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In the formula $$\int \frac {d^{4-2\epsilon} \ell} {(2\pi)^{4-2\epsilon}} \frac 1 {(\ell^2-\Delta)^2} = \frac i {(4\pi)^{2-\epsilon}} \Gamma(\epsilon) \left(\frac 1 \Delta\right)^\epsilon,$$ 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.

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If $\epsilon>0$, you can start with a small imaginary part in the denominator of the integrand, $\Delta - i\delta$. After finishing the calculation, you take the limitation $\delta \rightarrow 0^+$. Then the result will be

$\displaystyle{\int \dfrac{d^{4-\epsilon}l}{(2\pi)^{4-\epsilon}}\dfrac{1}{(l^2 - \Delta + i\delta)^2} = \dfrac{i}{(4\pi)^{2-\epsilon}}\Gamma(\epsilon)\left(\dfrac{1}{\Delta -i\delta}\right)^{\epsilon}, \delta \rightarrow 0^+}.$

Note that the sign of $i\delta$ changes because of Wick rotation.The reason why we have to put $i\delta$ to the denominator can see easily from

$\left(\dfrac{1}{\Delta -i\delta} \right)^{\epsilon} = 1 - \epsilon \ln \left(\Delta - i\delta \right) + O(\epsilon^2), \delta \rightarrow 0^+.$

If $\Delta>0$, we can ignore $i\delta$ at the beginning. But if $\Delta<0$, $i\delta$ needs to be kept in the denominator because there is no definition of the logarithm with the negative argument. For further calculation, some useful relations will be needed

$ln(z_1.z_2)= ln z_1 + ln z_2 + \eta(z_1, z_2)$, $\eta (z_1, z_2)= 2 \pi i [Im(-\theta (z_1))Im(-\theta (z_2))Im(\theta (z_1 z_2))-Im(\theta (z_1))Im(\theta (z_2))Im(-\theta (z_1 z_2))]$

where $z_1, z_2$ are the complex numbers and $\theta$ is a sign function. In the multi-legs one-loop computations, such as three-point and four-point scalar integral, the result will depend on Spence function.

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