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[itex] x \frac{d}{dx} [\delta (x)] = -\delta (x) [/itex]

this is problem 1.45 out of griffiths book by the way.

## Homework Equations

I attempted to use integration by parts as suggest by griffiths using [itex] f = x , g' = \frac{d}{dx}[/itex]

This yields [itex] x [\delta (x)] - \int \delta (x)dx[/itex]

next I tried taking the derivative of both sides so that I would get my original

[itex] x \frac{d}{dx} [\delta (x)] = x \frac{d}{dx}[\delta (x)] [/itex]

and so I'm back where I started. I have also tried using a definite integral so that

[itex]\int _{-\infty }^{\infty }x \frac{d}{dx}[\delta (x)] \text{dx} = \int_{-\infty }^{\infty } x[\delta (x)] \, dx - \int_{-\infty }^{\infty } [\delta (x)] \, dx [/itex]

but we know that [itex]\int_{-\infty }^{\infty } x[\delta (x)] \, dx = 0 [/itex] so our equation simplifies.

However this didn't get me any closer to solving the problem either.

I also have the second part of the problem regarding the step function which is defined as

[itex] \theta (x)

\begin{array}{ll}

\{ &

\begin{array}{ll}

1 & x>0 \\

0 & x\leq 0 \\

\end{array}

\\

\end{array}

[/itex] Show that [itex] \frac{d}{dx}\theta (x) =\delta (x)[/itex]

This something I can grasp by stating that

[itex]

\frac{\Delta \theta }{\text{$\Delta $x}}(x=0 )\longrightarrow 1/0, \text{ as } \text{$\Delta $x}\longrightarrow 0.[/itex]

which is clearly an undefined(or infinite) slope, and at every other point on this graph the derivative is going to be 0. However I want to know if this is sufficiently rigorous, and if it isn't what a step in the right direction might be.