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Is there a way of showing that for a monotonic (not necessarily strictly monotonic) function $f(x)$ defined on $[a,b]$ which has a simple zero at $c\in (a,b)$ that $\int\limits_a^b g(x)\delta(f(x))dx={g(c)\over |f'(c)|}$? I know that I could get there by using the property $\delta(f(x))={\delta(x-c)\over |f'(c)|}$, but how can I not quote this property but instead use the fact that $f$ is monotonic on $[a,b]$ to show the relation?

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Try change of variables, assuming $f'(x)>0$: $$ \begin{align} \int_a^bg(x)\delta(f(x))\;\mathrm{d}x &=\int_{f(a)}^{f(b)}g(f^{-1}(x))\delta(x)\;\mathrm{d}f^{-1}(x)\\ &=\int_{f(a)}^{f(b)}g(f^{-1}(x))\delta(x)\;\frac{\mathrm{d}x}{f'(f^{-1}(x))}\\ &=\frac{g(f^{-1}(0))}{f'(f^{-1}(0))}\\ &=\frac{g(c)}{f'(c)} \end{align} $$ If $f'(x)<0$, then the limits of integration need to be reversed, giving the form with absolute values.

To make this rigorous, we should really replace $\delta$ with an approximation, $\varphi_\epsilon(x)=\varphi(x/\epsilon)/\epsilon$ for a nice positive $\varphi$ whose integral is $1$, and let $\epsilon\to0$.

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Ouch. Luckily you mention how to make it rigorous :-). –  Jonas Teuwen Nov 11 '11 at 18:57
    
Thanks, @robjohn. –  delta Nov 11 '11 at 22:11
    
What happens if $f'(x)\geq 0$ i.e. we are not guaranteed strict monotonicity? Then $f$ won't be invertible... –  delta Nov 11 '11 at 22:30
    
Since the support of $\delta$ is $\{0\}$, the only place we need to worry about $f'(x)$ is at $c$. If $f'(c)=0$, then $\dfrac{g(c)}{f'(c)}$ doesn't exist. –  robjohn Nov 12 '11 at 0:05

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