homework and exercises - Feynman Propagator in Position Space through Schwinger Parameter

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?