaluates to the very simple closed-form result

  4 m Pi^2 BesselK[1, m r]/r

When m = 0, this reduces to 4 Pi^2 / r^2 as you can see from the 
following series expansion:

  4 m Pi^2 BesselK[1, m r]/r + O[m]

Expanding the result as a series in r, one obtains

  4 m Pi^2 (1/r^2 + (m BesselI[1, m r]/r) Log[r] + ...)

exactly of the form you write above (check this by series expansion of 
the BesselI[1, m r] function). 

The integral with (p^2 + m^2)^2 in the denominator is even simpler. 
Using parametric differentiation, you obtain

  2 Pi^2 BesselK[0, m r]


Paul Abbott                                   Phone: +61 8 6488 2734
School of Physics, M013                         Fax: +61 8 6488 1014
The University of Western Australia      (CRICOS Provider No 00126G)         
35 Stirling Highway
Crawley WA 6009                      mailto:paul at 

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