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I want to derive the property of the Fourier Transform that states that if $X(j\omega) = \mathcal{F} (x(t))$ then $$\mathcal{F} \left( \int_{-\infty}^{t} x(\tau) \mathrm{d} \tau \right) = \frac{1}{j\omega} X(j\omega) + \pi X(0) \delta(\omega).$$ I started as follows, since $$x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) e^{j\omega t} \mathrm{d}\omega,$$ integrating both sides gives us $$\int_{-\infty}^{t} x(\tau) \mathrm{d} \tau = \int_{-\infty}^{t} \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) e^{j\omega\tau} \mathrm{d}\omega\mathrm{d} \tau$$ $$\int_{-\infty}^{t} x(\tau) \mathrm{d} \tau = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) \left(\int_{-\infty}^{t} e^{i\omega \tau} \mathrm{d}\tau \right) \mathrm{d}\omega.$$

How do I evaluate the inner integral inside the parantheses? My gut feeling says that it's related to the Fourier Transform of the constant function $y(t) = 1$ but I can't relate it to that since the upper limit is $t$ rather than $\infty$. How do I reach the result above?

Thank you in advance.

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The actual derivation might be involved, instead here I would like to take the following simpler (but equivalent nevertheles) approach:

Let's define the integral $$\int_{-\infty}^{t} x(\tau) d\tau $$ as $$ x(t) \star u(t)$$ where $u(t)$ is the unit step function. Then from the convolution (in time) - multiplication (in frequency) property of the CTFT, one can get the following :

$$ \mathcal{F}\{ x(t)\star u(t) \} = X(\omega)U(\omega) $$ where $U(\omega)$ is the Fourier transform of the unit step function. Therefore your problem now in effect is reduced to the CTFT of the unit step. As most textbooks on S&S do, I'll also simply state the CTFT of the unit step function and conclude the derivation. Since $$U(\omega) = \frac{1}{j\omega} + \pi \delta(\omega) $$ then the Fourier transform of the integral of $x(t)$ is

$$ \mathcal{F}\{ \int_{-\infty}^{t} x(\tau) d\tau \} = \frac{ X(\omega) }{j\omega} + \pi X(0) \delta(\omega)$$

The last term results from the sifting property of the dirac impulse.

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  • $\begingroup$ Which if course begs the question: "How did the textbook author come up with the CTFT of $u(t)$?" A cynic might answer "Copied it from an earlier text." $\endgroup$ – Dilip Sarwate Feb 9 '18 at 18:45
  • $\begingroup$ Not that deep I suppose! A related discusison was triggered even here by a question of MattL. But instead I prefer breaking $u(t)$ into an ac and dc part and consider adding the Fourier transforms separately. The ac part of $u(t)$ is the 0.5 sign(t) function whose CTFT is $1/jw$ where as the dc part is 0.5 whose CTFT is $\pi \delta(w)$. I find it enough for an intuitive understanding. The formalist could argue whether the decomposed terms whould be convergent individually or not. But that discussion belongs to generalized function theory then. $\endgroup$ – Fat32 Feb 9 '18 at 19:00
  • $\begingroup$ Your response just changes the question to finding the CTFT of $\frac 12 \sgn(t)$ which also must be found somehow. $\endgroup$ – Dilip Sarwate Feb 10 '18 at 2:32
  • $\begingroup$ @DilipSarwate, i know a lotta people just use "\text{sgn}", but the "correct" way to TeX a function that is not in the LaTeX library is with "\operatorname{.}". as with $$ \tfrac12 \operatorname{sgn}(t) $$ $\endgroup$ – robert bristow-johnson Feb 10 '18 at 2:43
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    $\begingroup$ @robertbristow-johnson I didn't use "text{sgn}" but rather "\sgn(t)" under the assumption that sgn was a function known to LaTeX (like det and gcd and lim) but I guess I was wrong. It must have been a newcommand declaration that I always include in LaTeX files for non-stackexchange use. $\endgroup$ – Dilip Sarwate Feb 10 '18 at 2:58
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As mentioned in Fat32's answer, the integration property can be derived directly from the Fourier transform of the unit step function.

I would like to show you how you can finish your derivation, even though you will also need the Fourier transform of the unit step. The integral

$$\int_{-\infty}^te^{j\omega \tau}d\tau\tag{1}$$

can be written as

$$\int_{-\infty}^{\infty}u(t-\tau)e^{j\omega\tau}d\tau=\int_{-\infty}^{\infty}u(t+\tau)e^{-j\omega\tau}d\tau=\mathcal{F}_{\tau}\{u(t+\tau)\}=e^{j\omega t}U(\omega)\tag{2}$$

where $\mathcal{F}_{\tau}$ denotes the Fourier transform with $\tau$ as the independent time variable (and not $t$), and $U(\omega)$ is the Fourier transform of the unit step function $u(\tau)$.

