# Question on Dirac Delta function

1. Aug 6, 2013

### 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

2. Aug 6, 2013

### micromass

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

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

3. Aug 6, 2013

Thanks

4. Aug 6, 2013

### micromass

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

5. Aug 7, 2013

Thanks