Srednicki - computing divergent piece of loop integral I was reading through Srednicki and didn't quite understand one of the paragraphs in Section $51$ on loop corrections in the Yukawa theory on P.$322$. It's the fermion loop correction to the local four point vertex interaction - the numerator is a horrendous trace but he says that to compute the divergent contribution one can send all of the external momenta to zero, $k_i \rightarrow 0$. Why is this so? Is it just because the divergency occurs in the far UV so at this energy scale, the internal route momentum greatly exceeds any other scale in the problem? 
Upon doing this, he says the only contribution is then $4(\ell^2)^2$ to the trace. Since $$\text{Tr}(( \not l + m)( \not l + m)( \not l + m)( \not l + m)) = 4(l^4 + 6m^2l^2 + m^4)$$ ignoring terms odd in $l$ which vanish under symmetric integration. So I get the $4l^2$ term he mentions but I have additional terms. Is he again just neglecting them because to get the divergent piece we can neglect all other scales? Assuming this is true, I get the integral $$4g^4 \int \frac{d^D l}{(2\pi)^d} \frac{l^4}{(l^2-m^2)^4}$$ Since $[g] = \mu^{\epsilon}$, I can rewrite the coupling and then use $$\mu^{2 \epsilon} \int \frac{d^D l}{(2\pi)^d} \frac{(l^2)^2}{(l^2-m^2)^4} = \frac{1}{(4\pi)^{D/2}} iD(D+2) \frac{\Gamma(4-D/2-2)}{\Gamma(4)} \left(\frac{1}{m^2}\right)^{4-D/2-2}$$ And in the text there are 6 possible permutations of the momenta so is it a case of just multiplying this result by 6?
Thanks!
 A: The first question asks why in the computation of the divergent part of MS 51.49, we can set $k_i=0$. 
To see this, we do an intuitive counting of the power of $l$ in the integrals. Notice each fermion propagator effectively contributes $\frac{1}{l^4}$ and the integral measure $d^4l\sim l^3dl$, so it is $\frac{1}{l}$ divergent at large $l$. Thus if one subtracts the leading $l$ piece from the numerator, one gets at least $\frac{1}{l^2}$ UV behaviour for large $l$ and thus is safe from UV divergence, i.e. (to illustrate the point, I omit the $\gamma_5$'s)
$$\begin{align}
&\small\int \frac{d^4l}{(2\pi)^4}\frac{Tr( \not l(\not l-k_1)(\not l+\not k_2+\not k_3)(\not l+\not k_2) )}{(l^2+m^2)((l-k_1)^2+m^2)((l+k_2+k_3)^2+m^2)((l+k_2)^2+m^2)}-\int\frac{d^4l}{(2\pi)^4}\frac{Tr( \not l\not l\not l\not l )}{(l^2+m^2)^4}
\\=&\int \frac{d^4l}{(2\pi)^4}\frac{N'}{(l^2+m^4)^4((l-k_1)^2+m^2)((l+k_2+k_3)^2+m^2)((l+k_2)^2+m^2)}
\end{align}$$
where $$\begin{align}
N'=&Tr( \not l(\not l-k_1)(\not l+\not k_2+\not k_3)(\not l+\not k_2) )(l^2+m^2)^3-\\&Tr( \not l\not l\not l\not l )((l-k_1)^2+m^2)((l+k_2+k_3)^2+m^2)((l+k_2)^2+m^2)=O(l^9).
\end{align}$$
Thus for large $l$ the above behaves as $\sim \int \frac{d^4l}{(2\pi)^4}\frac{l^9}{l^{14}}\sim\int dl \frac{l^{12}}{l^{14}}\sim\int dl\frac{1}{l^2}$ which is UV finite. 
Therefore, to compute the UV divergent piece, one only needs to compute the leading term in the numerator, i.e. 
$$\int\frac{d^4l}{(2\pi)^4}\frac{Tr( \not l\not l\not l\not l )}{(l^2+m^2)^4}$$, which contains the UV divergence. 
 Setting $k_i=0$, and picking the leading term in $l$ effectively gives one the leading $l$ term in the numerator. This also clarifies your second question. 
For the third question, yes, just multiply the result by 6, as they give identical contributions.
There is one comment that I would like to make. The above explanation should really be understood with proper UV cut off- it does not make sense to do algebraic manipulations on divergent integrals! To give a fully rigorous argument on why it is enough to just consider the leading $l$ terms in the numerator, one needs to use some UV cut off, and subtract the integral with only the leading term from the full expression and then show it is bounded as the cut off goes to infinity. 
