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?



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


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

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

  • $\begingroup$ @Dear Matt L., Thank you for reading and taking the time for the great responses. Really appreciate it! $\endgroup$ – user62498 Nov 29 '15 at 12:00

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