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So I am aware of a thread at Propagator of a scalar in position space but it does not answer my question, which is more about poles in position space.

Starting from

$$D_F(x_1-x_2) = \int \frac{d^4 k}{(2\pi)^4}\frac{i}{k^2-m^2+i\epsilon}e^{i k\cdot x}$$

I have been able to show that

$$D_F(x_1-x_2) = \frac{-i}{16\pi^2}\int_{0}^{\infty}\frac{ds}{s^2}\exp\left[-i\frac{X^2}{4s}\right]\exp\left[-i(m^2-i\epsilon)s\right]$$

which by change of variable can be written as

$$ D_F(x_1-x_2) = \frac{-i}{16\pi^2}[i(m^2-i\epsilon)]\int_0^\infty\frac{dt}{t^2}\exp\left[-t-\frac{[-(m^2-i\epsilon)X^2]}{4t}\right].$$

Using the integral representation of $K_1(z)$ (the modified Bessel function of the second kind) I can see that

$$D_F(x_1-x_2) = \frac{(m^2-i\epsilon)}{16\pi^2}\frac{4}{\sqrt{-(m^2-i\epsilon)X^2}}K_1(\sqrt{-(m^2-i\epsilon)X^2}).$$

But I know that the correct answer is

$$D_F(x_1-x_2) = -\frac{i}{4\pi^2}\frac{1}{\sqrt{-X^2 + i\epsilon}}K_1(im\sqrt{-X^2 + i\epsilon}).$$

What bothers me is how $\sqrt{-(m^2-i\epsilon)X^2}$ is equal to $\sqrt{im(-X^2 + i\epsilon)}$, because according to me

$$\sqrt{-(m^2-i\epsilon)X^2)} = \sqrt{-m^2(1-i\epsilon)X^2} = im\sqrt{X^2 - i\epsilon}$$

What's the error here?

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  • $\begingroup$ So my confusion in the last step stems from the book 'Analytic Tools for Feynman Integrals' by Vladimir A. Smirnov (Springer 2012). The equation just above equation (2.17) gives the "correct answer" which has a manifestly different form. $\endgroup$ – leastaction Mar 3 '15 at 13:32
  • $\begingroup$ Have you solved this problem? I have a similar one $\endgroup$ – Eden Harder Apr 29 '15 at 7:38
  • $\begingroup$ My solution appears in my post. $\endgroup$ – leastaction Apr 29 '15 at 10:10
  • $\begingroup$ I refer to the book you mentioned and get to the step before changing variable. Can you teach me how you change $s$ to $t$? Thanks! $\endgroup$ – Eden Harder May 2 '15 at 11:49
  • $\begingroup$ I took $t = i X^24s$ from the looks of it... $\endgroup$ – leastaction May 2 '15 at 11:51

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