1
$\begingroup$

$$f(\omega,u):=\frac1{\omega+iu}$$ where $i$ is the imaginary unit number. We see that the integral of a Fourier transform $$\int_1^\infty du\int_{-\infty}^\infty d\omega\,f(\omega,u)\,e^{-i\omega x}=2\pi i\int_1^\infty du\,e^{-ux}=2\pi i\frac{e^{-x}}x$$ for $u>0$, $x>0$, while interchanging the order of integration, $$\int_{-\infty}^\infty d\omega \,e^{-i\omega x}\int_1^\infty duf(\omega,u)$$ is not even integrable in the first integration. Is there a theory, say distribution or generalized function, to treat this kind of double integral involving the Fourier transform?

On the other hand, if we integrate by part the Fourier transform $$\int_{-\infty}^\infty d\omega\,f(\omega,u)\,e^{-i\omega x}=\frac1 {ix}\int_{-\infty}^\infty d\omega\,\partial_\omega f(\omega,u)\,e^{-i\omega x},$$ $$\partial_\omega f(\omega,u)=-\frac1{(\omega+iu)^2}.$$ We interchange the order of integration and obtain a finite value $$\int_1^\infty du\,\partial_{iu} f(\omega,u)=if(\omega,u=1)=\frac i{\omega+i}$$ which is already shown to be Fourier transformable. It is easy to check the two orders of integration agree.

How do we turn this trick into a coherent theory? Does, say tempered distribution, help?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Replace $\lim_{n \to \infty} \int_{-n}^nh(x)dx$ by $\lim_{n \to \infty} \int_{-\infty}^\infty h(x)e^{-x^2/n^2}dx$ to extend $\int$ to a larger class of functions. The theory of distributions gives a setting where the result doesn't depend on how we extended $\int$. $\endgroup$
    – reuns
    Commented Feb 19, 2019 at 5:24
  • $\begingroup$ @reuns: I am having a bit difficulty applying the Schwartz test functions, say $\varphi(x)$, to derive the interchangeability in the framework of the distribution theory. Can you write a detailed answer formulating and deriving the interchangeability? Thank you. $\endgroup$
    – Hans
    Commented Feb 19, 2019 at 9:36

1 Answer 1

Reset to default
2
$\begingroup$

The subject of definite integrals for distributions was investigated in some detail by several mathematicians in the 50’s—-they used an elementary approach based on the fact that distributions are (locally) higher derivatives of continuous functions. This allows a treatment at the level of a (european) freshman calculus course, without using Schwartzian duality. You can find a lucid version online by googling the title „Integration and orders of growth of distributions“ of an article by J. Sebastião e Silva. The good news is that the integral you are looking at converges in the distributional sense and can be calculated as an iterated one as you ask—-this is 14.4 on p. 365. (The article isn‘t that long—-the pagination is due to the fact that it is one of a collection).

Improve this answer
$\endgroup$
1
  • $\begingroup$ +1. Thank you. Let me have a look. $\endgroup$
    – Hans
    Commented Mar 11, 2019 at 18:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .