# Showing that this identity involving the Gamma function is true

Homework Statement:
Show that
$$-\frac{\lambda^4}{M^4}U_S \int\frac{d^4k}{(2\pi)^4} \frac{1}{k^2-\sigma^2} =-4 \frac{i \lambda^{4}}{(4 \pi)^{2}} \frac{\sigma^{2}}{M^{4}} U_{S}\left[3 \frac{1}{\bar{\epsilon}}+3 \log \left(\frac{\mu^{2}}{\sigma^{2}}\right)+1\right]+\ldots,$$
where ##\frac{1}{\bar\epsilon} = \frac{1}{\epsilon}-\gamma_{EM}+\ln(4\pi)##, ##U_S## is just a scalar number, ##D=4-2\epsilon## and ##\mu## is the renormalization scale.
Relevant Equations:
$$\Gamma(1-\frac{D}{2}) = \Gamma(-1+\epsilon) = -\frac{1}{\epsilon}-1+\gamma_{EM}+\mathcal{O}(\epsilon).$$
My attempt at this:
From the general result
$$\int \frac{d^Dl}{(2\pi)^D} \frac{1}{(l^2+m^2)^n} = \frac{im^{D-2n}}{(4\pi)^{D/2}} \frac{\Gamma(n-D/2)}{\Gamma(n)},$$
we get by setting ##D=4##, ##n=1##, ##m^2=-\sigma^2##
$$-\frac{\lambda^4}{M^4}U_S \int\frac{d^4k}{(2\pi)^4} \frac{1}{k^2-\sigma^2} = \frac{i\lambda^4}{M^4}U_S\frac{1}{(4\pi)^{D/2}} \frac{\Gamma(1-D/2)}{\Gamma(1)} (\sigma^2)^{D/2-1}.$$
We further have
$$\frac{1}{(4\pi)^{D/2}}(\sigma^{2})^{D/2-1}=\frac{1}{(4\pi)^{2-\epsilon}}(\sigma^{2})^{1-\epsilon}= \frac{\sigma^2}{(4\pi)^{2}}\left(\frac{4\pi}{\sigma^{2}}\right)^{\epsilon} = \frac{\sigma^2}{(4\pi)^{2}}e^{\epsilon \ln(4\pi/\sigma^2)}\approx \frac{\sigma^2}{(4\pi)^{2}}\left(1+\epsilon \ln(4\pi/\sigma^2)\right)$$

We are left with
$$\frac{\Gamma(1-D/2)}{\Gamma(1)} = {\Gamma(1-D/2)} = -\frac{1}{\epsilon}-1+\gamma_{EM}+\mathcal{O}(\epsilon).$$

Put everything together:
\begin{align*} \frac{i\lambda^4}{M^4}U_S\frac{1}{(4\pi)^{D/2}} \frac{\Gamma(1-D/2)}{\Gamma(1)} (\sigma^2)^{D/2-1} &=\frac{i\lambda^4}{M^4}U_S \frac{\sigma^2}{(4\pi)^{2}}\left(1+\epsilon \ln(4\pi/\sigma^2)\right) \left(-\frac{1}{\epsilon}-1+\gamma_{EM}+\mathcal{O}(\epsilon)\right)\\ &= \frac{i\lambda^4}{(4\pi)^{2}}U_S \frac{\sigma^2}{M^4}\left(1+\epsilon \ln(4\pi/\sigma^2)\right) \left(-\frac{1}{\epsilon}-1+\gamma_{EM}+\mathcal{O}(\epsilon)\right)\\ &= -\frac{i\lambda^4}{(4\pi)^{2}}U_S \frac{\sigma^2}{M^4}\left(-\gamma_{EM} +\log \left(\frac{4 \pi}{\sigma ^2}\right)+\frac{1}{\epsilon }+1\right) + \mathcal{O}(\epsilon)\\ &= -\frac{i\lambda^4}{(4\pi)^{2}}U_S \frac{\sigma^2}{M^4}\left(\frac{1}{\epsilon }-\gamma_{EM} +\log ({4 \pi}) -\log({\sigma ^2})+1\right) + \mathcal{O}(\epsilon)\\ &= -\frac{i\lambda^4}{(4\pi)^{2}}U_S \frac{\sigma^2}{M^4}\left[\frac{1}{\bar\epsilon }+\log\left(\frac{1}{\sigma ^2}\right)+1\right] + \mathcal{O}(\epsilon)\\ &\neq -4 \frac{i \lambda^{4}}{(4 \pi)^{2}}U_{S} \frac{\sigma^{2}}{M^{4}} \left[3 \frac{1}{\bar{\epsilon}}+3 \log \left(\frac{\mu^{2}}{\sigma^{2}}\right)+1\right]+\ldots. \end{align*}

Conclusion:

The general form seems to be right, but the details are wrong... I don't see where the prefactors of ##4## and ##3## are supposed to come from, nor do I see where the renormalization scale ##\mu## comes into play...

## Answers and Replies

anuttarasammyak
Gold Member
I observe if renormalization factor does not exist in numerator
$$\log( 1/\sigma^2)$$
is ambiguous in its physical dimension. I assume dimension of ##\mu## and ##\sigma## is same. How is ##\mu## introduced in your text ? What is its definition ?