Fourier Transform of $e^{-4t^2} $?

Question

What are the steps to calculate Fourier transform of $e^{-4t^2} $ ?

If it was a form of $|t|$ maybe it would be easier, but as it is i cannot find a way.

Answer 1

Let be $f(t) = \mathrm e^{-at^2}$ and $$\mathcal F\{f(t)\}=\hat{f}(\omega)= \int_{-\infty}^{\infty} f(t)\mathrm e^{- i\omega t} \,\mathrm dt = \int_{-\infty}^{\infty} \mathrm e^{-at^2}\mathrm e^{-i\omega t}\,\mathrm dt $$ Differentiating with respect to $\omega$ yields $$\frac{\mathrm d}{\mathrm d\omega} \hat{f}(\omega) = \int_{-\infty}^{\infty} \mathrm e^{-at^2}(-i t)\mathrm e^{-i\omega t}\,\mathrm dt= \frac{i}{2a} \int_{-\infty}^{\infty} \left(\frac{\mathrm d}{\mathrm dt} \mathrm e^{-at^2} \right) \mathrm e^{- i\omega t} \,\mathrm dt$$ Integrating by parts, we obtain

$$\hat{f}'(\omega) = - \frac{\omega}{2a} \int_{-\infty}^{\infty} \mathrm e^{-at^2}\mathrm e^{-i\omega t}\,\mathrm dt= - \frac{\omega}{2a} \hat{f}(\omega)$$

The unique solution to this ordinary differential equation is given by

$$\hat{f}(\omega) =\beta \cdot \exp \left(- \frac{\omega^2}{4a} \right)$$

where the constant $\beta$ is $$\beta=\hat{f}(0) = \int_{-\infty}^{\infty} \mathrm e^{-at^2}\,\mathrm dt=\sqrt{\frac{\pi}{a}}$$ It follows that $$\mathcal F\left\{\mathrm e^{-at^2}\right\}=\hat{f}(\omega) =\sqrt{\frac{\pi}{a}} \exp \left(- \frac{\omega^2}{4a} \right)$$ and for $a=4$ we have $$\mathcal F\left\{\mathrm e^{-4t^2}\right\}=\sqrt{\frac{\pi}{4}} \mathrm e^{- \left(\omega/4\right)^2}$$

Answer 2

Assuming $t \in \mathbb{R^1},$ the task is to compute $F(e^{-4t}) \equiv \int_{-\infty}^\infty e^{i\xi t} e^{-4t^2} \; dt = {\hat f}(\xi).$

Here are the steps:

0) If the problem was n-Dimensional, you could write it as the 1D problem multiplied together n times. So dealing only with the 1D problem,

1) Complete the square in the exponential.

2) Use a contour integral to simplify the problem.

The details, $(-4t^2 + i\xi t) = -4(t - \frac{i\xi}{8})^2 - \frac{\xi^2}{16}.$

Thus the problem becomes, ${\hat f} = e^{- \frac{\xi^2}{16}}\int_R e^{-4(t - \frac{i\xi}{8})^2} \; dt.$ Use the rectangle contour where the long sides lay on the real axis and on $z = x + i\frac{\xi}{8},$ and the short sides are $z = \pm R + iy$ as $R \to \infty$ and $y$ ranges between $[0, \frac{\xi}{8}]$ to connect the contour. The function is analytic in this region so the full contour integral is zero, the short sides vanish in the limit and one finds that, \begin{equation} \int_R e^{-4(t - \frac{i\xi}{8})^2} \; dt = \int_R e^{-4t^2}\; dt = \sqrt{\frac{\pi}{4}}. \end{equation} Thus, ${\hat f}(\xi) = \sqrt{\frac{\pi}{4}} e^{- \frac{\xi^2}{16}},$ when the Fourier transform is defined as $ \int_{-\infty}^\infty e^{i\xi t} e^{-4t^2} \; dt.$ The details such as generalizing it to higher dimension and doing the Gaussian integral at the end are not difficult to fill in yourself.