Fourier Transform Equation Can anyone help me by explaining how to answer the following:
Determine the fourier transform of: $f(x) = e^{-4x^2-4x-1}$
thanks,
Euden
 A: My guess is that this was meant to be an exercise in using tables of Fourier transforms. Re-write as 
$f(x) = e^{-4x^2}e^{-4x}e^{-1}$
From here we can find transforms of common functions. For each (and note that there are alternative definitions of the transform--I selected the second of three definitions arbitrarily)--using 206 and 205 from the table:
$F_1(e^{-4x^2}) = \sqrt \frac{1}{2*4} e^{\frac{-\omega^2}{4*4}} $  
$F_2(e^{-4x} )= \frac{1}{ \sqrt {2\pi}(4+i\omega)}$, in which the transform is valid only for positive values of x because of the way it's defined.
And $e^{-1}$ is a constant, call it A. 
Again referring to tables, we can use that the transform of a product is a convolution of the transforms in the $\omega$ domain.
$F ( A f_1(x)f_2(x)) =  A\frac{F_1 * F_2}{\sqrt 2\pi} $ 
A: You will find the Fourier transform of $\exp(-x^2/2)$ or $\exp(-x^2)$ listed in tables (hint: it is of the form $\exp(-\omega^2)$).  Your function is
$$\exp(-4x^2 - 4x - 1) = \exp(-(2x+1)^2).$$
Tables of Fourier transform properties often list how the Fourier transforms
of $f(ax)$ and $f(x-b)$  relate to the Fourier transform of $f(x)$.  Use these to get the result.
A: It is sufficient to find the following $f(t)=\int\limits_Re^{itx}e^{-x^2}\, dx$. So, write $f'(t)=i\int\limits_Rxe^{itx}e^{-x^2}\,dx$. Using integrating by parts, one can get $f'(t)=-0.5ie^{itx}e^{-x^2}\Big|^{\infty}_{-\infty}+0.5i^2t\int\limits_Re^{itx}e^{-x^2}\,dx=-0.5tf(t)$. Thus, $f'(t)=-0.5f(t)$ and $f(0)=\int\limits_Re^{-x^2}=\sqrt{\pi}$. Hence you are able to determine the function $f$.
