Can anybody integrate this: $$ \int_0^K e^{i(A\sqrt{\mathstrut k^2+m^2} - Bk)} dk,$$ where $K$, $A$, $B$ and $m$ are real constants?
Sorry folks, I didn't realise anybody would be interested in the context...
I was reading about the Klein Gordon equation and raised my eyebrows over the bit where the phase velocity doesn't contradict relativity because signals supposedly only travel at the packet/group velocity.
I didn't buy that so I set about proving the opposite. If my integral is merely non-zero, I think I managed it. But I'm sure you'll all punch holes in my logic.
Let's start by demystifying the whole relativistic quantum thing by reading the KG equation as nothing but a guitar string with restoring springs all along its length. To put that the other way around, it's a row of simple harmonic oscillators coupled together.
The dependent variable might as well be real because if it was complex, the real and imaginary parts would form two independent systems because there are no first order diffs to mix them up.
We've scaled the units to suit this particular guitar such that 1 is the wave velocity in the absence of restoring springs. Add the springs and it gets stiffer so the phase velocity increases, but more so for longer wavelengths where the curvature term is less important. With phase velocity v depending on $\lambda$ we have a packet velocity which turns out to be $1/v$.
I'll whack the origin with a delta function and observe the dependent variable $\Psi$ an event $(t, x)$ at a spacelike interval away, i.e. $t>0$, $x>0$, $x>t$. (This is a 1D KG equation.) At that event, a certain wavelength with $k=K$ will have only just arrived. So, choose any two of $x$, $t$ and $K$ and the other is determined. Unless this is zero in every case, the KG equation breaks relativity.
I think the delta function injects $$\Psi = \int dk.{e^{i(\omega t - kx)}}$$ although this is a bit odd when the relationship between $\omega$ and $k$ isn't trivial. If I did all the maths in terms of $\omega$ instead I think it would come out the same.
This needs cutting off for short waves that haven't arrived yet, so the integral runs from 0 to $K$. Plug in $\omega^2 = k^2 + m^2$ and $x=v(K).t$, collect some constants into A and B and you arrive at my integral.
Can I go off hold now please?
One more thing: I contradicted myself by saying that the dependent variable might as well be real and then modelling it with a complex exponential. What I mean is that the deflection of the string is the imaginary part of $\Psi$, because I'm imagining the delta function as a little bullet that bounces off the string and instantaneously imparts a velocity to it. In reals, the deflection would be $sin(\omega t - kx)$ for a particular wavelength.