Determine divergence/convergence of an integral?

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The discussion focuses on proving the equality of two integrals involving a function F(x) defined as the integral of f(s) from 0 to x, with the condition that F(∞) = 1. Participants express confusion about the mathematical formulation and the steps required to demonstrate convergence or divergence of the integrals. Key insights include the use of integration by parts and the relationship between the two sides of the equation, which ultimately leads to the conclusion that both integrals converge or diverge together. The proof is confirmed through a series of transformations and limits, clarifying the initial misunderstandings. The final consensus is that the equality of the two sides implies their convergence or divergence is linked.
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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.
 
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crazedbeat said:
Given that
F(x) = \int_{0}^{1} f(s) ds
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?

crazedbeat said:
and
F(\infty)=1
Has no mathematical significance either. Did you mean

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

?

crazedbeat said:
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).

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

?
 
crazedbeat said:
Given that
F(x) = \int_{0}^{1} 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.
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.}}

~~
 
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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
 
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?
 
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.
 
crazedbeat said:
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?
The two equation sides were shown to be equal; therefore, they will converge or diverge together.


~~
 
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