# Fourier Transform of $|t|$

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.

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

$$(f\star\delta')(\omega)=f'(\omega)$$

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

$$\int_{-\infty}^{\infty}g'(t)\phi(t)dt=-\int_{-\infty}^{\infty}g(t)\phi'(t)dt\tag{1}$$

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):

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

Hence,

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

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

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

• 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! Commented Jan 9 at 21:34
• 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! Commented Jan 9 at 21:36
• @AhsanYousaf: Note that $\delta'(t)$ is just the impulse response of an ideal differentiator: $j\omega\Longleftrightarrow\delta'(t)$. Hence, $(f\star\delta')(t)=f'(t)$. Commented Jan 9 at 21:38