5
$\begingroup$

for a real number we know that

$$ f(a)= \int_{-\infty}^{\infty}dx \delta (x-a)f(x) $$

but what happens for $$ \int_{-\infty}^{\infty}dx \delta (x-2i)f(x) $$ ?

is this equal to $ f(2i) $ or equal to $0 $ , of course $ i= \sqrt -1 $ a complex number :)

if i use the generalize funtion approach $ \delta (ix)= \frac{sinh(nx)}{x \pi} $ but in the limit $ n \to \infty $ this limit makes no sense

$\endgroup$
  • 2
    $\begingroup$ I'm personally only familiar with the definition over the reals, but if this has the usual properties of $\delta$ functions then this integral should be zero since $x \neq 2i$. $\endgroup$ – Spencer Feb 9 '14 at 11:19
  • $\begingroup$ I think the crux of the matter is that the first equation is in fact how we define $\delta(x-a)$, rather than being a consequence of the context. It's therefore left to us to define the second integral in whichever way makes sense in light of our specific needs. $\endgroup$ – Jonathan Y. Feb 9 '14 at 17:53
  • $\begingroup$ I'm not actually familiar with that property of real numbers, does anybody have a handy link discussing the properties of the delta function? $\endgroup$ – JacksonFitzsimmons Jun 29 '15 at 22:02
1
$\begingroup$

The integral $\int\limits_{-\infty}^{\infty}\delta(x-2i)\,f(x)\,dx$ doesn't make sense, but the integral illustrated in (1) below does make sense with the variable substitution $s=x+2\,i$ given certain assumptions on $f$ (e.g. complex analytic) .

(1) $\quad\int\limits_{-\infty+2\,i}^{\infty+2\,i}\delta(s-2i)\,f(s)\,ds=\int\limits_{-\infty}^{\infty}\delta(x)\,f(x+2\,i)\,dx=f(2\,i)$

An integral of $\delta(s)$ for $s\in\mathbb{C}$ makes sense if-and-only-if it can be mapped to an integral of $\delta(x)$ for $x\in\mathbb{R}$. This can be done for integrals along lines both parallel and perpendicular to the real axis which is illustrated in the following answer I posted to a similar question.

Delta function with imaginary argument

$\endgroup$
  • $\begingroup$ I have removed my comments. At least the added attention has woken up the OP. $\endgroup$ – robjohn Sep 4 '18 at 17:00
3
$\begingroup$

Depending on the application you can generalize the dirac delta distribution using its cauchy representation to get a reasonable analytic continuation on the complex plane. $$ f(z)=\oint_{C}\frac{f(\zeta)}{z-\zeta} $$ Here's a short paper that uses this perscription with a short description.look at equations 4.7 and 4.8 and the paragraph preceding them: http://arxiv.org/pdf/1212.2132v1.pdf

$\endgroup$
0
$\begingroup$

One precise way to define the $\delta$-function is as a unit point measure on $\mathbb{R}$, $${\delta}(x-a)dx=d{\mu_a}(x)$$ for real $a$. Given this, we want to extend it to have a meaning for non-real $a$. This looks problematic. Secondly, in your example, you have to require some analyticity to give a meaning to $f(2i)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.