Fourier transform of $f(x)=e^{i2\pi ax}$

Question

The Fourier Transformation that I made : $$ \Im(f(x))= \int_{-\infty} ^{\infty} f(x) e^{-i2\pi ux} dx = \int_{-\infty} ^{\infty} e^{i2\pi ax} e^{-i2\pi ux} dx = \int_{-\infty} ^{\infty} e^{-i2\pi x(u-a)} dx $$

And stuck here. I think it should be equal to $\delta(u-a)$, but how can I continue and show that?

Answer 1

We know that \begin{equation} \int_{-\infty}^{\infty} \delta(u-a) e^{2 \pi u x} du = e^{2 \pi a x} = f(x) \end{equation} So $F(u) = \delta(u-a)$ is the Fourier transform, i.e. \begin{equation} \int_{-\infty}^{\infty} f(x) e^{-2 \pi u x} dx = \delta(u-a) \end{equation}

Answer 2

First of all, your function is not integrable so you have to compute the Fourier transform in a distributional sense. On the other hand, for any function $\varphi$ in the Schwartz space we can apply the definition of distributional Fourier transform and the inversion formula to get

$$\mathcal{F}(e^{2\pi iax})(\varphi)=\int_\mathbb{R} e^{2\pi iax}\mathcal{F}(\varphi)(x)dx=\varphi(a)=\delta_a \left(\varphi\right)$$.

Then $\mathcal{F}(e^{2\pi iax})=\delta_a$ as distributions.