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.


First, recognise the integral of $x(t)$ $$\int_{-\infty}^{t} x(\tau) d\tau \tag{1}$$ as a convolution with $u(t)$ $$ \int_{-\infty}^{t} x(\tau) d\tau =x(t) \star u(t) \tag{2}$$ where $u(t)$ is the unit-step function.

Then from the convolution property of the CTFT, one gets the following :

$$ \mathcal{F}\{ \int_{-\infty}^{t} x(\tau) d\tau \} = \mathcal{F}\{ x(t)\star u(t) \} = X(\omega)U(\omega) \tag{3}$$ where $U(\omega)$ is the Fourier transform of the unit-step function which is $$U(\omega) = \frac{1}{j\omega} + \pi \delta(\omega) \tag{4}$$ Then the Fourier transform of the integral of $x(t)$ is

$$ \mathcal{F}\{ \int_{-\infty}^{t} x(\tau) d\tau \} = \frac{ X(\omega) }{j\omega} + \pi X(0) \delta(\omega) \tag{5}$$

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

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 $u(\tau)$.



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


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


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


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


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:


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


  • $\begingroup$ 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

In my opinion, this might be a valid derivation.

Figure: The real axis integral representation in the complex $\omega = \omega^{\prime}+j\omega^{\prime\prime}$ domain. Note that $ |e^{j\omega t}| = e^{-\omega^{\prime\prime}t} $ diminishes at infinite radius in the upper half space for $t>0$, and vice versa for $t<0$.

Considering the Fourier transform pairs

\begin{equation} \begin{split} X(\omega) &= \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \\ x(t) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega \\ \end{split} \label{equation1} \end{equation}

The integral property is described by

\begin{equation} \int_{-\infty}^{t} x(\tau) d\tau = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) \Big[\int_{-\infty}^{t} e^{j\omega\tau} d\tau\Big] d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega \end{equation}

The above integral is singular at $\omega = 0$ and it is incorrect to use it as it is unless you incorporate Jordan's lemma and Cauchy's integral theorem. In this case, you have to deal with the hassle of $t>0$ and $t<0$ according to the Figure, which is more rigorous than the following conventional treatment commonly found in textbooks. Alternatively, this integral can be spitted into three parts in the complex $ \omega $ domain because of the singularity at $ \omega=0 $ as shown in Figure.

\begin{equation} \begin{split} \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega = \frac{1}{2\pi} \lim_{r\to0} \Bigg[ & \underbrace{\int_{-\infty}^{re^{-j\pi}} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega}_{I} + \underbrace{\int_{re^{j0}}^{\infty} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega}_{II} \\ & + \underbrace{\int_{-\pi}^{0} X(re^{j\theta})e^{j re^{j\theta}t} d\theta}_{III} \Bigg] \end{split} \end{equation}

In the third integral above, we substituted $ \omega =re^{j\theta} $ and $ d\omega =jre^{j\theta} d\theta $ in order to implement the detour illustrated in Figure. After taking the limit, this results in

\begin{equation} \begin{split} \int_{-\infty}^{t} x(\tau) d\tau & = \frac{1}{2\pi} \Bigg[\mathcal{P.V.} \int_{-\infty}^{\infty} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega + \pi X(0)\Bigg] \\ & = \frac{1}{2\pi} \Bigg[\mathcal{P.V.} \int_{-\infty}^{\infty} \frac{X(\omega)}{j\omega} e^{j\omega t} d\omega + \int_{-\infty}^{\infty} \pi\delta(\omega) X(\omega) e^{j\omega t} d\omega \Bigg] \\ & = \frac{1}{2\pi} \int_{-\infty}^{\infty} \Big(\underbrace{\frac{1}{j\omega}}_{\omega\neq0}+ \pi \delta(\omega) \Big) X(\omega) e^{j\omega t} d\omega \\ \end{split} \end{equation}

Hence, the effect of the integral operator on $ x(t) $ in $ \omega $ domain becomes

\begin{equation} \int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \big[\underbrace{1/j\omega}_{\omega \neq 0}+ \pi \delta(\omega)\big] X(\omega) \end{equation}

It is easy to show that the effect of the derivative operator on $ x(t) $ in $ \omega $ domain is

\begin{equation} \frac{d}{dt}x(t) \Longleftrightarrow j\omega X(\omega) \end{equation}

Taking the derivative of the integral of $ x(t) $ (i.e., $ \frac{d}{dt} \int_{-\infty}^{t} x(\tau) d\tau = x(t)$) results in

\begin{equation} j\omega \big[1/j\omega+ \pi \delta(\omega)\big] X(\omega) = X(\omega) \end{equation}

which illustrates the reversibility of the integral operator in the $ \omega $ domain after applying the derivative operator. A typical example would be the unit step function $ u(t) $

\begin{equation} u(t)= \int_{-\infty}^{t} \delta(\tau) d\tau \end{equation}

From the previous, the Fourier transform is $ U(\omega) = 1/j\omega + \pi\delta(\omega) $.


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