Fdtxjr

Consider the integral

$$
I equiv int_-infty^infty dp frac1sqrtp^2 + m^2 sin(t sqrtp^2 + m^2 ) e^i p x.
$$
It appears to me that when $x > 0$ and $x > t$ the above integral is $0$. First, rewrite it as
$$
I = fraci2 int_-infty^infty dp frac1sqrtp^2 + m^2 (e^-itsqrtp^2 + m^2 - e^itsqrtp^2 + m^2) e^i p x.
$$

Now consider deforming the $p$ integral on the complex plane. If $p$ has a large positive imaginary part, then the integrand goes to zero as long as $x > t$. Because the integrand has no poles, this allows us to close the contour with a semi circle in the upper half plane. Because there are no poles, this integral is then $0$.

I am not asking for a rigorous proof of why $I = 0$ when $x > t$ using complex analysis, because I think the proof I sketched above is right. What does confuse me is that, from its definition, nothing special appears to happen to $I$ when $x$ goes from being less than $t$ to being greater than $t$. What am I looking for is a satisfying explanation for why the value of this integral changes dramatically in the $x > t$ and the $x < t$ cases, just by using the properties of the $p$ integral along the real axis.

First, let's use the symmetry of the integrand to remove the complex exponential
beginmultline
I = 2int_0^infty fracsinleft(tsqrtp^2+m^2right)sqrtp^2+m^2cosleft(xpright)dp \ = int_0^infty fracsinleft(xp+tsqrtp^2+m^2right)sqrtp^2+m^2dp+int_0^infty fracsinleft(tsqrtp^2+m^2- xpright)sqrtp^2+m^2dp \
=int_-infty^inftyfracsin(xp + tsqrtp^2+m^2)sqrtp^2+m^2dp
endmultline
Now, let's look at the argument of that sine function. If $|x| > |t|$, then it's a monotonic function. We can make a simple substitution $u = xp + tsqrtp^2+m^2$, and doing the algebra gives
$$
I = int_-infty^inftyfracsin(u)sqrtu^2+m^2(x^2-t^2) = 0,
$$
since the integrand is odd.

On the other hand, if $|t| > |x|$, the argument of the sine function is not monotonic. It achieves a minimum at $p_0 = -mx/sqrtt^2-x^2$ with value $u_0 = msqrtt^2-x^2$. So we have to split the integral at $p_0$, then do the substitution $u = xp + tsqrtp^2+m^2$. The algebra ends up being the same as before, but because of the different limits of integration, you get
$$
I = 2int_msqrtt^2-x^2^infty fracsin(u)sqrtu^2-m^2(t^2-x^2) = pi J_0left(msqrtt^2-x^2right)
$$
where $J_0$ is a Bessel function.

If we define the Fourier transform of a function 𝑓 and its inverse as
$$mathcal F(f)(omega)=int_mathbb Rf(x)e^-jomega xdx ,,,text and ,,, f(x)=frac 1 2piint_mathbb R mathcal F(f)(omega)e^jxomegadomega$$

Then, what you're looking for the inverse Fourier transform of $$g(omega)=frac2pisqrtomega^2 + m^2 sin(t sqrtomega^2 + m^2 )$$

It turns out that function has a compact support, more precisely its support is $[-|t|, |t|]$. Indeed, according to this question, the inverse Fourier transform of $g$ is
$$f(x)= left{beginarraycc
CJ_0left(msqrtt^2-x^2right) & text if 0 < |x| < |t| \
0 & text if |t| < |x| < +infty
endarrayright.$$
Where $C$ is a constant and $J_0$ is the first Bessel function of the first kind.