I was going through Papoulis' book (The Fourier Integral and its Applications) when I came across the Fourier Transform for $|t|$. To find it he writes $|t|$ as (I am not sure how):

$$|t| = -\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\cos(\omega t)}{\omega^2}d\omega \tag{1}$$

and then states the Inverse Fourier Transform formula to write:

$$\mathcal{F}\{|t|\} = -\frac{2}{\omega^2}$$

which checks out since:

\begin{align} \mathcal{F}^{-1}\{-2/\omega^2 \} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}-\frac{2}{\omega^2}e^{j\omega t}d\omega\\ & = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{e^{j\omega t}}{\omega^2}d\omega\\ & = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\cos(\omega t)}{\omega^2}d\omega \qquad 1/\omega^2 \: \text{is an Even Function} \end{align}

and also for the forward Fourier Transform we can use the definition in $(1)$ as follows:

\begin{align} \mathcal{F}\{|t|\} & = -\frac{1}{\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{\cos(\omega_0 t)}{\omega_0^2}e^{-j\omega t}dt \:d\omega_0\\ & = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{1}{\omega_0^2}\int_{-\infty}^{\infty}\cos(\omega_0 t)e^{-j\omega t}dt \:d\omega_0\\ & = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{1}{\omega_0^2}\left[\pi (\delta(\omega_0-\omega)+\delta(\omega_0 +\omega) \right] \:d\omega_0\\ & = -\int_{-\infty}^{\infty}\frac{1}{\omega_0^2}\left[(\delta(\omega_0-\omega)+\delta(\omega_0 +\omega) \right] \:d\omega_0\\ & = -\left(\frac{1}{\omega^2}+\frac{1}{\omega^2}\right) = -\frac{2}{\omega^2} \end{align}

(apologies for any confusion with the two different $\omega$'s) but why does the definition in $(1)$ hold? Also is there a better way to find this Fourier Transform? I was thinking perhaps using the fact that:

$$|t| = t \:\mathtt{sgn}(t)$$

I could apply the convolution theorem where:

$$|t| \overset{\mathcal{F}}\longrightarrow \frac{1}{2\pi} \left(\mathcal{F}\{t\}\ast\mathcal{F}\{\mathtt{sgn}(t)\}\right)$$

where $\mathcal{F}\{\mathtt{sgn}(t)\} = \frac{2}{j\omega}$ and $\mathcal{F}\{t\}$ is:

$$tx(t) \overset{\mathcal{F}}\longrightarrow j\frac{d}{d\omega}X(j\omega)$$

where $x(t) = 1$ and $X(j\omega) = 2\pi \delta(\omega)$ which gives us:

$$t \overset{\mathcal{F}}\longrightarrow 2\pi j \delta'(\omega)$$

Writing the convolution we have:

\begin{align} \frac{1}{2\pi} \left(\mathcal{F}\{t\}\ast\mathcal{F}\{\mathtt{sgn}(t)\}\right) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} 2\pi j \delta'(y)\cdot \frac{2}{j(\omega-y)}dy\\ &= \int_{-\infty}^{\infty} \delta'(y)\cdot \frac{2}{(\omega-y)}dy \end{align}

which should equal $-\frac{2}{\omega^2}$ but I don't know how since I am not exactly sure how the derivative of the Dirac-Delta Impulse behaves under the integral or otherwise.

This was my attempt at getting a reasonable and well-understandable solution for the Fourier Transform of $|t|$. I would appreciate it if anyone could either explain how we can obtain $(1)$ or complete/fix my attempt using the convolution theorem (or perhaps my solution is completely wrong since $t$ is not absolutely integrable but I treated it in the realm of tempered distributions which may or may not work, I am not sure). Better yet if someone could offer an alternative solution that involves some rigorous mathematics then that would be wonderful.


1 Answer 1


For the derivation of Eq. $(1)$ in your question, Papoulis refers to Eq. ($I$-$32$) in the appendix on distributions. That equality is basically a consequence of how the derivative of a distribution is defined (see below).

Concerning your suggestion to compute the Fourier transform of $|t|$ by noting that

$$|t|=t \operatorname{sgn}(t)$$

and using the well-known Fourier identities

\begin{align*} t &\Longleftrightarrow 2\pi j\delta'(\omega)\\ \operatorname{sgn}(t) &\Longleftrightarrow \frac{2}{j\omega} \end{align*}

you just need to realize that for any differentiable function $f(\omega)$ we have


where $\star$ denotes convolution. Hence,

\begin{align*} \mathcal{F}\{|t|\} &= \frac{1}{2\pi}2\pi j\delta'(\omega)\star\frac{2}{j\omega} \\ &= \delta'(\omega)\star\frac{2}{\omega} \\ &= \left(\frac{2}{\omega}\right)' \\ &= -\frac{2}{\omega^2} \end{align*}

What follows is not a rigorous proof but an attempt to make Eq. $(1)$ of the question plausible. Let $g(t)$ be a distribution and $\phi(t)$ a well-behaved test function. The derivative $g'(t)$ of a distribution is defined by


Note that this definition is consistent with the rule for integration by parts, assuming that the product $g(t)\phi(t)$ vanishes for $|t|\to\infty$.

With $g'(t)=1/t^2$ and $\phi(t)=\cos\omega t$ we have $g(t)=-1/t$ and $\phi'(t)=-\omega\sin\omega t$. From $(1)$ it follows that

$$\int_{-\infty}^{\infty}\frac{\cos\omega t}{t^2}dt=-\omega\int_{-\infty}^{\infty}\frac{\sin\omega t}{t}dt\tag{2}$$

The integral on the right-hand side of $(2)$ is $\pi$ times the DC value of the frequency response of an ideal lowpass filter with cut-off frequency $\omega$ (for negative $\omega$ we have to invert the sign):

$$\int_{-\infty}^{\infty}\frac{\sin\omega t}{t}dt=\pi\operatorname{sgn}\omega\tag{3}$$


$$\int_{-\infty}^{\infty}\frac{\cos\omega t}{t^2}dt=-\omega\,\pi\operatorname{sgn}\omega=-\pi|\omega|\tag{5}$$

Exchanging the variables $\omega$ and $t$, we obtain

$$|t|=-\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\cos\omega t}{\omega^2}d\omega\tag{6}$$

Note that from $(5)$ we also obtain the Fourier transform pair

$$|\omega|\Longleftrightarrow-\frac{1}{\pi t^2}\tag{7}$$

  • $\begingroup$ Thank you for you answer. I am not sure why $(f\star\delta')(\omega)=f'(\omega)$ holds but I will try and work up a proof and add it as an addendum to my question or as a separate supporting answer! $\endgroup$ Jan 9 at 21:34
  • $\begingroup$ Also, as for the appendix reference, my copy of the Papoulis book is quite old and shabby so the reference to what you say is $\textbf{I-}32$ was too disfigured for me to understand. But thanks for pointing it out! $\endgroup$ Jan 9 at 21:36
  • 2
    $\begingroup$ @AhsanYousaf: Note that $\delta'(t)$ is just the impulse response of an ideal differentiator: $j\omega\Longleftrightarrow\delta'(t)$. Hence, $(f\star\delta')(t)=f'(t)$. $\endgroup$
    – Matt L.
    Jan 9 at 21:38

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