Query on 8757 Q63
Posted: Mon Jan 09, 2012 11:14 pm
Q63. Let f be a continous, strictly decreasing, real-valued function such that $$\int_0^\infty{f(x)}dx$$ is finite and f(0)=1.
In terms of $$f^{-1}$$ (the inverse function of f), $$\int_0^{\infty}{f(x)}dx$$ is:
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I tried using the "Inverse Function Integration method":
$$\int_{0}^{\infty}{f(x)}dx={[xf(x)]}_0^\infty-\int_0^\infty{f^{-1}(f(x))f'(x)}dx=0-\int_1^0{f^{-1}(y)}dy=\int_0^1{f^{-1}(y)}dy$$
Is the method valid?
Also, how do we conclude that xf(x)=0, when x is infinity? (x would be infinity while f(x)=0)
Lastly, is there any faster method?
Thank you very much.
In terms of $$f^{-1}$$ (the inverse function of f), $$\int_0^{\infty}{f(x)}dx$$ is:
---------------
I tried using the "Inverse Function Integration method":
$$\int_{0}^{\infty}{f(x)}dx={[xf(x)]}_0^\infty-\int_0^\infty{f^{-1}(f(x))f'(x)}dx=0-\int_1^0{f^{-1}(y)}dy=\int_0^1{f^{-1}(y)}dy$$
Is the method valid?
Also, how do we conclude that xf(x)=0, when x is infinity? (x would be infinity while f(x)=0)
Lastly, is there any faster method?
Thank you very much.