I have confusion on the Fourier representation of complex Dirac function, recently. As $t$ is real value, we have \begin{align} \delta(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{iwt}\text{d}w \end{align} However, the goal is to obtain Fourier representation of complex Dirac function.

What is I know. For complex $t$, there is \begin{align} \delta(t)=\delta(\text{Re}(t))\delta(\text{Im}(t)) \end{align} For simplification, we use $t_r=\text{Re}(t)$ and $t_m=\text{Im}(t)$. We then have \begin{align} \delta(t)&=\frac{1}{(2\pi)^2}\left({\int_{-\infty}^{+\infty}e^{iw_1t_r}\text{d}w_1}\right)\left({\int_{-\infty}^{+\infty}e^{iw_2t_m}\text{d}w_2}\right)\\ &=\frac{1}{(2\pi)^2}\int_{-\infty}^{\infty} e^{i\boldsymbol{w}^T\boldsymbol{t}}\text{d}\boldsymbol{w} \end{align} where $\boldsymbol{w}=\left(\begin{array}{ccc}{w_1\\w_2}\end{array}\right)$ and $\boldsymbol{t}=\left(\begin{array}{ccc}{t_r\\t_m}\end{array}\right)$.

But, how do I use 1-dimension complex integral to represent $\delta(t)$ rather than 2-dimension real integral.

  • $\begingroup$ It is probably better to just identify $\mathbb C$ and $\mathbb R^2$ here. That is, $\delta(z)=\delta(x)\delta(y)$, where $z=x+iy$. You are not using the complex multiplication in any way. $\endgroup$ – Giuseppe Negro Sep 21 '18 at 9:18

Your final expression is incorrect because it is written as an integral over $\mathbb{R}$ rather than $\mathbb{R}^2$; obviously, since you're integrating over both $w_1$ and $w_2$, your integration range should be two-dimensional. And now we just identify $\mathbb{R}^2$ with $\mathbb{C}$, so $w^Tt$ becomes $\operatorname{Re}(w^\ast t)$. Thus $\delta(t)=(2\pi)^{-2}\int_{\mathbb{C}}e^{i\operatorname{Re}w^\ast t} dt.$

  • $\begingroup$ Thank you very much for your answer. The integral is w.r.t vector $\boldsymbol{w}$ in my final expression. There is a clerical error. The variable of integration is w.r.t. complex variable $w$ rather than $t$. However, I don't know whether $\text{d} (w_1+iw_2)$ is equal to $\text{d}[w_1,w_2]^T$. Your response is expected. Thanks a lot. $\endgroup$ – Qiuyun.Zou Sep 21 '18 at 14:06

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