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.

## Query on 8757 Q63

### 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).

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.

### Re: Query on 8757 Q63

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.

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.