# 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.

• Hm, the authors do skip a few steps don't they? Jan 20, 2023 at 17:45
• You could try the Euler substitution with something like$$t = \frac{\sqrt{c^2x^2+(a^2-b^2-c^2)x+b^2-i\epsilon} - \sqrt{b^2-i\epsilon}}x$$ Jan 20, 2023 at 18:19

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 $$\frac{2\pi}{ab}$$.

• Excellent. You are also right with the prefactor. Indeed, I had noticed that in eq 24 of the paper there was a missing $2$ if eq 23 was correct. Thank you very much.
– AFG
Jan 20, 2023 at 20:13