# Delta function with imaginary argument

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?

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## migrated from mathematica.stackexchange.comMar 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.

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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.

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