91Ó°ÊÓ

The Dirac Delta function which is represented as $\mathit{δ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, is defined as $\mathit{θ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{ll}\mathbf{1}& \mathbf{}\mathbf{x}\mathbf{≠}\mathbf{0}\\ \mathbf{0}\mathbf{,}& \mathbf{}\mathbf{x}\mathbf{=}\mathbf{0}\end{array}$, .the Dirac delta function has the property ${\mathbf{∫}}_{\mathbf{-}\mathbf{∞}}^{\mathbf{∞}}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{δ}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{d}\mathit{x}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$, where $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is a continuous containing$\mathit{x}\mathbf{=}\mathbf{0}$.

(a)

Let function $u\left(x\right)=\mathrm{δ}\left(x\right)$and $v\left(x\right)=x$. Differentiate $u\left(x\right)=\mathrm{δ}\left(x\right)$and $v\left(x\right)=x$with respect to $x$.

$\begin{array}{l}u\text{'}\left(x\right)=\frac{d}{dx}\mathrm{δ}\left(x\right)\\ v^{\text{'}}\left(x\right)=1\end{array}$

Substitute $\mathrm{δ}\left(x\right)$for $u$, $x$for $v\left(x\right)$into $∫udv=uv-∫vdu$as,

localid="1657365897288" $$\begin{gathered} {∫}_{-∞}^{∞}\mathrm{δ}\left(x\right)dx={∫}_{-∞}^{∞}c\left(x\right)×1dx \\ ={\overline{)x\mathrm{δ}\left(x\right)}}_{-∞}^{∞}-{∫}_{-∞}^{∞}x\frac{d}{dx}\left(\mathrm{δ}\left(x\right)\right)dx \end{gathered}$$ ….. (1)

Define $x\mathrm{δ}\left(x\right)$as,

$x\mathrm{δ}\left(\mathrm{x}\right)=\{\begin{array}{ll}0×∞& \mathrm{x}=0\\ x=0& \mathrm{x}≠0\end{array}$

Thus the value of $x\mathrm{δ}\left(x\right)$is 0 for all x.

Hence, $x\mathrm{δ}{\overline{)\left(x\right)}}_{-∞}^{∞}=0$.

Now equation (1) can be rewritten as,

$$\begin{gathered} {∫}_{-∞}^{∞}\mathrm{δ}\left(x\right)dx={∫}_{-∞}^{∞}×\frac{d}{dx}\left(\mathrm{δ}\left(x\right)\right)dx \\ ={∫}_{-∞}^{∞}x\frac{d}{dx}\left(\mathrm{δ}\left(x\right)\right)dx \end{gathered}$$ ……. (2)

Equation (2) can be rewritten as $x\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{δ}\left(\mathrm{x}\right)\right)=-\mathrm{δ}\left(\mathrm{x}\right)$

(b)

According to equation (1), ${∫}_{-∞}^{∞}\mathrm{δ}\left(x\right)dx={\overline{)x\mathrm{δ}\left(x\right)}}_{-∞}^{∞}-{∫}_{-∞}^{∞}×\frac{d}{dx}\left(\mathrm{δ}\left(x\right)\right)dx$. The second property of Dirac Delta function can be written as follows

$$\begin{gathered} {∫}_{-∞}^{∞}f\left(x\right)\left(\frac{d\mathrm{θ}}{dx}\right)dx=f\left(x\right)\mathrm{θ}{\overline{)\left(x\right)}}_{-∞}^{∞}-{∫}_{-∞}^{∞}\frac{df}{dx}\left(\mathrm{θ}\left(x\right)\right)dx \\ =f\left(x\right)\mathrm{θ}{\overline{)\left(x\right)}}_{-∞}^{∞}-\left[{∫}_{-∞}^0\frac{df}{dx}\left(\mathrm{θ}\left(x\right)\right)dx+{∫}_0^{∞}\frac{df}{dx}\left(\mathrm{θ}\left(x\right)\right)dx\right] \\ =f\left(x\right)\mathrm{θ}{\overline{)\left(x\right)}}_{-∞}^{∞}-\left[0+{∫}_0^{∞}\frac{df}{dx}dx\right] \\ =f\left(x\right)(∞)-{∫}_0^{∞}\frac{df}{dx}dx \end{gathered}$$

Solve further as,

$$\begin{gathered} {∫}_{-∞}^{∞}f\left(x\right)\left(\frac{d\mathrm{θ}}{dx}\right)dx=(f\left(∞\right)-f\left(0\right)) \\ =f\left(0\right) \\ ={∫}_{-∞}^{∞}f\left(x\right)\mathrm{δ}\left(x\right)dx \end{gathered}$$

Thus, it can be written as $\left(\frac{d\mathrm{θ}}{dx}\right)=\mathrm{δ}x$.