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While reading a paper published at Physical Review Letters, I came across an integral, Eqns. (6) and (7) $$\text{Im} \int_{0}^{\omega} \int_{2M}^{2(M-\omega')}{{\frac{{\rm d}r}{1-\sqrt{\frac{2(M-\omega')}{r}}}}(-{\rm d}\omega')}=4 \pi \omega(M-\frac{\omega}{2})$$ The paper mentions this integral was performed using contour integration. Please explain how does one arrive at the answer and what are the implicit assumptions.

NB. The paper has already received more than 1460 citations.

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    What are the parameters $M$ and $w$ and the limits $r_1$ and $r_2$? How can the answer be purely imaginary? – 2017-01-13
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    Can it be imaginary due to the fact that a complex residue may have been used? – 2017-01-13
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    @Dr.MV. thanks. i have made the necessary edits. $w$ is the energy of a particle which is tunneling out of a black hole of mass $M$. – 2017-01-13
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    I haven't had time to look at it properly, but here is what I think is going on: the integral as written is likely divergent and what the authors do is to change $\omega \to \omega - i\epsilon$ to avoid the singularity in the integrand when $r = 2(M-\omega)$ and $\omega = \omega'$, perform the integral and take $\epsilon\to 0$ in the end. This turns a real integral into a complex (contour) integral. Also according to the paper the upper limit is $2(M-\omega)$ not $2(M-\omega')$. – 2017-01-13

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