Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We have an integral representation for the Dirac delta function as

$\delta(x) = \frac{1}{2\pi} \int_{-\infty}^\infty dk e^{ikx}$ .

On the other hand, we have for delta function the property: $\delta(\alpha x) = \frac{\delta(x)}{\vert \alpha \vert}$ , which, I believe, should work also for complex values of $\alpha$.

Putting these two together gives

$\delta (i x) = \frac{1}{2\pi} \int_{-\infty}^\infty dk e^{-kx}$ .

This formula doesn't make sense since the integral clearly diverges for $k\rightarrow -\infty$. On the other hand, one would expect that

$\int_{-\infty}^\infty \delta(x) f(x) dx = \int_{-\infty}^\infty \delta(ix) f(x) dx = f(0)$ .

What is going on here? How should one write the integral representation for the delta function with imaginary/complex arguments?

share|cite|improve this question

migrated from Mar 17 '14 at 16:45

This question came from our site for users of Mathematica.

If I attempted to answer this cold I would butcher it but I know where I would go for help: the book "Generalized Functions" by Gelfand and Shilov. Also if you google for the book you'll find various pdfs that might be faster than a trip to your library. – user1688949 Mar 17 '14 at 20:56
Why do you say $\delta(\alpha x) = \delta(x)/|\alpha|$ should work for complex $\alpha$? – user7530 Mar 17 '14 at 22:28
Well, according to the answer below, it doesn't so this was a false assumption from me. – Echows Mar 18 '14 at 16:07

I think the confusion comes the false generalisation to complex $\alpha$. Let's look at were this identity comes from.

By definition, for a test function $\phi$ $$\langle \delta(x),\phi(x)\rangle = \phi(0).$$ If we represent $\delta(x)$ as a limit of sequence of $C^\infty_c$ functions in the sense of distributions, let's note it $g_n(x)$, then we can write an integral

$$\int_{\Bbb R}g_n(x)\phi(x) dx\to\langle \delta(x),\phi(x)\rangle=\phi(0).$$ If we make the change of variables,

$$\int_{\Bbb R}g_n(\alpha x)\phi(x) dx=\int_{\Bbb R}g_n(y)\phi(y/\alpha) dy/|\alpha|$$ $$\to\frac{1}{|\alpha|}\phi(0) = \left\langle \frac{\delta(x) }{|\alpha|},\phi(x)\right\rangle.$$ On the other hand, it's clear that $g_n(\alpha x)$ converges to $\delta(\alpha x)$ in the sense of distributions, so we can write $$\delta(\alpha x)=\frac{\delta(x) }{|\alpha|}.$$

However, this trick doesn't work for comlpex $\alpha$ at all, because $\phi(x)$ is defined only for real $x$, so $\delta (\alpha x)$ has no sense.

share|cite|improve this answer
So what would be the equivalent formula for complex $\alpha$ then? – Echows Mar 18 '14 at 16:06
@Echows I don't know if such formula exists. What's your motivation to study $\delta (\alpha x)$ for complex $\alpha$? – TZakrevskiy Mar 18 '14 at 16:08
Actually I'm interested in the divergent integral $\int dk e^{-kx}$. My original thinking was that maybe there would be a way to assign a value to that integral in terms of a delta function. Kind of in the same spirit as in zeta regularization scheme one assigns finite values to diverging sums. – Echows Mar 19 '14 at 13:09

When considering a complex number $z$ , you can introduce the product of the delta function of the real and imaginary parts of $z$ , and if you wish you can call that $\delta (z) = \delta (\Re z) \delta( \Im z)$.

The fundamental object (the delta function) remains the delta function of a real number.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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