Imaginary part of this integral I'm reading a physics paper and there is one step involving the imaginary part of an integral that I cannot reproduce (eq 23). The integral is
$$I=\int_0^1 dx\frac{1}{\Big[xa^2+(1-x)b^2-x(1-x)c^2-i\epsilon\Big]^{3/2}},$$
where all parameters are real and $a,\,b,\,\epsilon>0$. They claim that the imaginary part of this integral is
$$\text{Im}\{I\}=\frac{2\pi}{ab}\Big[\delta(c-a-b)+\delta(c+a+b)\Big].$$
I have no clue how to get that, any help would be appreciated.

You can just plug this integral into some computer algebra system and get
$$I = \frac{2}{a^4 + b^4 + c^4 - 2(a^2 b^2 + a^2 c^2 + b^2 c^2) + 4i\epsilon c^2}\left[\frac{b^2-a^2-c^2}{\sqrt{a^2-i\epsilon}} + \frac{a^2-b^2-c^2}{\sqrt{b^2-i\epsilon}}\right].$$
Since $a$ and $b$ are nonzero and $\epsilon$ is infinitesimal, the imaginary part is provided by the first factor. Its denominator can be written as
$$[c^2-(a+b)^2][c^2-(a-b)^2]+4i\epsilon c^2.$$
If $c^2 = (a-b)^2$, the terms in square brackets in $I$ cancel each other. In the vicinity of $c^2 = (a+b)^2$
$$I = -2\frac{a+b}{ab}\frac{1}{c^2 - (a+b)^2 + i0},$$ and you get two delta functions by the Sokhotski–Plemelj theorem. Although it seems the prefactor will be $\frac{\pi}{ab}$, not $\frac{2\pi}{ab}$.
