Fourier transform of this? I need to prove:
$$\frac 1 {\Delta x} \int_0^\infty dk \, e^{-ik(\Delta t-i\epsilon)} \sin k\,\Delta x=\frac 1 {\left(\Delta t-i\epsilon\right)^2 -\Delta x^2}
$$
A possible hint could be: $-ik(\Delta t-i\epsilon)=-k(\epsilon+i\,\Delta t)$
I would appreciate some help. Thank you.
 A: I would start with writing the sin as sum of 2 exponential functions. Then the integrand becomes the sum of exponential functions, so is easy.
A: I don't want to write $\Delta$ so I will use $\Delta x=a$ and $\Delta t-i\epsilon=t$. The expression becomes:
$$\frac 1 {a} \int_0^\infty dk \, e^{-ikt} \sin k\,a=\frac 1 {t^2 -a^2}$$
This integral does not appear to converge at infinity so I will do a trick. I will multiply it by $e^{-bk}$ assuming $b>0$, and then insert $b\rightarrow 0$ at the very end.
$$\frac 1 {a} \int_0^\infty dk \, e^{-ikt}\,e^{-bk} \sin k\,a=\frac 1 {t^2 -a^2}$$
You can use partial integration (I don't want to write the whole thing here. If this step is where you get stuck, show your progress and I'll correct it) to get the antiderivative as:
$$-\frac 1 {a}\frac{e^{-k (b+i t)} (a \cos (a k)+(b+i t) \sin (a k))}{a^2+(b+i t)^2}$$
At $k\rightarrow \infty$ everything is zero because $e^{-kb}\rightarrow 0$. At $k=0$ we have:
$$-\frac{1}{a^2+(b+i t)^2}$$
We take $b\rightarrow 0$, and square $i$ at this point:
$$-\frac{1}{a^2+-t^2}=\frac 1 {t^2 -a^2}$$
Which is the result.
