 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2020/1/vi/solution

After a Wick rotation, the term in question has radial behavior

I_{m} = m^{2} \int \frac{d ^{4} k}{( 2 π ) ^{4}} \frac{1}{( k ^{2} + i 0 ) ( k ^{2} - m ^{2} + i 0 ) ^{2}} = - i \int \frac{d ^{4} k _{E}}{( 2 π ) ^{4}} \frac{m ^{2}}{k _{E}^{2} ( k _{E}^{2} + m ^{2} ) ^{2}} .

The radial integrand is proportional to

k_{E}

at the origin and to

k_{E}^{- 3}

I_{m} = - \frac{i m ^{2}}{8 π ^{2}} \int_{0}^{\infty} \frac{r d r}{( r ^{2} + m ^{2} ) ^{2}} = - \frac{i}{16 π ^{2}} .

This also follows from the Feynman parameter identity

\frac{1}{A B ^{2}} = 2 \int_{0}^{1} d x \frac{1 - x}{[ x A + ( 1 - x ) B ] ^{3}}

with

A = k^{2} + i 0

B = k^{2} - m^{2} + i 0

. Although the parameterized denominator becomes massless at

x = 1

1 - x

cancels the apparent endpoint singularity. Since

β \to - 2

β

preserves finiteness. No artificial photon mass is needed.

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