Quark scattering pion $t$-channel pole I'm going through the Nambu Jona-Lasinio model as outlined in their paper  'Dynamical Model of Elementary Particles Based on an Analogy
with Superconductivity I'. I cannot do the following manipulation, which is meant to be simple. It corresponds to the $t$-channel scattering of two quarks by the exchange of a pion.
I want to show that
\begin{align}
J_p(q)&=-\frac{2ig_0}{(2\pi)^4}\int d^4p\frac{4(m^2+p^2)-q^2}{[(p+\frac{1}{2}q)^2+m^2][(p-\frac{1}{2}q)^2+m^2]}\\
&= \frac{g_0}{4\pi}\int_{4m^2}^{\Lambda^2}\frac{\kappa^2(1-4m^2/\kappa^2)^{1/2}}{q^2+\kappa^2}d\kappa^2.
\end{align}
If the limits on my $p$-integral extend all the way to infinity, then I can use the Feynman parametrisation, Wick rotate, and then do the angular and $x$ integrals. However if I impose a cut-off, then I cannot do the $x$-integral anymore, as the limits of the $p$-integral now depend on $x$. Does anyone have any suggestions?
 A: This look like a dispersion relation integral. If I'm correct (I have not done any of the algebra so this is just a guess based on first appearances) you compute the discontinuity ${\rm Im}J(q^2)$ in across the normal threshold cut that starts from $s=q^2=4m^2$ and us it in a "slot" integral. Some subtraction may be necessary. The discontinuity comes from unitarity.
A: Feynman parametrisation is NOT permitted for an integral which is more than logarithmically divergent.
More specifically, the change of integration order after the Feynman parametrisation (or proper-time regularisation) is problematic for a quadratically divergent integral in your case. 
A hint: never attempt performing the quadratically divergent integration. Instead, you should leverage the gap equation (which is also quadratically divergent) to cancel out the quadratically divergent portion in the t-channel. And then you are left with a logarithmically divergent integral which can be tackled using the usual QFT toolkits. 
And for that matter, one should steer away from any change of order of integration or differentiation for an integral which is more than logarithmically divergent. And in this sense, the Wilson-Polchinski -Wetterich functional renormalization group approach is questionable, since the differentiation with respect to the cutoff $\Lambda$ is freely moving inside/out of the functional integrals regardless of the divergence degree.
A: To fill in the details of mike stone's post, this does come from a dispersion relation.
Since the Hilbert space is spanned by a complete set of physical states, we can introduce a cut-off and write the propagator as
$$
J_p(q^2) = \int^{\Lambda^2}_0d\kappa^2\frac{\rho(\kappa^2)}{q^2+\kappa^2-i\epsilon},
$$
where $\rho$ is real. If we use the Poisson kernel represenation of the Dirac delta, we can see that
$$\rho(p^2) = \frac{1}{\pi}\text{Im}[J(p^2)].$$
We can compute the first expression in the question for $J_p(q)$ to get
$$
J_p(q) \sim \int_0^1dx[q^2+m^2x(1-x)]\ln[\frac{q^2+m^2x(1-x)+i\epsilon}{\Lambda^2}] + \text{real pieces},
$$
with $x$ a Feynman parameter. As mike said, the imaginary part comes from crossing the branch cut, so
\begin{align*}
\text{Im}[J_p(q)] &\sim \int_0^1dx[q^2+m^2x(1-x)]\theta(-[q^2+m^2x(1-x)]) \\
&\sim q^2\sqrt{1-4m^2/q^2}\theta(q^2-4m^2) +\text{subleading},
\end{align*}
giving us the result.
The IR cut-off is then not imposed by hand as I thought, so we have no issues with lower limits when integrating out momenta.
