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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!

Adrian Keister
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Caprikuarius
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  • I think $\displaystyle\int_0^{\infty}f(x),dx<\infty$ implies $\displaystyle\lim_{x\to\infty}f(x)=0.$ The converse is certainly not true. – Adrian Keister Sep 06 '19 at 18:36
  • @AdrianKeister we also need the 'strictly decreasing' criterion, correct? – user1952500 Sep 06 '19 at 19:48
  • @user1952500: I don't think so, actually. I think the 'strictly decreasing' criterion is there to ensure $f$ has an inverse. – Adrian Keister Sep 06 '19 at 20:25
  • @AdrianKeister then if we use $f(x) = \sin(x)$, we will have the integral bounded but the limit may not exist right? (The integral also may not exist but it will be bounded by $1$) – user1952500 Sep 06 '19 at 21:43
  • I'm not so sure that $\sin(x)$ provides a good counterexample. It might make intuitive sense that $\int_0^{\infty}\sin(x),dx<0,$ but how would you prove it? The problem is that you're integrating on an infinite interval. The other issue is that "strictly decreasing" is more than you need to ensure the limit exists. After all, $\sin(x)/x$ has a limit of $0$ at infinity, but it's not strictly decreasing. – Adrian Keister Sep 08 '19 at 01:32
  • @AdrianKeister yes after thinking more what you say is correct. As a non-rigorous thought, if it doesn't tend to zero and has a minimum value after some point, the integral would diverge. – user1952500 Sep 08 '19 at 06:03

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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

$$\int_0^{\infty} f(x)dx=\int_{0}^{f(0)} f^{-1}(x)dx$$

Mostafa Ayaz
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