The Dirac Delta Function

The Dirac Delta Function Swapnil Sunil Jain December 27, 2006

Definition

The Dirac delta function can be defined as

\displaystyle\delta(t)\triangleq\lim_{\epsilon\to 0}\frac{1}{\epsilon}\Pi\Big{% (}\frac{t}{\epsilon}\Big{)}

where $\Pi(t)$ is the pulse function. The delta function can also be defined as

\displaystyle\delta(t)=\begin{cases}0&\mbox{for }t\neq 0\\ \infty&\mbox{for }t=0,\end{cases}\qquad\mbox{s.t }\int_{-\infty}^{+\infty}% \delta(t)dt=1

or it can also be defined as an operator s.t.

\displaystyle\delta(t)[f(t)]\triangleq\lim_{\epsilon\to 0}\frac{1}{\epsilon}% \int_{-\infty}^{+\infty}\Pi\Big{(}\frac{t}{\epsilon}\Big{)}f(t)dt

\displaystyle\mbox{1. }\int_{-\infty}^{\infty}f(t)\delta(t-a)dt=f(a)

Proof:

	$\displaystyle\int_{-\infty}^{\infty}f(t)\delta(t-a)dt$	$\displaystyle=\lim_{\epsilon\to 0}\Big{(}\int_{-\infty}^{a-\epsilon}f(t)\delta% (t-a)dt+\int_{1-\epsilon}^{a+\epsilon}f(t)\delta(t-a)dt+\int_{a+\epsilon}^{% \infty}f(t)\delta(t-a)dt\Big{)}$
		$\displaystyle=\lim_{\epsilon\to 0}\Big{(}\int_{a-\epsilon}^{a+\epsilon}f(t)% \delta(t-a)dt\Big{)}$
		$\displaystyle=f(a)\lim_{\epsilon\to 0}\Big{(}\int_{a-\epsilon}^{a+\epsilon}% \delta(t-a)dt\Big{)}=f(a)(1)=f(a)$

\displaystyle\mbox{2. }\int_{-\infty}^{\infty}f(t)\delta(t)dt=f(0)

Proof: Readily seen if we set $a=0$ in Property #1.

\displaystyle\mbox{3. }\int_{-\infty}^{\infty}f(t)\delta(at)dt=\frac{1}{a}f(0)

Proof: Set $u=at$ , then

\displaystyle\int_{-\infty}^{\infty}f(t)\delta(at)dt

\displaystyle=\int_{-\infty}^{\infty}f(\frac{u}{a})\delta(u)\frac{du}{a}=\frac% {1}{a}f(0)

\displaystyle\mbox{4. }\int_{-\infty}^{\infty}f(t)\delta(at-t_{0})dt=\frac{1}{% a}f(\frac{t_{0}}{a})

Proof: Set $u=at-t_{0}$ , then

\displaystyle\int_{-\infty}^{\infty}f(t)\delta(at-t_{0})dt

\displaystyle=\int_{-\infty}^{\infty}f(\frac{u-t_{0}}{a})\delta(u)\frac{du}{a}% =\frac{1}{a}f(\frac{t_{0}}{a})

\displaystyle\mbox{5. }\int_{-\infty}^{\infty}f(t)\delta(g(t))dt=\sum_{\{x_{i}% \;|\;g(x_{i})\;=\;0\}}\frac{1}{|g^{\prime}(x_{i})|}f(x_{i})

\displaystyle\mbox{6. }\int_{-\infty}^{\infty}f(t)\delta^{\prime}(t)dt=-f^{% \prime}(0)

Proof: If we let $u=f(t)$ and $dv=\delta^{\prime}(t)dt$ , then, using integration by parts,

\displaystyle\int_{-\infty}^{\infty}f(t)\delta^{\prime}(t)dt

\displaystyle=f(t)\delta(t){\Big{|}}_{-\infty}^{\infty}-\int_{-\infty}^{\infty% }f^{\prime}(t)\delta(t)dt=-f^{\prime}(0)