# Determine divergence/convergence of an integral?

1. Apr 9, 2005

### crazedbeat

Determine divergence/convergence of an integral??

Given that
$$F(x) = \int_{0}^{x} f(s) ds$$
and
$$F(\infty)=1$$
prove that for any
$$\alpha \geq 1$$,

$$\int_{0}^{\infty} x^{\alpha} dF(x) = \alpha \int_{0}^{\infty} x^{\alpha - 1}(1 - F(x))dx$$
where two sides either converge or diverge together.

note: some of it didn't come out correctly. on the left hand side it means x raised alpha, and on the left x raised to (alpha - 1).

I have NO idea how to even START the problem. I've been doing convergence/divergence by finding limits, ratio test, etc. But these are not even close-- these are integrals and they don't make any sense. Any guidance into what i should be looking at would be greatly appreciated.

Last edited: Apr 9, 2005
2. Apr 9, 2005

### quasar987

This makes no sense to me. Where is the x in the righthand side?

Did you mean

$$F(x) = \int_{0}^{x}f(s)ds$$?

Has no mathematical significance either. Did you mean

$$\lim_{x\rightarrow \infty}F(x) = 1$$

?

You need to enclose the power of x with {} intead of ().

Correctly written, it does it look like

$$\int_{0}^{\infty} x^{\alpha} dF(x) = \alpha \int_{0}^{\infty} x^{\alpha - 1}(1 - F(x))dx$$

?

3. Apr 9, 2005

### xanthym

From the problem statement, we have:

$$1: \ \ \ \ F(x) \ = \ \int_{0}^{x} f(s) \, ds \hspace{0.8cm} \mbox{where:} \ \ \lim_{x \to \infty} F(x) \ = \ \left(1\right) \ \ \mbox{and} \ \ \ F(0) \ = \ \left(0\right)$$

Define the following:

$$2: \ \ \ \ g\left(x\right) \ \, = \, \ x^{\displaystyle \alpha} F\left(x\right)$$

Thus:

$$3: \ \ \ \ \frac{dg(x)}{dx} \ \ \, = \, \ \ \alpha x^{\displaystyle (\alpha - 1)} F(x) \ \ \, + \ \, \ x^{\displaystyle \alpha} \left ( \frac{dF(x)}{dx}\right )$$

$$4: \ \ \ \ \Longrightarrow \ \ x^{\displaystyle \alpha} \left (\frac{dF(x)}{dx}\right) \ \ \, = \, \ \ \frac{dg(x)}{dx} \ \ \, - \ \, \ \alpha x^{\displaystyle (\alpha - 1)} F(x)$$

$$5: \ \ \ \ \Longrightarrow \ \ x^{\displaystyle \alpha} \left (\frac{dF(x)}{dx}\right ) \ \ \, = \, \ \ \frac{d}{dx} \left \{ x^{\displaystyle \alpha} F(x) \right \} \ \ \, - \ \, \ \alpha x^{\displaystyle (\alpha - 1)} F(x)$$

$$6: \ \ \ \ \Longrightarrow \ \ \int_{0}^{w} x^{\displaystyle \alpha} \left (\frac{dF(x)}{dx}\right ) \, dx \ \ \, = \, \ \ \color{blue}\int_{0}^{w} \frac{d}{dx} \left \{ x^{\displaystyle \alpha} F(x) \right \} \, dx \color{black} \ \ - \ \ \int_{0}^{w} \alpha x^{\displaystyle (\alpha - 1)} F(x) \, dx$$

$$7: \ \ \ \ \Longrightarrow \ \ \int_{0}^{w} x^{\displaystyle \alpha} \, dF(x)\right ) \ \ \, = \, \ \ \color{blue}\left [ x^{\displaystyle \alpha} F(x) \right ]_{0}^{w}\color{black} \ \ \, - \ \, \ \int_{0}^{w} \alpha x^{\displaystyle (\alpha - 1)} F(x) \, dx$$

$$8: \ \ \ \ \Longrightarrow \ \ \int_{0}^{w} x^{\displaystyle \alpha} \, dF(x)\right ) \ \ \, = \, \ \ \color{blue}\alpha \int_{0}^{w} x^{\displaystyle (\alpha - 1)} F(w) \, dx \color{black} \ \ \, - \ \, \ \alpha \int_{0}^{w} x^{\displaystyle (\alpha - 1)} F(x) \, dx$$

$$9: \ \ \ \ \Longrightarrow \ \ \int_{0}^{w} x^{\displaystyle \alpha} \, dF(x)\right ) \ \ \, = \, \ \ \alpha \int_{0}^{w} x^{\displaystyle (\alpha - 1)} \left \{ F(w) \, - \, F(x) \right \} \, dx$$

$$10: \ \ \ \Longrightarrow \ \ \lim_{w \rightarrow \infty} \, \int_{0}^{w} x^{\displaystyle \alpha} \, dF(x)\right ) \ \ \, = \, \ \ \lim_{w \rightarrow \infty} \, \left ( \alpha \int_{0}^{w} x^{\displaystyle (\alpha - 1)} \left \{ F(w) \, - \, F(x) \right \} \, dx \right )$$

$$11: \ \ \ \color{red} \Longrightarrow \ \ \int_{0}^{\infty} x^{\displaystyle \alpha} \, dF(x)\right ) \ \ \, = \, \ \ \alpha \int_{0}^{\infty} x^{\displaystyle (\alpha - 1)} \left \{ 1 \, - \, F(x) \right \} \, dx \hspace{1.5cm} \mbox{\LARGE \textbf{Q.E.D.}}$$

~~

Last edited: Apr 9, 2005
4. Apr 9, 2005

### quasar987

Basically, what you need to do is simply to use the fact that dF(x) = f(x)dx to rewrite the left handside integral. Use also the definition of improper integral, i.e. that

$$\int_{a}^{\infty} = \lim_{M\rightarrow \infty}\int_{a}^{M}$$

to "allow" the use of the integration by part formula:

$$\int_{0}^{M}udv = [uv]_{0}^{M} - \int_{0}^{M}vdu$$

with u = x^alpha and dv = f(x)dx

5. Apr 9, 2005

### crazedbeat

Oh sorry, it is in fact,
$$F(x) = \int_{0}^{x}f(s)ds$$

xanthym, I have no idea what you did. Though I do understand that
$$\ \ \ \ g\left(x\right) \ \, = \, \ x^{\displaystyle \alpha} F\left(x\right)$$.
That was cool. I didn't realize that. How did you figure that out so fast? Was it obvious or experience?

I know the integration by parts, but I still don't understand what I am to do? What happens once I integrate? Am I trying to show that both sides are the same? How will I ever prove that they converge/diverge together?

Is there any way to put the question in simpler terms?

6. Apr 9, 2005

### Imo

xanthym, that was quite an amazing proof. Very hard to see it because it is so simple. Except for going from step 7 to 8. That is a very easy step to miss.

7. Apr 9, 2005

### xanthym

The two equation sides were shown to be equal; therefore, they will converge or diverge together.

~~