Query on 8757 Q63

Forum for the GRE subject test in mathematics.
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Query on 8757 Q63

Postby yoyobarn » 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:

I tried using the "Inverse Function Integration method":


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.

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Re: Query on 8757 Q63

Postby owlpride » Tue Jan 10, 2012 1:42 am

Fastest method: draw a picture. No need for any formal reasoning here.

Your chain of reasoning can be made rigorous (throw in a couple of limits) but requires stronger assumptions than you are given. Notice for example that you wrote down f'(x) but you don't know that f is differentiable.

The formal version of "draw a picture" is Fubini's Theorem, which states that the area you care about can be computed either by integrating along horizontal or vertical lines:

\int_0^\infty f(x)dx = \int_0^\infty \int_0^{f(x)}  dy dx = \int_0^1 \int_0^{f^{-1}(y)}  dx dy = \int_0^1 f^{-1}(y)dy

In the middle step, we used that f ranges from 0 to 1 and that f^{-1} is a single valued function (since f is 1-1).
Last edited by owlpride on Tue Jan 10, 2012 1:57 am, edited 2 times in total.

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Re: Query on 8757 Q63

Postby ns2675 » Tue Jan 10, 2012 1:47 am

Regarding your first question: xf(x)\to0 as x\to\infty because 0<xf(x)\leq 2\int_{x/2}^{x}f(t)\,dt by monotonicity; the last expression tends to 0 because the integral converges.

I agree about the second part; just draw a picture.
Last edited by ns2675 on Tue Jan 10, 2012 11:42 am, edited 1 time in total.

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Re: Query on 8757 Q63

Postby yoyobarn » Tue Jan 10, 2012 6:57 am

Thank you very much.

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