Fourier transform of the integral with variable upper limit

Question

I have the following function

$$g(t)=\int_{-\infty}^t f(t')dt'.$$

And I want to calculate the Fourier transform

$$G(\omega)=\int_{-\infty}^\infty g(t)e^{-i\omega t}dt.$$

Is there any way to proceed?

Answer 1

Noting that $g(t)=(f*H)(t)$, where $H$ is the Heaviside function, and using the convolution theorem, we have

$$\begin{align} \mathscr{F}\{g\}&=\mathscr{F}\{f*H\}\\\\ &=\mathscr{F}\{f\}\mathscr{F}\{H\}\\\\ &=F(\omega)\left(\pi \delta(\omega)-\frac{i}{\omega}\right)\\\\&=\pi F(0)\delta(\omega)+\frac{F(\omega)}{i\omega} \end{align}$$

And we are done.

Answer 2

$$g(t)=\int_{-\infty}^{t}f(t')dt' \implies g(t)=F(t)-\lim_{n \rightarrow -\infty} F(n)$$

$$\therefore g(\omega)=\int_{-\infty}^{\infty}F(t)e^{-i\omega t}dt -\lim_{n \rightarrow -\infty} F(n) \int_{-\infty}^{\infty}e^{-i\omega t}dt$$