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 $f^{-1}$, $\int_{0}^{+\infty}f(x)\,dx$ is?
The answer is "equal to $\int_{0}^{1}f^{-1}(y)\,dy$"
Okay here comes my question. If there is a way to guarantee that $\displaystyle\lim_{x \rightarrow +\infty} f(x) = 0$, then I totally agree with the answer. However, suppose the $f(x)$ converges to somewhere greater or less than $0$, then how can this answer still be true?
Or, is there a way to prove that $f(x)$ will definitely converge to $0?$
Thanks!
By using the substitution $x=f^{-1}(u)$ and integration by parts we obtain $$\int_0^\infty f(x)dx{=\int_{f(0)}^{f(\infty)}u{d\over du}f^{-1}(u)du\\=uf^{-1}(u)\Bigg|_{f(0)}^{f(\infty)}-\int_{f(0)}^{f(\infty)} f^{-1}(u)du\\=\infty\times f(\infty)-\int_{f(0)}^{f(\infty)} f^{-1}(u)du}$$so the integral exists only if $\lim_{x\to \infty}xf(x)=0$ which leads to