Find the Fourier transform of $f(x)=\alpha, ~\alpha\in\mathbb{C}$.

Question

Solution:\begin{align*} T(u)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{iux}f(x) dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 e^{iux}f(x)dx+\frac{1}{\sqrt{2\pi}}\int_{0}^\infty e^{iux}f(x)dx\\ &=\alpha\frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 e^{iux}dx+\alpha\frac{1}{\sqrt{2\pi}}\int_{0}^\infty f(x)dx\\ &=\frac{\alpha}{\sqrt{2\pi}}\lim_{a\to\infty}\left[\frac{e^{iux}}{iu}\right]_{-a}^0+\frac{\alpha}{\sqrt{2\pi}}\lim_{a\to\infty}\left[\frac{e^{iux}}{iu}\right]_{0}^a\\ &=\frac{\alpha}{\sqrt{2\pi}}\lim_{a\to\infty}\left[\frac{1}{iu}-\frac{e^{-iua}}{iu}+\frac{e^{iua}}{iu}-\frac{1}{iu}\right]\\ &=\frac{\alpha}{\sqrt{2\pi}}\lim_{a\to\infty}\frac{2i\sin (ua)}{iu}\\ &=\sqrt{\frac{2}{\pi}}\frac{\alpha}{u}\lim_{a\to\infty}\sin (ua) \end{align*}

But in the last line limit doesn't exist. So how do we find the Fourier transform of this function. Please help me out!

The Fourier transforms of constant functions are non-function distributions. For example, $f(x) \equiv 1$ are mapped to a multiple of the Dirac $\delta$ function. Use the linearity to deduce what $f(x) \equiv \alpha$ is mapped to. — Zerox, Oct 29 '22 at 07:34
Does this answer your question? Derivation of Fourier Transform of a constant signal — Anne Bauval, Oct 31 '22 at 15:02

score 2 · Accepted Answer · answered Oct 29 '22 at 07:47

As @Zerox commented.

\begin{align*} T(u)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{iux}f(x) dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{iux}\alpha dx\\ &=\sqrt{2\pi}\alpha\frac{1}{2\pi}\int_{-\infty}^\infty e^{iux} dx\\ &=\sqrt{2\pi}\alpha\delta(u) \end{align*}

Find the Fourier transform of $f(x)=\alpha, ~\alpha\in\mathbb{C}$.

1 Answers1