How to evaluate the coordinate transformation of Dirac delta function? I have come across the following calculation while evaluating path integrals.How do one determine the coordinate transformation of delta function? For instance,If the delta function $\delta(X_{N}-X_{f})$ (where $X_{N}=X(t_{N})$ and $X_{f}=X(t_{f})$) is transformed to to another set variables $Z_{N}$ and $Z_{f}$ from X coordinate. How does $\delta(X_{N}-X_{f})$ transforms to $\delta(Z_{N}-Z_{f})$. Also how does the following transforms transform to $Z$ coordinate? $\int dX_{N} \delta(X_{N}-X{f})f(X_{N})$. The transformation of $X$ to $Z$ is defined to be $X(t)=Z(t)exp(-\omega t)$
 A: Of course we can evaluate the given integral without transforming the Dirac $\delta-$functions since :
\begin{equation}
\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left(x_{_{\rm{N}}}\!-\!x_{_{\rm{F}}}\right) f\left(x_{_{\rm{N}}}\right)\mathrm dx_{_{\rm{N}}}= f\left(x_{_{\rm{F}}}\right)=f\left[e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z\left(t_{_{\rm{F}}}\right)\right]=  f\left(e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}}\right)
\tag{01}
\end{equation}
But if for other purposes we want to see what is happening via transformation of the Dirac $\delta-$functions, we could proceed as follows. Let the transformation between $\:x,z\:$
\begin{equation}
x\left(t\right)=e^{\boldsymbol{-}\omega t}z\left(t\right)
\tag{02}
\end{equation}
We express the variables under the integral in (01) from $x$-coordinate to $z$-coordinate
\begin{align}
x_{_{\rm{N}}} & =x\left(t_{_{\rm{N}}}\right)=e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z\left(t_{_{\rm{N}}}\right)=e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}
\tag{03a}\\
x_{_{\rm{F}}} & =x\left(t_{_{\rm{F}}}\right)=e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z\left(t_{_{\rm{F}}}\right)=e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}}
\tag{03b}\\
\mathrm dx_{_{\rm{N}}} & = e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}\mathrm d z_{_{\rm{N}}}
\tag{03c}
\end{align}
and so
\begin{align}
\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left(x_{_{\rm{N}}}\!-\!x_{_{\rm{F}}}\right) f\left(x_{_{\rm{N}}}\right)\mathrm dx_{_{\rm{N}}} & =
\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\!-\!e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}}\right) f\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\right) e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}\mathrm d z_{_{\rm{N}}}
\nonumber\\
& =e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\!-\!e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}}\right) f\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\right) \mathrm d z_{_{\rm{N}}}
\nonumber\\
& =\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left[z_{_{\rm{N}}}\!-\!e^{\omega \left(t_{_{\rm{N}}}\boldsymbol{-}t_{_{\rm{F}}}\right)} z_{_{\rm{F}}}\right] f\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\right) \mathrm d z_{_{\rm{N}}}
\tag{04}
\end{align}
The last equality in (04) is valid since(1) 
\begin{equation}
\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\delta\left(\alpha x+\beta\right)\mathrm{h}\left(x\right)\mathrm dx=\dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right)\mathrm{h}\left(x\right)\mathrm dx
\tag{05}
\end{equation}
written symbolically
\begin{equation}
\delta\left(\alpha x+\beta\right) \doteq \dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right)
\tag{06}
\end{equation}
The equality is proved by making in the 2nd term of (04) the following substitutions 
\begin{align}
x & \quad \boldsymbol{-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow} \quad z_{_{\rm{N}}}
\tag{07a}\\
\alpha & \quad \boldsymbol{-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow} \quad e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}
\tag{07b}\\
\beta & \quad \boldsymbol{-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow} \quad \boldsymbol{-}e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}}
\tag{07c}
\end{align}
So
\begin{equation}
\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left(x_{_{\rm{N}}}\!-\!x_{_{\rm{F}}}\right) f\left(x_{_{\rm{N}}}\right)\mathrm dx_{_{\rm{N}}} 
=\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\!\!\delta\left[z_{_{\rm{N}}}\!-\!e^{\omega \left(t_{_{\rm{N}}}\boldsymbol{-}t_{_{\rm{F}}}\right)} z_{_{\rm{F}}}\right] f\left(e^{\boldsymbol{-}\omega t_{_{\rm{N}}}}z_{_{\rm{N}}}\right) \mathrm d z_{_{\rm{N}}}=  f\left(e^{\boldsymbol{-}\omega t_{_{\rm{F}}}}z_{_{\rm{F}}}\right)
\tag{08}
\end{equation}
a result identical to that given in (01).
Now, I don't think it would be useful to write (08) symbolically
\begin{equation}
\delta\left(x_{_{\rm{N}}}\!-\!x_{_{\rm{F}}}\right)  \doteq \delta\left[z_{_{\rm{N}}}\!-\!e^{\omega \left(t_{_{\rm{N}}}\boldsymbol{-}t_{_{\rm{F}}}\right)} z_{_{\rm{F}}}\right] 
\tag{09}
\end{equation}
since this is valid in the special case of transformation (02).

(1)
Equation (05) is proved as equation (A-07) in  Example A of my answer here : Physical meaning of the Jacobian in relation to Dirac delta function. 
 
