How Do I Prove Symmetry and Scaling Properties of the Dirac Delta Function?

[yungman]

I want to proof (1)$\delta(x)=\delta(-x)$ and (2) $\delta(kx)=\frac{1}{|k|}\delta(x)$

(1) let $u=-x\Rightarrow\;du=-dx$
\int_{-\infty}^{\infty}f(x)\delta(x)dx=(0)
but
\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{-\infty}^{\infty}f(-u)\delta(u)du=-f(0)

I cannot proof (1) is equal as I don't know if $f(u)=-f(-u)$. Please help.

(2)Let $u=kx\Rightarrow\;\frac{du}{k}=dx$. The sign of the limit reverse depends on the polarity of k.
\int_{-\infty}^{\infty}f(x)\delta(kx)dx=\frac{1}{k}\int_{-\infty}^{\infty}f(\frac{u}{k})\delta(u)du=^+_-\frac{1}{k}f(0)
Since sign is +ve if k is positive. Sign is -ve if k is -ve. This means the result is always $\frac{1}{|k|}f(0)$. Am I correct?

Thanks

 

[micromass]

I want to proof (1)$\delta(x)=\delta(-x)$ and (2) $\delta(kx)=\frac{1}{|k|}\delta(x)$

(1) let $u=-x\Rightarrow\;du=-dx$
\int_{-\infty}^{\infty}f(x)\delta(x)dx=(0)
but
\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{-\infty}^{\infty}f(-u)\delta(u)du=-f(0)

I cannot proof (1) is equal as I don't know if $f(u)=-f(-u)$. Please help.

The bounds of the integral change too, so you would have

\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{\infty}^{-\infty}f(-u)\delta(u)du

intead of what you had. Does that change anything?

 

[yungman]

The bounds of the integral change too, so you would have

\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{\infty}^{-\infty}f(-u)\delta(u)du

intead of what you had. Does that change anything?

Got it. How about (2)

Thanks

 

[micromass]

Got it. How about (2)

Thanks

You should switch the bounds in (2) too, if $k$ is negative. Aside from that, it's alright.

 

[yungman]

You should switch the bounds in (2) too, if $k$ is negative. Aside from that, it's alright.

Thanks