## Query on 8757 Q63

Forum for the GRE subject test in mathematics.
yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

### Query on 8757 Q63

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":

$\int_{0}^{\infty}{f(x)}dx={[xf(x)]}_0^\infty-\int_0^\infty{f^{-1}(f(x))f'(x)}dx=0-\int_1^0{f^{-1}(y)}dy=\int_0^1{f^{-1}(y)}dy$

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.

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

### Re: Query on 8757 Q63

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.

ns2675
Posts: 6
Joined: Tue Jan 10, 2012 1:32 am

### Re: Query on 8757 Q63

Regarding your first question: $xf(x)\to0$ as $x\to\infty$ because $0 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.

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

### Re: Query on 8757 Q63

Thank you very much.