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I want to prove [This comes from Appendix A.22 in Smirnov, Evaluating Feynman integrals.]

\begin{align} \int \frac{d^{d}k}{(-k^2+m^2)^{\lambda_1} (2p\cdot k)^{\lambda_2}} = \frac{i \pi^{\frac{d}{2}}}{(p^2)^{\frac{\lambda_2}{2}} (m^2)^{\lambda_1 + \frac{\lambda_2}{2} + \epsilon -2}} \frac{\Gamma(\frac{\lambda_2}{2}) \Gamma(\lambda_1 + \frac{\lambda_2}{2} + \epsilon -2)}{2 \Gamma(\lambda_1) \Gamma(\lambda_2)} \end{align}


First, to prove this, it seems we need

\begin{align} I_1 &= \frac{1}{\Gamma(\alpha)} \int_{0}^{\infty} dt \phantom{1} t^{\alpha-1} \int d^{d}k e^{-(-k^2+m^2) t} \\ &= i\frac{1}{\Gamma(\alpha)} \int_{0}^{\infty} dt \phantom{1} t^{\alpha-1} e^{-m^2 t} \int d^{d}k e^{-k^2 t} \\ & = i \frac{\pi^{\frac{d}{2}}}{\Gamma(\alpha)} \int_0^{\infty} dt \phantom{1} t^{\alpha - \frac{d}{2} -1} e^{-m^2t} \\ & = \frac{i \pi^{\frac{d}{2}}}{\Gamma(\alpha)} (m^2)^{\frac{d}{2}-\alpha} \int_0^{\infty} \phantom{1} t^{\alpha-\frac{d}{2}-1} e^{-t} \\ & = \frac{i \pi^{\frac{d}{2}}}{\Gamma(\alpha)} (m^2)^{\frac{d}{2}-\alpha}\Gamma(\alpha-\frac{d}{2}) \end{align} Using $\epsilon=\frac{4-d}{2}$, we have \begin{align} \int \frac{d^d k}{(-k^2+m^2)^{\lambda}} = i \pi^{\frac{d}{2}} \frac{\Gamma(\lambda+\epsilon-2)}{\Gamma(\lambda)} \frac{1}{(m^2)^{\lambda+\epsilon-2}} \end{align} In the process we used the integeral expressions for Gamma functions


The followings are my trials using Scwhinger trick \begin{align} &\int \frac{d^{d}k}{(-k^2+m^2)^{\lambda_1} (2p\cdot k)^{\lambda_2}} \\ &= \frac{1}{\Gamma(\lambda_1) \Gamma(\lambda_2)} \int d^{d} k \int_0^1 dx \frac{x^{\lambda_1} (1-x)^{\lambda_2}}{[x(-k^2+m^2) + (1-x) (2p\cdot k)]^{\lambda_1+\lambda_2}} \\ &= \frac{1}{\Gamma(\lambda_1) \Gamma(\lambda_2)} \int d^{d} k \int_0^1 dx \frac{x^{\lambda_1} (1-x)^{\lambda_2}}{x^{\lambda_1+\lambda_2}[-(k-\frac{1-x}{x} p)^2 + \frac{(1-x)^2}{x^2} p^2+m^2]^{\lambda_1+\lambda_2}} \\ & = \frac{i \pi^{\frac{d}{2}} \Gamma(\lambda_1 + \lambda_2 + \epsilon-2)}{\Gamma(\lambda_1) \Gamma(\lambda_2) \Gamma(\lambda_1+\lambda_2)} \int_0^1 dx \left( \frac{1-x}{x} \right)^{\lambda_2} \left( \left( \frac{1-x}{x} \right)^2 p^2+m^2\right)^{2 - \epsilon-\lambda_1 - \lambda_2} \end{align} In the process we used $x(-k^2+m^2) + (1-x) (2p\cdot k) = x ( - (k-\frac{1-x}{x}p)^2 + \frac{(1-x)^2}{x^2} p^2+m^2)$. And this seems not good.


How one can obtain the above formula?

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2 Answers 2

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Start by using Schwinger parameters. You'll arrive to $$\int d^dk \frac{1}{\beta (\lambda_1,\lambda_2)}\int_0^{\infty}ds\frac{s^{\lambda_2}-1}{(-k^2+m^2+2p\cdot ks)^{\lambda_1+\lambda_2}}$$ Completing the square, shifting away the additive constant in the square bracket and calling the rest Delta we get:$$\int d^dk \frac{1}{\beta (\lambda_1,\lambda_2)}\int_0^{\infty}ds\frac{s^{\lambda_2}-1}{(\Delta -k^2)^{\lambda_1+\lambda_2}}$$Wick rotating and computing the d-dimensional solid angle integral gives us $$\frac{i}{\beta(\lambda_1 ,\lambda_2)}\int ds (s^{\lambda_2}-1) \frac{2\pi^{d/2}}{\Gamma(\frac{d}{2})}\int_0^{\infty} dk \frac{k^{d-1}}{(k^2+\Delta)^{\lambda_1+\lambda_2}}$$ Finding the correct substitution one can find the k integral to be $$\frac{\Delta^{d/2-\lambda_1-\lambda_2}}{2}\beta(d/2, \lambda_1+\lambda_2-d/2)$$Finally evaluating the s integral and again finding the correct substitution can bring it to the form of a beta function and rewriting a bit gives you the correct result.

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Starting from Alternative Feynamn trick for $\frac{1}{AB}$, and the $I_1$ results we have \begin{align} \int \frac{d^{d}k}{(-k^2+m^2)^{\lambda_1} (2p\cdot k)^{\lambda_2}} &= \frac{\Gamma(\lambda_1 + \lambda_2)}{\Gamma(\lambda_1) \Gamma(\lambda_2)}\int_0^1 \int_0^{\infty} \frac{x^{\lambda_2-1} d^d k}{(-(k-xp)^2 + x^2 p^2 +m^2)^{\lambda_1+\lambda_2}} \\ &= \frac{i \pi^{\frac{d}{2}} \Gamma(\lambda_1 + \lambda_2 +\epsilon-2)}{\Gamma(\lambda_1) \Gamma(\lambda_2)}\int_0^{\infty} \frac{x^{\lambda_2 -1} dx}{(x^2 p^2 +m^2)^{\lambda_1 + \lambda_2 + \epsilon-2}} \end{align} Note that \begin{align} \int_0^{\infty} \frac{x^{\lambda_2 -1} dx}{(x^2 p^2 +m^2)^{\lambda_1 + \lambda_2 + \epsilon-2}} = (m^2)^{-\lambda_1 - \frac{\lambda_2}{2} - \epsilon + 2} (p^2)^{-\frac{\lambda_2}{2}} \int_0^{\infty} x^{\lambda_2-1} (1+x^2)^{-\lambda_1 - \lambda_2 - \epsilon +2} dx \end{align} Using $x^2=t$, one can reduce the integral as $\frac{1}{2} \int_0^{\infty} t^{\frac{\lambda_2}{2}-1} (1+t)^{-\lambda_1 -\lambda_2 - \epsilon +2} dt = \frac{ \Gamma(\frac{\lambda_2}{2}) \Gamma(\lambda_1+ \frac{\lambda_2}{2} + \epsilon -2)}{2 \Gamma(\lambda_1+\lambda_2 + \epsilon-2)}$ and so we obtain desired result.

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