How would you show $\delta(\xi-x)=\delta(x-\xi)$ if you know that $$\int _{-\infty}^{\infty}\delta(x)h(x)=h(0)$$
Also how would you then show more generally that if $f(\xi)$ is a monotonic increasing or decreasing fuction of $\xi$ which vanishes for $\xi=x$ then,
$$\delta[f(\xi)]=\frac{\delta(\xi-x)}{|f'(x)|}$$
Thanks in advance.
For the first question: let $u = \xi-x$, for any test function $h$ $$\int _{-\infty}^{\infty}\delta(\xi - x)h(\xi) d\xi = \int _{-\infty}^{\infty}\delta(u)h(u+x)du = h(u+x)\big|_{u=0} = h(x)$$ Try using the change of variable for $$\int _{-\infty}^{\infty}\delta(x - \xi)h(\xi) d\xi$$ See if you get the same result.
For the second question, here I just show the general result for $f$ is monotonic increasing: For any smooth test function $\varphi$ $$ \int^{\infty}_{-\infty}\delta[f(\xi)]\varphi(\xi)d\xi = \int^{\infty}_{-\infty}\delta[f(\xi)]\frac{\varphi(\xi)}{f'(\xi)} f'(\xi)d\xi $$ Apply changing variable $u = f(\xi)$, notice that the limits of the integral do not change, i.e., $f(-\infty)=-\infty$, and $f(\infty)=\infty$ (for decreasing function limits are reversed): $$ \int^{\infty}_{-\infty}\delta[f(\xi)]\frac{\varphi(\xi)}{f'(\xi)} f'(\xi)d\xi = \int^{\infty}_{-\infty}\delta[u]\frac{\varphi(f^{-1}(u))}{f'(f^{-1}(u))} du $$ $u=f(\xi) =0$ when $\xi= x$, hence: $$ \int^{\infty}_{-\infty}\delta[u]\frac{\varphi(f^{-1}(u))}{f'(f^{-1}(u))} du = \frac{\varphi(f^{-1}(0))}{f'(f^{-1}(0))} = \frac{\varphi(x)}{f'(x)} $$ this tells us: $$ \int^{\infty}_{-\infty}\delta[f(\xi)]\varphi(\xi)d\xi = \frac{\varphi(x)}{f'(x)} $$ Also notice $$ \int^{\infty}_{-\infty}\frac{\delta(\xi-x)}{f'(x)}\varphi(\xi)d\xi = \frac{\varphi(x)}{f'(x)} $$ for any test function $\varphi$ too. Therefore $$ \delta[f(\xi)] = \frac{\delta(\xi-x)}{f'(x)} $$ when $f$ is monotonic increasing. When $f$ is decreasing, the proof is exactly the same except there is an extra minus sign in front, but you know $f'<0$, so an absolute value is added.