With

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

and with $(1)$ and $(2)$ we get

$$\int_{-\infty}^te^{j\omega \tau}d\tau=e^{j\omega t}\left(\frac{1}{j\omega}+\pi\delta(\omega)\right)\tag{4}$$

and the last integral in your question becomes

$$\begin{align}\int_{-\infty}^{t} x(\tau) \mathrm{d} \tau &= \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) \left(\int_{-\infty}^{t} e^{i\omega \tau} \mathrm{d}\tau \right) \mathrm{d}\omega\\&=\frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega)e^{j\omega t}\left(\frac{1}{j\omega}+\pi\delta(\omega)\right)d\omega\\&=\mathcal{F}^{-1}\left\{X(j\omega)\left(\frac{1}{j\omega}+\pi\delta(\omega)\right)\right\}\tag{5}\end{align}$$

from which it follows that

$$\mathcal{F}\left\{\int_{-\infty}^{t} x(\tau) \mathrm{d} \tau\right\}=X(j\omega)\left(\frac{1}{j\omega}+\pi\delta(\omega)\right)=X(j\omega)\frac{1}{j\omega}+\pi X(0)\delta(\omega)\tag{6}$$


What follows is a rather straightforward derivation of the Fourier transform of the unit step $u(t)$. If we accept without any further proof that the (generalized) derivative of the unit step is the Dirac delta impulse

$$u'(t)=\delta(t)\tag{7}$$

we get the following relation for the Fourier transform of $u(t)$:

$$j\omega U(\omega)=1\tag{8}$$

From $(8)$ we can conclude that $U(\omega)$ must have the form

$$U(\omega)=\frac{1}{j\omega}+c\delta(\omega)\tag{9}$$

Multiplying $(9)$ with $j\omega$ gives

$$1+cj\omega\delta(\omega)=1+0\cdot\delta(\omega)=1$$

satisfying $(8)$ (because for any function $f(\omega)$ that is continuous at $\omega=0$ we have $f(\omega)\delta(\omega)=f(0)\delta(\omega)$.)

It only remains to determine the constant $c$ in $(9)$. This can be done by considering the value of the inverse Fourier transform of $U(\omega)$ at $t=0$. Since the inverse Fourier transform gives the average of the left-sided and of the right-sided limit, and since $u(0^-)=0$ and $u(0^+)=1$, we have

$$\frac12=\frac{1}{2\pi}\int_{-\infty}^{\infty}U(\omega)d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\frac{1}{j\omega}+c\delta(\omega)\right]d\omega\tag{10}$$

The Cauchy principal value of the integral of the first term on the right-hand side of $(10)$ vanishes because it is an odd function of $\omega$, so we're left with the second term:

$$\frac12=\frac{1}{2\pi}\int_{-\infty}^{\infty}c\delta(\omega)d\omega=\frac{c}{2\pi}\tag{11}$$

from which $c=\pi$ follows. So from $(9)$ we finally get

$$U(\omega)=\frac{1}{j\omega}+\pi\delta(\omega)\tag{12}$$

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  • $\begingroup$ That's convincing except for the Fourier Transform of the step function. How do we derive the Fourier Transform of the step function then? I believe Oppenheim derives the Fourier Transform of the step function using the very property that I asked about (the integration property), so it seems like a circular argument. I can "derive" the Fourier Transform of the constant function by considering its Fourier Series representation, but I can't do the same for $u(t)$ since it's not periodic. $\endgroup$ – user33568 Feb 9 '18 at 19:26
  • $\begingroup$ @0MW: There's a rather simple derivation similar to the one here (which is for the discrete-time case). I'll add it to my answer as soon as I have the time to do so. $\endgroup$ – Matt L. Feb 9 '18 at 21:41
  • $\begingroup$ @0MW: I added the derivation of the Fourier transform of $u(t)$ to my answer. $\endgroup$ – Matt L. Feb 10 '18 at 9:36
  • $\begingroup$ How can we conclude that equation (9) is true based on equation (8)? Why not just divide both sides by $j\omega$? $\endgroup$ – user33568 Feb 10 '18 at 15:38
  • $\begingroup$ I get that (9) satisfies (8) but how do we conclude (9) based on (8)? It's a bit counter-intuitive. $\endgroup$ – user33568 Feb 10 '18 at 15:44

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