A definition of the Fourier transform commonly used is (I always forget which convention of normalization to use) \begin{align}f(\omega)=\int_{-\infty}^\infty e^{i \omega t}f(t) dt\end{align} For a constant function $f(t)=1$, this evaluates to the Dirac delta function, \begin{align}\int_{-\infty}^\infty e^{i \omega t}dt = 2\pi\delta(\omega)\end{align} Question:

What is the one-sided Fourier transform of a constant function\begin{align}\int_{0}^\infty e^{i \omega t}dt = ?\end{align}


2 Answers 2


Put in a cutoff
$$ \int_0^\infty e^{i\omega t}e^{-\epsilon t}dt= \frac 1{i\omega -\epsilon} $$ So the Fourier transform of the Heaviside step function $\theta(t)$ is the $\epsilon\to 0$ limit $$ \mathcal{F}[\theta](\omega)= \lim_{\epsilon\to 0} \left\{\frac{-i}{\omega-i\epsilon}\right\}= P\left(\frac{1}{i\omega}\right)+\pi \delta(\omega), $$ where "$P$" is the principal part distribution. (I added edit to show connection with @Thomas Fritsch's answer)


You look for the Fourier transform of the step function: $$f(t)=\begin{cases} 1, &\text{for }t > 0 \\ 0, &\text{for }t < 0 \end{cases}$$

The solution is well known (see for example at Fourier transform of typical signals):

$$F(\omega)=\frac{1}{i\omega} + \pi\delta(\omega)$$

Depending on your sign and normalization convention you may need to modify this by a minus sign or by a $2\pi$ factor.


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