$ \mathscr{F}\Big\{e^{-(a-jb) t^2}\Big\}$

Question

By using the definition of continuous-time-Fourier-transform: $$ \mathscr{F}\Big\{x(t)\Big\} \triangleq X(ω) = \int\limits_{-\infty}^{\infty}x(t)\, e^{-j 2 \pi ω t} \,\mathrm{d}t$$

and solving the differential equation I ended up with after derivation of the Fourier equation :$${X'(ω)}=\frac{-ω}{2(a-jb)} \cdot X(ω)$$ I calculated

$$X(ω)=\mathscr{F} \Big\{x(t)=\exp(-(a-jb)t^2)\Big\}= K \cdot \exp\left ( \frac{-ω^2}{4(a-jb)}\right) $$

Which is the value of this constant K?

Answer 1

Let $\zeta = a - i b$. To see how the transform is related to the Gaussian integral, make the substitution $t = z/\sqrt \zeta - \pi i \hspace{1px} \omega/\zeta$: $$\int_{\mathbb R} e^{-\zeta t^2 - 2 \pi i \omega t} dt = \frac {e^{-(\pi \omega)^2/\zeta}} {\sqrt \zeta} \int_\gamma e^{-z^2} dz.$$ Then prove that the integral over the line $\gamma$ is the same as the integral over the real line, with the principal value of the square root giving $\int_{-\infty}^\infty e^{-z^2} dz$.

Answer 2

Note that $$ X(0) = \int_{-\infty}^\infty x(t)\,dt, $$ so if you can do this integral you should be able to find $K$.