If the Fourier transform of f(x) is G(w), then what is the inverse Fourier transform of f(w)? I have the Fourier Transform of $e^{-ax^2}$, which is $\sqrt{\frac{{\pi}}{a}}$$e^{\frac{-w^2}{4a}}$.
I now want to find the inverse Fourier Transform of $e^{-t(w-ib)^2}$.
I can see that this is similar to my original function, and I can see where the first shift theorem would be used (for the shift $-ib$).
However, I don't fully understand the relationship between a function and it's Fourier transform, and the reverse. I noticed that choosing $a$ as $t$, using the first shift theorem and swapping variables from $w$ to $x$ was a factor of $\frac{1}{2{\pi}}$ out.
The answer should be $\frac{1}{2\sqrt{{\pi}t}}$$e^{\frac{-x^2}{4t}}$.
Any help would be greatly appreciated! Exam is on Monday!
 A: Here's what I get, working from the definition of the Inverse Fourier Transform
$$\begin{align*}\mathscr{F}^{-1}\left\{e^{-t(\omega-ib)^2}\right\} &= \dfrac{1}{2\pi} \int_{-\infty}^{\infty}e^{-t(\omega-ib)^2}e^{i\omega x}\space d\omega\\
\\
& \Omega = \omega -ib \\
\\
&= \dfrac{1}{2\pi} \int_{-\infty-ib}^{\infty-ib}e^{-t\Omega^2}e^{i(\Omega+ib) x}\space d\Omega\\
\\
&= e^{-bx}\sqrt{\dfrac{1/4t}{\pi} } \dfrac{1}{2\pi}\int_{-\infty-ib}^{\infty-ib}\sqrt{\dfrac{\pi}{1/4t} }e^{-t\Omega^2}e^{i\Omega x}\space d\Omega\\
\\
&\mbox{using an infinite rectangular contour in the complex plane,}\\
&\mbox{which has the real axis and the line parallel to the real axis}\\
&\mbox{at $-ib$ as two of the four segments, we can then show}\\
\\
&= e^{-bx}\sqrt{\dfrac{1}{4\pi t} } \dfrac{1}{2\pi}\int_{-\infty}^{\infty}\sqrt{\dfrac{\pi}{1/4t} }e^{-t\Omega^2}e^{i\Omega x}\space d\Omega\\
\\
&=e^{-bx}\sqrt{\dfrac{1}{4\pi t} } \mathscr{F}^{-1}\left\{\sqrt{\dfrac{\pi}{1/4t} }e^{-t\Omega^2}\right\}\\
\\
&=\dfrac{e^{-bx}}{2\sqrt{\pi t} } e^{-\frac{x^2}{4t}}\\
\\
\end{align*}$$
This is slightly different from what you state should be the answer (which suspiciously excludes $b$ entirely).
