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This question is about heuristics to understand with the tempered distribution given by the oscillatory integral $\int xe^{ix\xi} d\xi$ (all integrals are over $\mathbb R$). Source is M. Zworski's Semiclassical Analysis book, page 54.

Maybe it bears mentioning: the author is "not a crank" and this was supposed to be a simple example as more calculations are yet to come in later chapters. I am aware how to compute things rigourously, but it seems that the author has a way to guess the right answer beforehand that I cannot understand.

PS $\int x e^{ix\xi} d\xi= x\int e^{ix\xi} d\xi = 2\pi x\delta = 0$ as tempered distributions (obviously just heuristically but do it with Fourier inversion starting from \eqref{B} below). In contrast with the previous form of reuns' answer below which computed $\int i\xi e^{ix\xi}d\xi =D_x\int e^{ix\xi}d\xi =2\pi \delta'$.

First, its alleged that "integration by parts" (quotation marks are from the book) shows $$ \int x e^{ix\xi} d\xi = \int D_\xi(e^{ix\xi})d\xi = 0.$$ (Nevermind the missing $-i$ since things should be zero) I suppose the author intended me to envision the following? $$ \int D_\xi (e^{ix\xi}) d\xi = e^{ix\xi}|_{-\infty}^{+\infty} -\int(D_\xi1)e^{ix\xi}d\xi\tag{IBP}\label{IBP}$$ but I don't have the 'feeling' that this 'should' evaluate to zero.

Second, the author tries to justify this as follows: for some large $k$, $$ u(\psi) := \lim_{\epsilon\to 0+} \iint xe^{ix\xi} \psi(x) \chi(\epsilon\xi) dxd\xi = \iint \frac x{(1+\xi^2)^k} e^{ix\xi} (1-\partial_x^2)^k \psi(x) dxd\xi\tag{A}\label{A}$$ From here, it is said that integration by parts in $\xi$ gives the result. But I think that there's an error here and I cannot commute $x$ with the $x$ derivatives, $[(1-\partial_x^2), x]\psi = 2\psi'$? Instead what I get is either $$ u(\psi) = \iint \frac 1{(1+\xi^2)^k} e^{ix\xi} (1-\partial_x^2)^k \Big(x\psi(x)\Big) dxd\xi\tag{B}\label{B}$$ or $$ u(\psi) = \iint \frac x{(1+\xi^2)^k} e^{ix\xi} L\psi(x) dxd\xi\tag{C}\label{C}$$ where $L\neq(1-\partial_x^2)^k$ is some other differential operator. From \eqref{B}, I can conclude with the Fourier inversion formula, but I see no integration by parts. Whereas if I start from \eqref{C} and try to follow the steps in \eqref{IBP}, I'm led to $$ \int L\psi(x) \left(\int (1+\xi^2)^{-k} D_\xi e^{ix\xi}d\xi \right) dx = \int L\psi(x) \left( \left.\frac{e^{ix\xi}}{(1+\xi^2)^k}\right|_{\xi=-\infty}^\infty - \int D_\xi((1+\xi^2)^{-k}) e^{ix\xi} d\xi\right)d x$$ and while there is now decay to say $\left.\dfrac{e^{ix\xi}}{(1+\xi^2)^k}\right|_{\xi=-\infty}^\infty=0$, its not immediately obvious to me that the other term should be zero.

Questions:

  • Should \eqref{IBP} be intuitive? If it should, how far is this intuition correct, is it e.g. correct only if I see power law growth in $\xi$?

  • Did I make a mistake?

  • What is the intended calculation?

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  • $\begingroup$ I don’t want to bump the question for a minor edit but there is no missing $-i$ because in this book $D:=\partial/i$ $\endgroup$ – Calvin Khor Mar 7 at 5:12
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There is no heuristic and you should not try to "evaluate" divergent integrals that converge only in the sense of distributions.

$$h_A(x)=x\int_{-A}^A e^{ixt}dt =2\sin(A x)$$ For $\phi \in C^\infty_c$ $$\lim_{A\to \infty}\int_{-\infty}^\infty h_A(x)\phi(x)dx= \lim_{A\to \infty}\int_{-\infty}^\infty \frac{2\cos(Ax)}{A} \phi'(x)dx=0$$ ie. $$\lim_{A\to \infty} h_A = 0 \text{ in the sense of distributions}$$

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  • $\begingroup$ I agree that it’s dodgy but I still want to know what point that the author is making. Also I believe you misread, it is indeed zero because i wrote a $d\xi$ integral so it eventually simplifies to $x\delta=0$? $\endgroup$ – Calvin Khor Mar 5 at 23:22
  • $\begingroup$ Yes but it doesn't change the idea, forget about dodgy calculations, there are no such, if the book says it does then pick another one. $\endgroup$ – reuns Mar 6 at 10:25
  • $\begingroup$ I do not plan to rely on it, or necessarily even use it, but I would like to know what the author wanted to say and why he thinks it is useful. $\endgroup$ – Calvin Khor Mar 6 at 10:46
  • $\begingroup$ Nothing, people are just wrong when they are trying to give heuristics on distributions. $\endgroup$ – reuns Mar 6 at 10:49
  • $\begingroup$ See how my answer is trivial (once the definition of the distribution $x\int_{-\infty}^\infty e^{ixt}dt$ is known, ie. $\lim h_A$), zero reason to ask for an heuristic. $\endgroup$ – reuns Mar 6 at 10:52

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