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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?

Mark Viola
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user342314
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  • This is not correct, because it does not take into account the discontinuity at $t=0$. – Math1000 Jun 19 '18 at 19:57
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    Since $xe^{-\varepsilon x} H(x) \to xH(x)$ for $\varepsilon \downarrow 0$ in the sense of tempered distributions: yes, you can. – Daniel Fischer Jun 19 '18 at 20:05
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    @danielfischer A naive application of that limit neglects the contribution of the Dirac Doublet (i.e., $\delta'(\omega)$). – Mark Viola Jun 19 '18 at 20:59
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    @MarkViola Yes, but hopefully the knowledge that one is working with tempered distributions reduces one's naïveté. – Daniel Fischer Jun 19 '18 at 21:11

3 Answers3

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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 $(2)$.

Mark Viola
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  • There's a typo in $1,1/\omega^2$. $PV(1/\omega^2)$ hasn't been defined, you do not say how it acts on an arbitrary test function. Suppose you have a functional $g$ that gives zero for every test function $f$ s.t. $f(0)=0$ (I assume that's what you mean by $f(\omega)=0$). That information is not sufficient, the $\delta$ term in $g$ is arbitrary. Incidentally, $PV$ is a rather misleading notation for a regularization of $1/\omega^2$. Also, unless you define that regularization in a non-standard way to include a $\delta'(\omega)$ term, the sign at $\delta'(\omega)$ in the answer is wrong. – Maxim Jun 20 '18 at 00:30
  • I don't see any edits though. If you know only how the functional acts on test functions that are vanishing at zero, how can you say that you've fully defined the functional? Do you agree with what I said about the $\delta$ component being arbitrary? In the example from my first comment, can you say that $g$ is identically zero? The sign is wrong because of the error in the evaluation of the integral of $\omega/(\omega-i\epsilon)^2$. – Maxim Jun 20 '18 at 13:59
  • Then it's simply not a well-defined distribution at all, if it doesn't map every test function to $\mathbb C$. You lose all the nice properties of distributions, e.g., you cannot compute the convolution of this object with a distribution, because the convolution requires the action on $f(x+y)$, and now you cannot restrict $f$ in a suitable way. Look at it this way: you can add $\delta(\omega)$ to your answer and say it is also an answer; there's nothing in your definition that rules out a $\delta$ term. You keep ignoring this point. – Maxim Jun 20 '18 at 15:32
  • There's a typo in the second line of (1) now, the minus sign before $1/\omega^2$ is missing. A quick way to see that the sign of $\delta'$ was wrong is to consider the transform of $H(x)$. – Maxim Jun 20 '18 at 15:33
  • For the record, I believe the corrected answer still is effectively indeterminate. Let me present one more argument. The object that you denote by $PV(1/\omega^2)$ is not the distributional derivative of $-PV(1/\omega)$, because the distributional derivative is a well-defined functional and therefore doesn't return any signed infinities, which you call complex numbers for some reason. Therefore all Fourier transform identities involving derivatives break down. – Maxim Jun 21 '18 at 12:39
  • $t H(t)$ certainly has a Fourier transform, see my answer. I'm saying that what you denote by $PV(1/\omega^2)$ either isn't a distribution (if you choose to introduce the infinities) or, at best, is a distribution that has an indeterminate $C \delta(\omega)$ term (see my comment above that mentions that you can add $\delta(\omega)$ to your answer), that's why it's not correct. – Maxim Jun 21 '18 at 13:15
  • Remember that we're dealing with singular functionals, whose action isn't given by $\int g f$. $w^{-2}$ is fully defined by saying that it's the distributional derivative of $-w^{-1}$: $$(w^{-2}, f) = ((-w^{-1})', f) = (w^{-1}, f') = \ \operatorname{v.!p.} \int_{-\infty}^\infty \frac {f'(w)} w dw = \int_0^\infty \frac {f(w) + f(-w) - 2f(0)} {w^2} dw.$$ This is well-defined for any test function $f$. This is why $FP(1/w^2)$ ($FP$ standing for finite part) is preferable to $PV$ when just $w^{-2}$ can be ambiguous $-$ generally speaking, it's not the p.v. integral of $f(w)/w^2$. – Maxim Jun 21 '18 at 14:17
  • @Maxim I removed the unnecessary and inappropriate constraint that $f(0)=0$. Now, there is an extra term $2f(0)/\delta=\int_{|\omega|\ge\delta}\frac{f(0)}{\omega^2},d\omega$ that combines nicely with the integral $\int_{|\omega|\ge\delta}\frac{f(\omega)}{\omega^2},d\omega$. All is right with the world now. Thank you for your patience and your helping with getting it correct. I've deleted all of my prior comments to reduce the clutter and noise. – Mark Viola Jun 21 '18 at 21:58
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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.

paul garrett
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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.

Maxim
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  • (+1) for the clear and concise post. – Mark Viola Jun 21 '18 at 22:00
  • It might be useful and instructive to define how $-\omega^{-2}$ operates on a test function. That is, it would benefit future readers to discuss the "finite part" integral by folding the principal value of $\frac{f(\omega)-f(0)}{\omega^2}$. – Mark Viola Jun 22 '18 at 13:49