i need to find the fourier transform of $\displaystyle f(x)=e^{-ax^2} sin(bx)$

i get $\displaystyle F(\lambda) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} e^{-ax^2} sin(bx) e^{i \lambda x} dx$ (by definition)

now $\displaystyle sin(bx) = \frac{e^{ibx} - e^{-ibx}}{2i}$

so $\displaystyle F(\lambda) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} e^{-ax^2} \[\frac{e^{ibx} - e^{-ibx}}{2i}\] e^{i \lambda x} dx$

$\displaystyle = \frac{1}{\sqrt{2\pi}} \frac{1}{2i} \int^{\infty}_{-\infty} e^{-ax^2 + ibx + i\lambda x} - e^{-ax^2 - ibx + i\lambda x} dx$

im stuck as to where to go from here! judging by our notes the next step should be

$\displaystyle = \frac{1}{\sqrt{2\pi}} \frac{1}{2i} \sqrt{\frac{\pi}{a}} \int^{\infty}_{-\infty} e^{-\frac{(b+\lambda)^2}{4a}} - e^{-\frac{(b-\lambda)^2}{4a}}$

but i have no idea why this should be (and not even sure its the right step!)