How to change integration bounds

Question

i solved the integration $\frac{1}{ab}\int_{-\infty}^{\infty}e^{\frac{-(x-z)^2}{2\sigma^2}}\cdot K_0(\frac{|z|}{b})\text{d}z$ and reach upto the following step: $\frac{e^{\frac{-x^2}{2\sigma^2}}}{ab}\int_{-\infty}^{\infty}K_0(\frac{|z|}{b})\cdot e^{\frac{-z^2}{2\sigma^2}}e^{\frac{xz}{\sigma^2}}\text{d}z$. But in research paper it is derived as: $\frac{e^{\frac{-x^2}{2\sigma^2}}}{ab}\int_{0}^{\infty}K_0(\frac{|z|}{b})\cdot e^{\frac{-z^2}{2\sigma^2}}(e^{\frac{xz}{\sigma^2}}+e^{\frac{-xz}{\sigma^2}})\text{d}z$. Where $K_0$ is modified bessel function of second kind. I am not getting how it is coming. Any help please. Thank you.

Answer 1

$$\int_{-\infty}^\infty f(x) \, dx = \int_0^\infty f(x) \, dx + \int_{-\infty}^0 f(x) \, dx = \int_0^\infty f(x) \, dx + \int_0^\infty f(-x) \, dx = \int_0^\infty (f(x) + f(-x)) \, dx$$ where $f(x) = K_0(|z|/b) e^{-z^2/(2\sigma^2)} e^{xz/\sigma^2}$.