TITLE: Alternative ways to show $\int_{0}^{\infty}f(x)\, dx = \int_{0}^{1}f^{-1}(y)\, dy$
QUESTION [7 upvotes]: Let $f$ be a continuous, strictly decreasing, real-valued function such that $\int_{0}^{\infty}f(x)\, dx$ is finite and $f(0) = 1$. In terms of $f^{-1}$, we see that $$\int_{0}^{\infty}f(x)\, dx = \int_{0}^{1}f^{-1}(y)\, dy.$$ The way I saw this was by drawing an $f(x)$ and noticing that if I integrate in $y$ instead, the left handed integral becomes the right handed integral. Is there a way to do this with change of variables/$u$-substitution or other ways (namely ones that don't require drawing a picture of $f$)?

REPLY [6 votes]: \begin{align}
I&=\int_{y=0}^{y=1}f^{-1}(y)dy\\
\text{substitute $y\to f(x)$ and $dy\to \frac{df}{dx}dx$}\\
&=\int_{f^{-1}(0)}^{f^{-1}(1)}f^{-1}(f(x))\frac{df}{dx}dx\\
&=\int_\infty^0x\frac{df}{dx}dx\\
\text{apply integration by parts}\\
&=[xf(x)]_\infty^0-\int_\infty^0f(x)dx\\
\text{switching the bounds negates the integral}\\
&=(0f(0)-\lim_{x\to\infty}xf(x))+\int_0^\infty f(x)dx\\
\text{$\lim_{x\to\infty}xf(x)$ must be 0 if the integral converges, hence}\\
&=\int_0^\infty f(x)dx
\end{align}