Delta functions live under integral signs, and two expressions $\left(D_1\left(x\right)and{D}_2\left(x\right)\right)$involving delta functions are said to be equal if

${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{+}\mathbf{\infty }}\mathit{f}\left(x\right){\mathit{D}}_{\mathit{1}}\left(x\right)\mathit{d}\mathit{x}\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{+}\mathbf{\infty }}\mathit{f}\left(x\right){\mathit{D}}_{\mathit{2}}\left(x\right)\mathit{d}\mathit{x}$for every (ordinary) function f(x).

(a) Show that

$\mathit{\delta }\left(cx\right)\mathbf{=}\frac{\mathbf{1}}{\left|c\right|}\mathit{\delta }\left(x\right)$(2.145)

where c is a real constant. (Be sure to check the case where c is negative.)

(b) Let $\mathit{\theta }\left(x\right)$ be the step function:

$\mathit{\theta }\left(x\right)\mathbf{\equiv }\{\begin{array}{l}1,x>0\\ 0,x>0\end{array}$(2.146).

(In the rare case where it actually matters, we define $\mathit{\theta }\left(0\right)$ to be 1/2.) Show that $\mathit{d}\mathit{\theta }\mathit{}\mathit{l}\mathit{}\mathit{d}\mathit{x}\mathit{}\mathit{=}\mathit{}\mathit{\delta }$

Delta function under integral signs is given as:

${\int }_{-\infty }^{+\infty }f\left(x\right)D_1\left(x\right)dx={\int }_{-\infty }^{+\infty }f\left(x\right)D_2\left(x\right)dx$

The step function is given as:

$\theta \left(x\right)=\left\{\begin{array}{l}1,x>0\\ 0,x<0\end{array}\right.$

The Dirac delta distribution is a distribution, where

• It is distributed over real numbers.
• Its value is zero, where x is not zero. Else value is infinity at all the other values of x .

Let $y=cx$

Then,

$dx=\frac{1}{c}dy.\left\{\begin{array}{l}Ifc>0,y:-\infty \to \infty \\ Ifc<0,y:\infty \to -\infty \end{array}\left.\begin{array}{r}\\ \end{array}\right\}\right.$

localid="1658205343693" ${\int }_{-\infty }^{\infty }f\left(x\right)\delta \left(cx\right)dx=\left\{\begin{array}{cc}\frac{1}{c}{\int }_{-\infty }^{\infty }f(y/c)\delta \left(y\right)dy=\frac{1}{c}f\left(0\right)(c>0);or& \\ \frac{1}{c}{\int }_{\infty }^{-\infty }f(y/c)\delta \left(y\right)dy=-\frac{1}{c}{\int }_{-\infty }^{\infty }f(y/c)\delta \left(y\right)dy=-\frac{1}{c}f\left(0\right)& (c<0)\end{array}\right.$

In either case,

$$\begin{gathered} {\int }_{-\infty }^{\infty }f\left(x\right)\delta \left(cx\right)dx=\frac{1}{\left|c\right|}f\left(0\right) \\ {\int }_{-\infty }^{\infty }f\left(x\right)\frac{1}{\left|c\right|}\delta \left(x\right)dx=\frac{1}{\left|c\right|}f\left(0\right) \end{gathered}$$

So,$\delta \left(cx\right)=\frac{1}{\left|c\right|}\delta \left(x\right)$

Thus, $\delta \left(cx\right)=\frac{1}{\left|c\right|}\delta \left(x\right)$.

Using method integration by parts to solve the integral as:

$$\begin{gathered} {\int }_{-\infty }^{\infty }f\left(x\right)\frac{d\theta }{dx}dx=f{\overline{)\theta }}_{-\infty }^{\infty }-{\int }_{-\infty }^{\infty }\frac{df}{dx}\theta dx \\ {\int }_{-\infty }^{\infty }f\left(x\right)\frac{d\theta }{dx}dx=f(\infty )-{\int }_0^{\infty }\frac{df}{dx}dx \\ =f\left(\infty \right)-f\left(\infty \right)+f\left(0\right) \\ =f\left(0\right) \\ ={\int }_{-\infty }^{\infty }f\left(x\right)\delta \left(x\right)dx \end{gathered}$$

So,

$d\theta /dx=\delta \left(x\right)$ .

[Makes sense: The θ function is constant (so the derivative is zero) except at x = 0, where the derivative is infinite.].

Thus, $d\theta /dx=\delta \left(x\right)$.