I have this integral:
$$\int_{-\infty}^\infty e^{-(t²+2t)/2}e^{-i\omega t}dt$$ I don't know how to solve it, but I have tried, like this:
$$ \int_{-\infty}^\infty e^{(-(t²+2t)/2)+(-i\omega t)}dt $$ Then I substitute $$ u = \frac{t²-2t}{2}-i\omega t = \frac{1}{2}t²-t-i\omega t\\du = t-1-i\omega dt \\ dt = \frac{1}{t-1-i \omega}$$
I'm stuck here. I don't think that there should be a t left in the last formula, but I don't know how to get rid of it. If I put this with the integral I still have an annoying t to handle...
How do I get from here?
Because
$$(t²+2t)/2+i\omega t=t^2+2t\left(\frac{1}{2}(1+i\omega)\right)=t^2+2tu=(t+u)^2-u^2$$ $$u=(1/2)(1+i \omega)$$ We have
$$\int_{-\infty}^\infty \exp((-(t²+2t)/2)+(-i\omega t))dt=\exp(u^2)\int_{-\infty}^\infty e^{-(t+u)^2}dt=\exp(u^2)\sqrt{\pi}$$ $$=\exp(-(iu)^2)\sqrt{\pi}=\sqrt{\pi}\exp(-(1/4)(-i+ \omega)^2)$$