# Determine divergence/convergence of an integral?

• crazedbeat
In summary, the given problem is to prove that for any value of alpha greater than or equal to one, the integral from 0 to infinity of x raised to the power of alpha with respect to F(x) is equal to alpha times the integral from 0 to infinity of x raised to the power of alpha minus one multiplied by the difference of 1 and F(x) with respect to x, where both sides either converge or diverge together. This can be shown by using the fundamental theorem of calculus and the definition of improper integrals.
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:
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:

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

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

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

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

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

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

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

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

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

~~

Last edited:
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.

~~

## 1. How do you determine the convergence or divergence of an integral?

To determine the convergence or divergence of an integral, you can use various tests such as the Comparison Test, Limit Comparison Test, Ratio Test, Root Test, or the Integral Test. These tests involve evaluating the limit of the integral and comparing it to a known convergent or divergent series.

## 2. What is the Integral Test?

The Integral Test is a method used to determine the convergence or divergence of an integral by comparing it to a known convergent or divergent series. It involves evaluating the limit of the integral and comparing it to the limit of the series. If the limits are equal, then the integral and the series have the same convergence or divergence behavior.

## 3. When should the Comparison Test be used?

The Comparison Test should be used when the integral being evaluated contains a term that is similar to a known convergent or divergent series. This test involves comparing the given integral to the known series and determining if the integral has the same convergence or divergence behavior.

## 4. What is the difference between convergence and absolute convergence?

Convergence refers to the behavior of a series or integral when the terms or values are added together. If the sum of the terms approaches a finite value, the series or integral is said to converge. Absolute convergence refers to the behavior of a series or integral when the absolute values of the terms are added together. If the sum of the absolute values approaches a finite value, the series or integral is said to have absolute convergence.

## 5. How can you determine the convergence or divergence of an improper integral?

To determine the convergence or divergence of an improper integral, you can use the same tests used for determining the convergence or divergence of a regular integral. However, in the case of an improper integral, you will need to evaluate the limits of integration as well as the limit of the integral. Additionally, you may need to split the integral into multiple parts and evaluate each part separately before determining the overall convergence or divergence behavior.

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