91Ó°ÊÓ

The given expressions are $\frac{d}{dx}sgnx=2\mathrm{δ}\left(x\right)$.

A function operator is a function that takes one (or more) functions as input and returns a function as output. In some ways, function operators are similar to functional

Let $G_1\left(x\right)=\frac{d}{dx}sgnx$and $G_2\left(x\right)=2\mathrm{δ}\left(x\right)$.

To verify the operator equation

$G_1\left(x\right)=G_2\left(x\right)$,

show that the

$∫\mathrm{ϕ}\left(x\right)G_1\left(x\right)dx=∫\mathrm{ϕ}\left(x\right)G_2\left(x\right)dx$

For any arbitrary test function $\mathrm{Ï•}\left(x\right)$which is zero for $\left|x\right|>R$for some finite $R$.

Assuming $\mathrm{Ï•}\left(x\right)$is such a decent test function, and then do the (integrating by parts)

$$\begin{gathered} ∫\mathrm{ϕ}\left(x\right)\frac{d}{dx}sgnxdx=\left(x\right)sgnx∫-∫sgnx\frac{d}{dx}\mathrm{ϕ}\left(x\right)dx \\ ={∫}_{-∞}^{∞}\frac{d\mathrm{ϕ}}{dx}dx-\underset{-∞}{\overset{∞}{∫}}\frac{d\mathrm{ϕ}}{dx}dx \\ =\mathrm{ϕ}\left(0\right)+\mathrm{ϕ}\left(0\right) \\ =2\mathrm{ϕ}\left(0\right) \end{gathered}$$

At the same time, the property of the delta function implies that,

$∫\mathrm{ϕ}\left(x\right)2\mathrm{δ}\left(x\right)dx=2\mathrm{ϕ}\left(0\right)$

Therefore, it is the case that $∫\mathrm{ϕ}\left(x\right)\frac{d}{dx}\sin xdx=∫\mathrm{ϕ}\left(x\right)2\mathrm{δ}\left(x\right)dx$for any test function $\mathrm{ϕ}\left(x\right)$, which means that $\frac{d}{dx}\sin x=2\mathrm{δ}\left(x\right)$

It is shown in figure which is given below