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I have a question regarding calculating the following integral with cutoff. $$ \int_{-\infty}^{\infty} \frac{d\omega}{|\omega|} \cos(\omega(\tau_i-\tau_j)-1)$$

How should I set up the correct cutoff so that the result of the integral becomes: $$-2\ln \frac{\tau_i-\tau_j}{\tau_c}$$ where $\tau_c$ is the short time cutoff~$E_F^{-1}$.

This calculation is from certain resonant tunneling problem of Luttinger liquid, for which you can obtain renormalization group flow equation.

rewrite the cosine in terms of exponenials, and add a tiny real part to the exponents.

In more detail, rewrite the integral as twice the integral from 0 to $\infty$, abbreviate $\tau_i-\tau_j$ by $t$, differentiate with respect to $t$ to get rid of the denominator, use $cos x = (e^{ix}+e^{-ix})/2$, and change $e^{iz}$ to $e^{iz-\epsilon\omega}$ to be able to perform the integration. Then set $\epsilon=0$ and integrate the result from $t=\tau_c \approx 0$ to $t=\tau_i$-$\tau_j$.

Could you explain a little bit more in detail?

I added some details.

So how do you handle the 0 for integral of (1/t), wouldn't that produce -infinity there?

In the final integration w.r. to $t$, one needs the cutoff mentioned.

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