I am trying to evaluate a diagram of $\phi^3$ theory to practice one-loop amplitudes, but I am stuck at a certain point. The amplitude is given by the integral,

$$\mathcal{M} = \frac{(-ig)^2}{2} \int \frac{d^4k}{(2\pi)^4} \frac{i}{k^2-m^2} \frac{i}{(k-p)^2-m^2}$$

where the $1/2$ is the symmetry factor. Using Feynman parametrization I was able to write it in the form,

$$\frac{g^2}{2}\int_0^1 dx \, \int \frac{d^d \ell}{(2\pi)^d} \frac{1}{[\ell^2-\Delta]^2}$$

where $\Delta \equiv m^2 +p^2x(x-1)$, shifting momenta $\ell = k-px$. I also completed the square, and went to arbitrary $d$ dimensions. Using the formula provided in Peskin and Schroeder's appendix (the non Wick rotated version), I found,

$$=i\frac{g^2}{4}\frac{\Gamma(2-d/2)}{(4\pi)^{d/2}} \int_0^1 dx \frac{1}{[m^2+p^2x(x-1)]^{2-d/2}}$$

If I evaluate the $dx$ integral in Mathematica I get a horrible expression with hypergeometric functions. I don't recall seeing this in other loop amplitudes, it should be a relatively simple expression. How should I proceed? Have I erred somewhere?

  • 1
    $\begingroup$ By the way, this is not a tadpole diagram. This one is called bubble. A tadpole looks like a lollipop. $\endgroup$
    – Kagaratsch
    Jun 25, 2019 at 0:28

1 Answer 1


I think you have done everything correctly so far. Now while the integral would be perfectly finite in $d=4$ (which I assume is the dimension you want your result in), $\Gamma(0)=\infty$.

Therefore we expand $I=\Gamma(2-d/2)[m^2+p^2x(x-1)]^{d/2-2}$ around $d=4$ and then do the integration. I actually replaced $d\rightarrow 4-2\epsilon$ and expanded around $\epsilon = 0$.

The result is then divergent, namely

$\mathcal{I}=\int_0^1 dx\ I = \frac{1}{\epsilon} + 2 - \gamma_E - \frac{2}{z} \arctan z -\log m^2 \ ,$

where $z=\frac{p}{\sqrt{4m^2-p^2}}$ and $\gamma_E$ is the Euler-Mascheroni constant. For $p^2>4m^2$ we can use the identity $\arctan z = \frac{1}{2i}\log \frac{1+iz}{1-iz}$ to express the $\arctan$.

I hope this helps you to carry on your calculation.


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