Find the Fourier Transform of $e^{i x t}$.

Question

Find the Fourier Transform of $e^{ixt}$, where $x$ is a real parameter, $t\in \mathbb R$. I started writing: $$\int_{-\infty}^\infty e^{ixt} e^{-i\omega t}dt$$ but I do not know how to go on!

Answer 1

$$\int_{-\infty}^{+\infty} e^{ixt} e^{-i \omega t} dt=\int_{-\infty}^{+\infty} e^{ixt-i \omega t} dt=\int_{-\infty}^{+\infty} e^{i(x- \omega )t} dt=2 \pi \delta(x- \omega)$$

EDIT: It is known that:

$$\delta(x-a)= \frac{1}{2 \pi}\int_{-\infty}^{\infty} e^{ip(x-a)} dp$$

Answer 2

Using the inverse Fourier transform, you may write (everything in distributional sense)

$f(x)=\frac{1}{2\pi}\int_\mathbb{R}e^{ixt}\int_\mathbb{R} e^{-i\omega t} dt d\omega =\frac{1}{2\pi}\int_\mathbb{R}(\int_\mathbb{R}e^{ixt} e^{-i\omega t}dt) d\omega=\frac{1}{2\pi}\int_\mathbb{R}\delta(x-\omega)f(\omega)d\omega $

And therefore your Integral equals $2\pi \delta(x-\omega)$ in the distributional sense.