# Calculating exactly the divergent part of amplitudes at all loop order

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Suppose we have the $L$-loop amplitude of the form

$$\mathcal{I}_L=\int \prod_{i=1}^L \frac{d^D q_i}{(2 \pi)^D} \frac{1}{q_i^2} \frac{1}{(p-\sum_{i=1}^L q_i)^2}.$$

Introducing Feynman parameters to merge the denominators as usual, we may write the amplitude in the form

$$\mathcal{I}_L = (N-1)! \int_0^1 \prod_{j=1}^{L+1} dx_j \delta \left( \sum_{i=1}^{L+1} x_i-1\right) \int \prod_{i=1}^L \frac{d^D q_i}{(2 \pi)^D} \left[ q_i q_j M_{ij}-2 q_j K_j+J \right]^{-(L+1)}$$

where $M$ is a symmetric matrix. We can evaluate the $q$ integrals using the formula

$$\int \frac{d^D l}{(2 \pi)^D}\frac{1}{(l^2-\Delta)^N} = \frac{(-1)^Ni}{(4 \pi)^{D/2}}\frac{\Gamma(N-D/2)}{\Gamma(N)}\left( \frac{1}{ \Delta}\right)^{N-D/2},$$

It can be shown that we can write $\mathcal{I}_L$ as

$$\mathcal{I}_L=\frac{(-1)^L \Gamma(L+1-D)}{(4 \pi)^D} \int_0^1 \prod_{i=1}^{L+1} dx_i \delta \left( \sum_{i=1}^{L+1} x_i-1\right) \frac{\mathcal{U}^{L+1-3D/2}}{\mathcal{F}^{L+1-D}},$$

where we have defined  $\mathcal{U}=\det M$ and $\mathcal{F}=\det M \left( K_i M^{-1}_{ij} K_j -J \right).$

By setting $D=4-2 \epsilon$, my goal is to calculate the divergent part

$$\mathcal{I}_L = \frac{c_L}{\epsilon} + \mathcal{O}(\epsilon^0)$$

for any $L$. With that aim in mind, let's first look at $L=2$. Doing the above computations and calculating $\mathcal{U}$ and $\mathcal{F}$ explicitly, amounts to

$$c_L \propto \int dx_1 dx_2 dx_3 \delta \left( x_1+x_2+x_3-1\right) \frac{x_1 x_2 x_3}{(x_1 x_2 +x_2 x_3 + x_1 x_3)^3}.$$

The form for larger $L$ follows a similar pattern (product of $x_i$ in the numerator to some power, and sum of products of $x_i$ with one missing in each term, just like above), so evaluating the integral for $L=2$ would probably lead the way for a more general result. However, how does one treat integrals such as this? Any ideas about how one might try to evaluate it completely?

The $\delta$-function as such is pernicious evil, obscuring often what is meant. If the above is indeed a volume integral, then the result is 0 (taking the $\delta$-function as what it is usually described to be, infinity at one point and zero everywhere else, and using Lebesgue-integrals). If you take it as a distribution then things should be expressed differently, or one could employ a limiting procedure (which would point a way to a solution, as it makes clearer what is meant).
What probably is meant, however, is that there is a contribution to the integral only if $x_1+x_2+x_3-1=0$, which then suggests that the integral is not intended as a volume integral but as a surface integral. Why not substitute $x_3=1-x_1-x_2$ and integrate over $x_1$and $x_2$ only?
Thanks for the comment! Indeed, I mean the latter. Of course, substituting for $x_3$ is the first thing I tried, but it is still not very tractable. I left it as is because it has a symmetric form that is similar to a more general expression for all $L$. But yes, it is essentially a surface integral.
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