# Dimensional Regularization Integral Formula

In the formula $$\int \frac {d^{4-2\epsilon} l} {(2\pi)^{4-2\epsilon}} \frac 1 {(l^2-\Delta)^2} = \frac i {(4\pi)^{2-\epsilon}} \Gamma(\epsilon) \left(\frac 1 \Delta\right)^\epsilon,$$ how should I deal with the case when $\Delta<0$?

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In complex calculus, one may compute powers – and other functions – for all negative or complex values of the argument. But it would lead to confusions and ambiguities and indeed, it's not needed.

Whenever the theory is stable, it's guaranteed that $\Delta$ in these d.r. integrals ends up positive. A typical value of $\Delta$ for Feynman-parameterized integrals is an example from a 4-point diagram (imagine 2 to 2 scattering) $$\Delta = m^2 - x(1-x)q^2$$ Now, $m^2$ is positive because we don't have tachyons and $x(1-x)$ is between $0$ and $1/4$. The only risk how the second term could beat $m^2$ and make $\Delta$ negative is that $q$ is hugely timelike, with length over $2m$. But it really can't happen because $q$ is the sum of two external momenta. Timelike $2m$ length is really the maximum and in most cases, $q$ will end up shorter or spacelike. The UV region we're integrating corresponds to very long spacelike $q$ (all directions in the Euclideanized spacetime are spacelike) for which $\Delta$ is high and positive because the spacelike vectors have negative signature above.

I can't provide you with a general proof (it may be easy, however) but I am pretty sure that the negative-$\Delta$ problem doesn't arise in any loop diagrams.

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I have an example where $\Delta$ is negative. Consider the one-loop EM vertex correction for fermion-fermion scattering in QED. Ignoring fermion masses, we have $\Delta=-x(1-x)q^2$, where $q^2$ is the photon virtuality. Now $q^2$ is -ve for t-channel elastic scattering, but +ve for s-channel scattering, so $\Delta$ may have either signs. –  felix Dec 17 '11 at 8:21
I point out that I agree with the preceding answer but the question here is just a mathematical one. The value of the integral will depend on the sign of $\Delta$. So, if $\Delta<0$ the integral value is not the one you wrote but
$$\int dx\frac{x^{3-2\epsilon}}{(x+\Delta)^2}=x^{2-2\epsilon}\left(\frac{1}{x-\Delta^2}\right)^{2\epsilon-2}\ _2F_1(2\epsilon-2, 2\epsilon-3, 2\epsilon-1, -\Delta^2/(x - \Delta^2))\times$$ $$\frac{1}{2-2\epsilon}$$
The beast that you see on the rhs is a hypergeometric function that does not converge when you take the limit of $x$ going to infinity. So, physical arguments given above are sound.