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 $f$)?


1 Answer 1


\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}

  • $\begingroup$ I don't understand the change of variables here. How do you know that $f^{-1}(0)=\infty$ and $f^{-1}(1)=0$?? $\endgroup$ Oct 21, 2015 at 16:04
  • $\begingroup$ @tattwamasiamrutam I know that $f(0)=1$ because OP said so. That implies $f^{-1}(1)=0$. I assume that $f(\infty)=0$ because otherwise the integral of $f(x)$ wouldn't converge very easily. So if our integral is relatively sane, we can also assume that $f^{-1}(0)=\infty$ :) $\endgroup$
    – Navin
    Oct 21, 2015 at 23:12

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