# Deriving the integration property of the Fourier Transform

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.

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

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

The integral property is described by

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

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.

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

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

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

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

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

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

$$$$\frac{d}{dt}x(t) \Longleftrightarrow j\omega X(\omega)$$$$

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

$$$$j\omega \big[1/j\omega+ \pi \delta(\omega)\big] X(\omega) = X(\omega)$$$$

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 $$u(t)$$

$$$$u(t)= \int_{-\infty}^{t} \delta(\tau) d\tau$$$$

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

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

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

• 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." Commented Feb 9, 2018 at 18:45
• 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. Commented Feb 9, 2018 at 19:00
• Your response just changes the question to finding the CTFT of $\frac 12 \sgn(t)$ which also must be found somehow. Commented Feb 10, 2018 at 2:32
• @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 $$\tfrac12 \operatorname{sgn}(t)$$ Commented Feb 10, 2018 at 2:43
• @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. Commented Feb 10, 2018 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 $u(\tau)$.

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

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

• 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.
– user33568
Commented Feb 9, 2018 at 19:26
• @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. Commented Feb 9, 2018 at 21:41
• @0MW: I added the derivation of the Fourier transform of $u(t)$ to my answer. Commented Feb 10, 2018 at 9:36
• How can we conclude that equation (9) is true based on equation (8)? Why not just divide both sides by $j\omega$?
– user33568
Commented Feb 10, 2018 at 15:38
• I get that (9) satisfies (8) but how do we conclude (9) based on (8)? It's a bit counter-intuitive.
– user33568
Commented Feb 10, 2018 at 15:44