2

How to prove that

$$\mathcal{F}\left\{\int_{-\infty}^{t}f(\tau)\, d\tau\right\}=\frac{1}{i\omega}F(\omega)+\pi F(0)\delta(\omega),$$

where $F(\omega)$ is Fourier transform of $f(t)$. Could anyone explain to me how to prove this?

thanks.

Matt L.
  • 10,636
user62498
  • 3,556
  • 1
    In general, we cannot assign meaning to $\mathscr{F}{u*f}$.

    For example, if $f(t)=u(t)$, then $u*u=tu(t)$. I showed in THIS ANSWER that in distribution

    $$\mathscr{F}{ u*u}=\left(-\frac1{\omega^2}\right)+i\pi \delta'(\omega)\tag1$$

    where $\left(-\frac1{\omega^2}\right)$ is the distributional derivative of $\text{PV}\left(\frac1{\omega}\right)$.

    The right-hand side is *not* equal to $\pi F(\omega)\delta(\omega)+\text{PV}\left(\frac{F(\omega)}{i\omega}\right)$.

     – Mark Viola Jun 15 '21 at 18:00

1 Answers1

3

One way to look at it is to note that

$$\int_{-\infty}^tf(\tau)d\tau=(f\star u)(t)\tag{1}$$

where $\star$ denotes convolution, and $u(t)$ is the unit step function. Consequently, the Fourier transform of $(1)$ is

$$\mathcal{F}\left\{\int_{-\infty}^{t}f(\tau)\, d\tau\right\}=F(\omega)U(\omega)\tag{2}$$

where $F(\omega)$ and $U(\omega)$ are the Fourier transforms of $f(t)$ and $u(t)$, respectively. The Fourier transform of $u(t)$ is

$$U(\omega)=\pi\delta(\omega)+\frac{1}{i\omega}\tag{3}$$

A discussion of $(3)$ can be found here and here. With $(3)$, $(2)$ can be written as

$$\mathcal{F}\left\{\int_{-\infty}^{t}f(\tau)\, d\tau\right\}=\pi F(\omega)\delta(\omega)+\frac{F(\omega)}{i\omega}=\pi F(0)\delta(\omega)+\frac{F(\omega)}{i\omega}\tag{4}$$

Community
  • 1
Matt L.
  • 10,636
  • @Dear Matt L., Thank you for reading and taking the time for the great responses. Really appreciate it! – user62498 Nov 29 '15 at 12:00
  • @MattL. Hi Matt. In general, we cannot assign meaning to $\mathscr{F}{u*f}$.

    For example, if $f(t)=u(t)$, then $u*u=tu(t)$. I showed in THIS ANSWER that in distribution

    $$\mathscr{F}{ u*u}=\left(-\frac1{\omega^2}\right)+i\pi \delta'(\omega)\tag1$$

    where $\left(-\frac1{\omega^2}\right)$ is the distributional derivative of $\text{PV}\left(\frac1{\omega}\right)$.

    The right-hand side is *not* equal to $\pi F(\omega)\delta(\omega)+\text{PV}\left(\frac{F(\omega)}{i\omega}\right)$.

     – Mark Viola Jun 15 '21 at 17:59
  • @mattl. I'm not sure if you had a chance to read my comment. Please let me know. – Mark Viola Jun 26 '21 at 03:33
  • @MarkViola: Yes, I need to check, but one should add certain conditions on $f(t)$ for this relation to be true. Obviously, $F(0)$ must exist, and moreover, $F(\omega)$ must exist in the conventional sense, not as a distribution. – Matt L. Jun 26 '21 at 12:22