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Dimensional regularization is a method of isolating divergencies in scattering amplitudes.
3
votes
0
answers
109
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Why can I not asymptotically expand a Feynman integral this way?
I would like to asymptotically expand a series of Feynman diagrams in Euclidean space, and as a toy I started with the following integral, for which I know the full solution in $4d$ ($\omega \to 2$):
…
2
votes
1
answer
64
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Is it possible to expand the measure in dimensional regularization?
In the dimensional regularization scheme, four-dimensional integrals are analytically continued from their $d$-dimensional counterparts, i.e.,
$$\int d^4 x\, f(x) \longrightarrow d^d x\, f(x)\,, \tag{ …
1
vote
1
answer
251
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When can I set $d=4$ in dimensional regularization?
I am using dimensional regularization to extract the divergence of some complicated integral. I work in $d=2\omega$ dimensions, with $\omega\approx 2$. After I extract the divergence, I have an expres …
1
vote
0
answers
296
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Can I use dimensional regularization with this integral?
I would like to extract the divergence of this integral in 4d Euclidean space:
$$\int d^4z \frac{1}{(x-z)^4}\tag{1}$$
This divergence is expected to cancel with other divergences, which I got using …
1
vote
1
answer
143
views
How can I relate this integral to dimensional regularization?
In the paper "Scattering into the Fifth Dimension of $\mathcal{N}=4$ Super Yang-Mills", the authors give the following result for an integral:
$$\begin{align} I^{(1)}(x_{13}^2,x_{24}^2,m) =& \left( x …
1
vote
0
answers
38
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What is the Fourier transform of this expression in $2\omega$ dimensions?
I would like to perform the following Fourier transform in $2\omega$ (Euclidean) dimensions:
$$A(x_1,q) = \int d^{2\omega} p_1\ e^{i p_1 \cdot x_1} \frac{\delta^{(1)}(v \cdot (p_1 + q))}{p_1^2 (p_1^2 …
0
votes
1
answer
295
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How to take the Fourier transform of the propagator of a vector field?
In the paper Wilson Loops in N=4 Supersymmetric Yang--Mills Theory, the authors give the following generalized Fourier transform for a propagator in $d=2\omega$ dimensions:
$$\int \frac{d^{2\omega}p} …