Fourier transform of $t H(t)$

I want to calculate the Fourier transform of $f(t):=t\cdot H(t)$ ($H$ denotes the Heaviside function). For the integral I got

$$\hat{f}(y)=\left[\left(\frac{1}{y^2}+\frac{ix}{y}\right)e^{-i\omega x}\right]_{x=0}^{\infty}$$

Now I'm having trouble getting handling the boundaries, since $|e^{-i\omega x}|=1$.

Could I maybe compute $\hat{g}$ for $g(x):=xe^{-\varepsilon x}H(x)$ for $\varepsilon>0$ and then send $\varepsilon \rightarrow 0$ to get the transform I want?

Solution 1:

If we proceed using the limiting procedure discussed at the end of the question, we find that for $\omega\ne0$

$$\begin{align} \lim_{\epsilon\to0^+}\int_{-\infty}^\infty xH(x)e^{-\epsilon x}e^{-i\omega x}\,dx&=\lim_{\epsilon\to0^+}\int_{0}^\infty xe^{-(\epsilon+i\omega) x}\,dx \\\\ &=\lim_{\epsilon\to 0^+}\frac1{(\epsilon+i\omega)^2}\\\\ &=-\frac1{\omega^2} \end{align}$$

But this is not the Fourier Transform of $tH(t)$.

We now proceed to evaluate the distributional limit $\displaystyle \lim_{\epsilon\to 0^+}\frac1{(\epsilon+i\omega)^2}$ by applying it to a suitable test function.

Let $f$ be a smooth function of compact support. We can write for any $\delta>0$

$$\begin{align} \lim_{\epsilon\to 0^+}\int_{-\infty}^\infty \frac{f(\omega)}{(\epsilon+i\omega)^2}\,d\omega&=\lim_{\epsilon\to 0^+}\int_{|\omega|\ge \delta}\frac{f(\omega)}{(\epsilon+i\omega)^2}\,d\omega+\lim_{\epsilon\to 0^+}\int_{|\omega|\le \delta}\frac{f(\omega)}{(\epsilon+i\omega)^2}\,d\omega\\\\ &=-\int_{|\omega|\ge \delta}\frac{f(\omega)}{\omega^2}\,d\omega-\lim_{\epsilon\to 0^+}\int_{|\omega|\le \delta}\frac{f(\omega)}{(\omega -i\epsilon)^2}\,d\omega\\\\ &=-\int_{|\omega|\ge \delta}\frac{f(\omega)}{\omega^2}\,d\omega-\lim_{\epsilon\to 0^+}\int_{|\omega|\le \delta}\frac{f(0)+f'(0)\omega}{(\omega -i\epsilon)^2}\,d\omega+O(\delta)\\\\ &=-\int_{|\omega|\ge \delta}\frac{f(\omega)}{\omega^2}\,d\omega+2\frac{f(0)}{\delta}-i\pi f'(0)+O(\delta)\\\\ &=-\int_{|\omega|\ge \delta}\frac{f(\omega)-f(0)}{\omega^2}\,d\omega-i\pi f'(0)+O(\delta)\tag1 \end{align}$$

Letting $\delta\to0$ in $(1)$ reveals

$$\begin{align}\lim_{\epsilon\to 0^+}\int_{-\infty}^\infty \frac{f(\omega)}{(\epsilon+i\omega)^2}\,d\omega&=-\text{PV}\int_{-\infty}^\infty\frac{f(\omega)-f(0)}{\omega^2}\,d\omega-i\pi f'(0)\\\\ &=\text{PV}\int_{-\infty}^\infty\left(- \frac{1}{\omega^2}\right)\left( f(\omega) - f(0) \right)\,d\omega+i\pi\int_{-\infty}^\infty \delta'(\omega)f(\omega)\,d\omega\tag2 \end{align}$$

We deduce from $(2)$ that in distribution

$$\lim_{\epsilon\to 0^+}\frac1{(\epsilon+i\omega)^2}=\left(-\frac1{\omega^2}\right)+i\pi \delta'(\omega)$$

where the distribution $\displaystyle \left(-\frac1{\omega^2}\right)$ is interpreted in the sense of $(2)$.

Solution 2:

In addition to @MarkViola's good answer, and his and @DanielFischer's good comments, one might also think in terms of degrees of homogeneity of distributions, plus "sign". That is, $t\cdot H(t)$ is a linear combination of $|x|$ and of $\hbox{sgn}(x)\cdot |x|$, which are (positive-) homogeneous of degree $1$ in both cases, one odd and one even.

It is an exercise to show that homogeneity (and parity) behave intelligibly under Fourier transform. The exact formulation depends on conventions about notation of "homogeneity"... This is in principle very well known, indeed, but beginners seem to chronically overlook it in favor of direct computation...

So, for example, one finds that up to constants the Fourier transform of $\hbox{sgn}(x)$ is (principal value integral attached to) $1/x$. And similarly in the cases at hand.

Solution 3:

The transform of $e^{−\epsilon x} H(x)$ is $1/(\epsilon + i w)$, which has a well-known distributional limit. We obtain $$\mathcal F[H] = \lim_{\epsilon \downarrow 0} \mathcal F[e^{-\epsilon x} H(x)] = -\lim_{\epsilon \downarrow 0} \frac i {w- i \epsilon} = -i w^{-1} + \pi \delta(w), \\ \mathcal F[x H(x)] = i \frac d {dw} \mathcal F[H] = -w^{-2} + i \pi \delta'(w),$$ where $w^{-1}$ is the p.v. functional and $-w^{-2}$ is its distributional derivative.