# Solving the quantum field propagator for 1+1 dimensional spacetime at $t=0$

I just started studying Quantum Field Theory in a Nutshell By Zee to quench my thirst for physics. But I got stuck on one of the early exercises. In exercise 1.3.1 (and 1.3.2), the question asks to work out the propagator $D(x)$ in 1+1 dimensional spacetime with $t=0$ where general $D(x)$ is defined as:

$$D(x) = −i \int{ \frac{d^3k}{(2π)^32ω_k} [e^{−i(ω_kt−k.x)}θ(x^0) + e^{i(ω_kt−k.x)}θ(−x^0)] }$$
where $k$ and $x$ are vectors. and $x^0$ is time dimension.

Using some help from the text itself, I figured for 1+1 Dimensional spacetime at t=0, the expression for propagator would be:

$$D(x) = -i \int^{-\infty}_{\infty}\frac{dk}{2\pi\sqrt{k^2+m^2}}e^{-ikx}$$
with all scalars.
Now, how do I further solve it? I have following two questions to answer:

1. To show that it decays exponentially
2. To study its behaviour for large $x$

Can the above questions be solved without further solving the integral?
I tried using $k=m*tan(z)$ substitution but that didn't work too well.
Any hint on further solving this would be helpful.

We begin by simply rewriting the integral that you have stated $$D(x)=-i\int^\infty_{-\infty}\frac{dk}{2\pi}\frac{e^{-ikx}}{\sqrt{k^2+m^2}} =-\frac{i}{m}\int^\infty_{-\infty}\frac{dk}{2\pi}\frac{e^{-ikx}}{\sqrt{(\frac{k}{m})^2+1}}$$ Next we substitute variables $k = m t$, where $t$ is our new variable, so $$D(x)=-i\int^\infty_{-\infty}\frac{dt}{2\pi}\frac{e^{-imtx}}{\sqrt{t^2+1}}$$ Now we use Eulers identity to expand the exponential into trigonometric functions $$D(x)=-i\int^\infty_{-\infty}\frac{dt}{2\pi}\frac{\cos(mtx)}{\sqrt{t^2+1}} -\int^\infty_{-\infty}\frac{dt}{2\pi}\frac{\sin(mtx)}{\sqrt{t^2+1}}$$ Note that the second term vanish because it's an odd function over an even interval. Likewise, the first term is an even function over an even interval so we can modify the integration limits and write $$D(x) =-2i\int^\infty_0\frac{dt}{2\pi}\frac{\cos(mtx)}{\sqrt{t^2+1}}$$ We identify this integral as a Modified Bessel Function of the Second Kind Reference eq. (6), hence $$D(x)=-2i\int^\infty_0\frac{dt}{2\pi}\frac{\cos(mtx)}{\sqrt{t^2+1}} =-i\frac{K_0(mx)}{\pi}$$ It is the last part that I find tricky. I hope this exercise doesn't scare you away from this beautiful field. I promise the upcoming exercises in the book aren't usually this difficult to solve.
Finally, to study the behavior for large $x$, you have too look up the asymptotic behavior of the $K_n(x)$ Bessel function. Me myself use a mathematics handbook to look it up $$x\rightarrow\infty:\\ K_n(x)=\sqrt{\frac{\pi}{2x}}e^{-x}\Big[1+\mathcal{O}\Big(\frac{1}{x}\Big)\Big]$$