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When we evaluate the Feynman diagram containing a loop, we commonly use the identity: \begin{align} \frac{1}{A_{1}^{m_{1}} A_{2}^{m_{2}} \cdots A_{n}^{m_{n}}}= \int_{0}^{1} d x_{1} \cdots d x_{n} \delta\left(\sum x_{i}-1\right) \frac{\prod x_{i}^{m_{i}-1}}{\left[\sum x_{i} A_{i}\right]^{\Sigma m_{i}}} \frac{\Gamma\left(m_{1}+\cdots+m_{n}\right)}{\Gamma\left(m_{1}\right) \cdots \Gamma\left(m_{n}\right)} \end{align} to combine denominators of propagators, and make shifts in the loop momentum so that the denominator has a form: \begin{align} \left[\ell^{2}-\Delta\right]^{m} \end{align} where $\ell$ is the loop momentum, and $\Delta$ doesn't depend on $\ell$.

However, I'm not sure if this procedure is always valid. Would someone know the conditions for shifts to be allowed or give me an example we cannot make the shifts?

Qmechanic
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tak
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2 Answers2

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If we use dimensional regularization, if $x$ denotes the Feynman parameters, and if the denominator $$\ell^2+ 2a(x)\cdot \ell+ b(x)=\ell^{\prime 2}-\Delta(x)$$ is a quadratic expression in the loop-momentum $\ell^{\mu}$, then clearly the shift of integration variable is $$ \ell^{\prime \mu}~=~\ell^{\mu}+a^{\mu}(x), \qquad \Delta(x)~=~a(x)^2-b(x).$$ This should in principle always work, but be aware that it may shift the integration contour, cf. e.g. this Phys.SE post.

Qmechanic
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As long as the integral is defined, we have $$\int_{-\infty}^\infty dx f(x) = \int_{-\infty}^\infty dx f(x+a) , \qquad a \in {\mathbb R}$$ In dimensional regularization, the loop integrals are all $UV$ finite (for sufficiently large $d$) so the shift in the loop momentum is ALWAYS ok to do.

Prahar
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