Given an integration like, $$ \int_{b_1}^{b_2} g(x^*(b))~ \mathrm{d}b $$ Where, $x^*(b)$ is a solution of $f(x)=b$. How to reduce the integration in terms of the $f(x)-b$?
My attempt
\begin{align} &\int_{b_1}^{b_2} g(x^*(b))~ \mathrm{d}b \tag{1}\label{1}\\ &=\int_{b_1}^{b_2} \left[ \int_{x^*(b)^-}^{{x^*(b)^+}} \mathrm{d}y ~ \delta(y - x^*(b)) \right] g(y)~ \mathrm{d}b \tag{2}\label{2}\\ &=\int_{b_1}^{b_2} \mathrm{d}b ~\left[ \int_{x^*(b)^-}^{{x^*(b)^+}} \mathrm{d}y ~ \frac{\delta(y - x^*(b))}{|{f'(x^*(b))}|} \right] |{f'(x^*(b))}|~ g(y) \tag{3}\label{3}\\ &=\int_{b_1}^{b_2} \mathrm{d}b \left[ \int_{x^*(b)^-}^{{x^*(b)^+}} \mathrm{d}y ~ \frac{\delta(y - x^*(b))}{|{f'(x^*(b))}|} |{f'(y)}| \right] g(y) \tag{4}\label{4} \\ &=\int_{b_1}^{b_2} \mathrm{d}b \left[ \int_{x^*(b)^-}^{{x^*(b)^+}} \mathrm{d}y ~ \delta(f(y) - b)|{f'(y)}| \right] g(y) \tag{5}\label{5} \end{align}
In Eq. \eqref{4} I have used the fact $\int dx~\delta(x-y) h(x) = \int dx~\delta(x-y) h(y)$. Is this the correct way to do this? Are there any alternative ways to do this?
Also try with $f(f^{-1}(x))=x$ and $x=f(y)$:
$$\int g(f^{-1}(x))dx=\int g(y)d(f(y))$$
Or rewrite as:
$$\int g(f^{-1}(x))dx=\int g(y)f’(y)dy,y=f^{-1}(x)$$
Now you have an integral with no inverse function in terms of $f(x)$ and $g(x)$. You could use integration by parts:
$$\int g(f^{-1}(x))dx\mathop=^{y=f^{-1}(x)}\int g(y)d(f(y))=f(y)g(y)-\int f(y)d(g(y))=f(y)g(y)-\int f(y)g’(y)dy$$
if one derivative is easier than the other.