# Determine divergence/convergence of an integral?

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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quasar987
Homework Helper
Gold Member
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

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

?

xanthym
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:

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

Thus:

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

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

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

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

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

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

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

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

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

~~

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quasar987
Homework Helper
Gold Member
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
$$\ \ \ \ g\left(x\right) \ \, = \, \ x^\alpha} F\left(x\right)$$.
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?

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.

xanthym
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
$$\ \ \ \ g\left(x\right) \ \, = \, \ x^\alpha} F\left(x\right)$$.
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.

~~