# Feynman Parameters¶

When integrating a denominator like , the idea is to introduce auxiliary parameters in order to make the denominator simpler. We start with the identity:

(1)

which can be proven easily:

>>> var("A B")
(A, B)
>>> integrate(1/(x*A + (1-x)*B)**2, (x, 0, 1))
1/(A*B - A**2) - 1/(-A*B + B**2)
>>> simplify(_)
1/(A*B)


By repeatedly differentiating with respect to B:

(2)

Then we prove:

(3)

For we get (1) and if it holds for it also holds for , because we multiply (3) by and get:

Where we used (2) and the fact, that , after the substituation we also restricted the limits of integration from 1 to , since , , ... are all positive.

## Example 1¶

where

In the last part we used . We now shift by introducing:

and we get:

thus:

## Example 2¶

where

In the last part we used . We now shift by introducing:

and we get:

where

thus:

This integral has an infrared divergence. We can cure this by pretending that the photon has a small nonzero mass , then in the denominator of the photon propagator we need to change:

This denominator is multiplied by later on, so at the end we need to do the change:

and we get:

for we get:

We can use the following integral:

that is equal to in the limit .

here are a few special cases for , and :

Code:

>>> from sympy import log, atan, var, sqrt, Eq, Integral, S
>>> var("z m mu")
>>> F = -log(z*(1 - 2/m) + 1/m + z**2/m)/2 + \
atan((1 - 2/m + 2*z/m)/sqrt(-1 + 4/m))/sqrt(-1 + 4/m)
>>> f = F.diff(z).simplify()
>>> print f
(1 - z)/(1 - 2*z + m*z + z**2)
>>> integ_f_0_1 = F.subs(z, 1) - F.subs(z, 0)
>>> e = Eq(Integral(f.subs(m, mu**2), (z, 0, 1)), integ_f_0_1.subs(m, mu**2))
>>> print e
Integral((1 - z)/(1 - 2*z + z**2 + z*mu**2), (z, 0, 1)) == log(mu**(-2))/2 + atan((-1 + 4/mu**2)**(-1/2))/(-1 + 4/mu**2)**(1/2) - atan((1 - 2/mu**2)/(-1 + 4/mu**2)**(1/2))/(-1 + 4/mu**2)**(1/2)
>>> print e.subs(mu, 1)
Integral((1 - z)/(1 - z + z**2), (z, 0, 1)) == pi*3**(1/2)/9
>>> print e.subs(mu, S(1)/2)
Integral((1 - z)/(1 - 7*z/4 + z**2), (z, 0, 1)) == log(4)/2 + 15**(1/2)*atan(15**(1/2)/15)/15 + 15**(1/2)*atan(7*15**(1/2)/15)/15
>>> print e.subs(mu, S(1)/3)
Integral((1 - z)/(1 - 17*z/9 + z**2), (z, 0, 1)) == log(9)/2 + 35**(1/2)*atan(35**(1/2)/35)/35 + 35**(1/2)*atan(17*35**(1/2)/35)/35


Then for and small we get: