# Decay of amplitude integral

Consider the function

$$f(\vec{x}) = \int_{\Bbb R^3} {\frac{ e^{-i\,\vec{x}\cdot\vec{k}}}{\sqrt{\vec{k}^2 + m^2}}} d^3 k$$

from Zee's Quantum Field Theory in a Nutshell. He argues like this: “the square root cut starting at $±im$ tells us that the characteristic value of $|\vec{k}|$ in the integral is of order $m$, leading to an exponential decay $\sim e^{−m|\vec{x}|}$”. I cannot understand what he means. It would be nice if someone can expand or suggest a reference for me. Thanks.

• What he likely means is that - to take a 1D analog - the integral may be expressed by Cauchy's Theorem as $$\int_m^{\infty} dk \, \frac{e^{-k x}}{\sqrt{k^2-m^2}}$$ which may be shown to be dominated by the contribution near $k=m$. – Ron Gordon May 11 '15 at 18:50
• I have been thinking about the residue stuff too, but how does it lead to the bound $e^{-m|\vec{x}|}$. And what is the precise definition of "be dominated" ? – Tuyet Nhi May 11 '15 at 18:57
• Note that I am keeping this to a comment because I do not wish to get into precise definitions at this point. I am using the language of Steepest Descents. Think about approximating the integral with a sum (avoiding the "pole"), what term or terms make the biggest contribution to the sum? That's what I mean by "dominating." – Ron Gordon May 11 '15 at 19:00
• Thanks, that Steepest Descent does make sense; although I still cannot see how to bound the integral by $e^{-m|\vec{x}|}$. – Tuyet Nhi May 11 '15 at 19:03

Letting \begin{align*} \vec{x}&=(0,0,x), \\ \vec{k}&=k(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi), \end{align*} and integrating over $\theta$ and $\phi$ we find \begin{align*} \int_{\Bbb R^3} {\frac{ e^{-i\,\vec{x}\cdot\vec{k}}}{\sqrt{\vec{k}^2 + m^2}}}\, d^3 k &= \frac{4\pi}{x} \int_0^\infty \frac{k\sin k x}{\sqrt{k^2+m^2}}\,dk = \frac{2\pi}{x} \int_{-\infty}^\infty \frac{k\sin k x}{\sqrt{k^2+m^2}}\,dk \\ &= \frac{2\pi}{x} \textrm{Im} \int_{-\infty}^\infty \frac{k e^{i k x}}{\sqrt{k^2+m^2}}\,dk. \end{align*} An integral of exactly this form is worked out in detail here.