# $\phi^4$ loop digram with six legs going out

I want to set up an integral for the following diagram in $$\phi^4$$ theory

my attempt is $$\begin{split} & 15(-i\lambda)^3\iint \frac{d^Dl}{(2\pi)^D}\frac{d^Dq}{(2\pi)^D}\frac{i}{l^2-m^2+i\epsilon}\frac{i}{q^2-m^2+i\epsilon}\frac{i}{(l+q)^2-m^2+i\epsilon} \end{split}$$

where there is zero external momentum.

Is the equation I wrote correct?

• What is your question? – mike stone Jun 16 at 12:47
• Is my approach to writing the integral representation of the loop correct? Are there other ways to write it? – sultana Jun 16 at 12:52
• @sultana Your external legs do not carry momentum? – Prahar Jun 16 at 12:53
• @PraharMitra yes, there is zero external momentum. – sultana Jun 16 at 12:55
• If there is no external momenta, then all the loop propagators have momentum $\ell$ so the denominator should be $\frac{i^3 }{ [ \ell^2 - m^2 + i \epsilon ]^3 }$ – Prahar Jun 16 at 13:20