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### arXiv:1005.1855 [hep-ph]AbstractReferencesReviews (1)Resources

#### Notes on Feynman Integrals

Zan Pan (Published 2014-09-03, edited 2016-04-30)

I will post my reading notes on the topic of Feynman integrals. It serves as supplementary materials for the introduction written by S. Weinzierl .

#### Alpha Reresentation

In this section, we will provide an direct example for Eqs. (2.25) and (2.26) in V.A. Smirnov’s book . Although it is not a strict proof, it helps to understand how Feynman integrals are related to the language of graph theory. For alpha parameters, one starts from the representation
\begin{equation} \frac{1}{(p_l^2 - m_l^2)^{a_l}} = \frac{(-i)^{a_l}}{\Gamma(a_l)} \int_0^{\infty}\mathrm{d}\alpha_l \, \alpha_l^{a_l - 1} e^{i(p_l^2-m_l^2) \alpha_l} \end{equation}
Consider the general form of scalar integral
\begin{equation} F_{\Gamma} (q_1, \ldots, q_n; d) = \int\mathrm{d}^d k_1 \cdots \int\mathrm{d}^d k_h \prod_{l=1}^L \frac{1}{(p_l^2 - m_l^2)^{a_l}} \end{equation}
where the internal momenta $p_l$ of the line $l$ is given by
\begin{equation} p_l = \sum_{i=1}^h e_{il} k_i + \sum_{i=1}^n d_{il} q_i \end{equation}
Please refer to the description below Eq. (2.5) for notation conventions. Then we have
\$p_l^2 = \sum_{i=1}^h \sum_{j=1}^h e_{il} e_{jl} k_i k_j + 2\sum_{i=1}^h \sum_{j=1}^n e_{il}d_{jl} k_i q_j + \sum_{i=1}^n \sum_{j=1}^n d_{il} d_{jl} q_i q_j \$
For the easy of writing, we introduce the matrices $A$, $B$ and the vector $Q$ as
\begin{equation} A_{ij} = \sum_{l=1}^L e_{il} e_{jl} \alpha_l, \quad Q_i = \sum_{l=1}^L \sum_{j=1}^n e_{il} d_{jl} q_j \alpha_l, \quad B_{ij} = \sum_{l=1}^L d_{il} d_{jl} \alpha_l \end{equation}
Using the formula for $4h$-dimensional Gauss integrals
\begin{equation} \begin{split} & \int\mathrm{d}^d k_1 \cdots \int\mathrm{d}^d k_h \, \exp\bigg[i\Big(\sum_{i,j}A_{ij} k_i k_j + 2 \sum_i q_i k_i \Big)\bigg] \\ & \qquad = e^{i\pi h(1-d/2)/2} \pi^{hd/2} (\det A)^{-d/2} \exp \Big({-i} \sum_{i,j}A_{ij}^{-1} q_i q_j\Big) \end{split} \end{equation}
we can proceed to combine the above relations as
\begin{equation} \begin{split} F_{\Gamma} (q_1, \ldots, q_n; d) & = \prod_{l=1}^L \frac{(-i)^{a_l}}{\Gamma(a_l)} e^{i\pi h(1-d/2)/2} \pi^{hd/2} \int_0^{\infty}\mathrm{d}\alpha_l \, \alpha_l^{a_l - 1} \\ & \qquad \times (\det A)^{-d/2} \exp\Big({-i}QA^{-1}Q^{\mathrm{T}} + i qBq^{\mathrm{T}} - i \sum_{l=1}^L m_l^2\alpha_l \Big) \end{split} \end{equation}

Let us take the two-loop Feynman diagram in Fig. 1 as an example, where $n=1$, $h=2$ and $L=5$:

According to the definition of $A_{ij}$, they are given by
\$A_{11} = \alpha_1 + \alpha_2 + \alpha_5, \quad A_{12} = A_{21} = -\alpha_5, \quad A_{22} = \alpha_3 + \alpha_4 + \alpha_5 \$
It is easy to obtain that
\$\det A = (\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4)\alpha_5 + (\alpha_1 + \alpha_2)(\alpha_3 + \alpha_4) \$
This is exactly the same with $\mathcal{U}$ in Eq. (3.45). For $Q$ and $B$, we have
\$Q_1 = \alpha_1 q, \quad Q_2 = \alpha_3 q, \quad B = \alpha_1 + \alpha_3 \$
With the expression of $\mathcal{V}$ in Eq. (3.46), we can verify that
\$QA^{-1}Q^{\mathrm{T}} - qBq^{\mathrm{T}} = -\mathcal{V} / \mathcal{U} \$

#### References

 Stefan Weinzierl, Introduction to Feynman Integrals. arXiv:1005.1855 [hep-ph].
 Vladimir A. Smirnov, Analytic Tools for Feynman Integrals. Springer Tracts in Modern Physics, 250 (2012). DOI:10.1007/978-3-642-34886-0.

#### Revision History

• Zan Pan30 Apr 2016 10:56
Remove unnecessary linebreaks (-10)
• Zan Pan14 Nov 2014 11:56
replaced embeded svg code with an external file (-3107)
• Zan Pan09 Sep 2014 13:38
updated the first section (+6537)
• Zan Pan03 Sep 2014 12:47
initial commit (+275)