Proving that a propagator decays exponentially for spacelike separations

[NanakiXIII]

I'm trying to show that the propagator for spacelike separation decays like $e^{-m r}$ and I'm stuck. At some point I hit the integral

$$\int_{-\infty}^{\infty} \frac{dk}{\sqrt{k^2 + m^2}} e^{i k r}.$$

Integration of complex functions not being my forte, I only managed to get to this point using provided answers, but I don't understand the next step. Apparently the above integral is equal to

$$2 \int_0^{\infty} dy e^{-(y+m) r} \frac{1}{\sqrt{(y+m)^2 - m^2}},$$

where the substitution $k = i(m+y)$ has been made. Now there are several things I don't understand. Firstly, I see how the integrand has changed, but since the integrand is not an even function of $k$ or $y$, how is it possible to change the integration limits to $0$ and $\infty$? Secondly, shouldn't the integration now be along the complex axis, i.e. shouldn't the upper limit be $i \infty$, due to the $i$ in the substitution? Finally, there is also a bit of text where the author says the integrand has a branch cut going from $i m$ to $i \infty$, and one has to fold the contour of the integral around this cut. I don't see what that cut has to do with anything whatsoever, or exactly what contour he is using.

Could someone help me along with this? I think I understand the rest of the derivation, this is the last link I need.

 

[Eynstone]

Consider Fa(k) = 1/sqrt(k^2 +m^2) for |k|<=a & 0 otherwise.
Applying the Riemann-Lebesgue lemma , the integral of exp(irk)Fa(k) =o(r) as r-> inf.
If you argue carefully & let a->inf. , you'll get the desired result.

 

[NanakiXIII]

I don't quite follow. I'm not familiar with this lemma, but it seems to say that the integral goes to 0 as $r$ goes to infinity, not to $O(r)$ (I'm assuming that's what you meant). Could you elaborate a little? This is not homework, I'm just trying to understand the author's derivation.

 

[NanakiXIII]

I figured it out. Thanks, though.

 

[ismaili]

Consider Fa(k) = 1/sqrt(k^2 +m^2) for |k|<=a & 0 otherwise.
Applying the Riemann-Lebesgue lemma , the integral of exp(irk)Fa(k) =o(r) as r-> inf.
If you argue carefully & let a->inf. , you'll get the desired result.

Could you elaborate on this a little bit more?
Since the integral of NanakiXIII's question is the same as
https://www.physicsforums.com/showthread.php?t=420759

But, there is a problem to use Jordan's lemma.
To perform the integral along the keyhole contour, we need some regularization as weejee did in that series of post.