1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Determine divergence/convergence of an integral?

  1. Apr 9, 2005 #1
    Determine divergence/convergence of an integral??

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

    [tex]\int_{0}^{\infty} x^{\alpha} dF(x) = \alpha \int_{0}^{\infty} x^{\alpha - 1}(1 - F(x))dx[/tex]
    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. jcsd
  3. Apr 9, 2005 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

    Did you mean

    [tex]F(x) = \int_{0}^{x}f(s)ds[/tex]?

    Has no mathematical significance either. Did you mean

    [tex]\lim_{x\rightarrow \infty}F(x) = 1[/tex]

    ?

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

    Correctly written, it does it look like

    [tex]\int_{0}^{\infty} x^{\alpha} dF(x) = \alpha \int_{0}^{\infty} x^{\alpha - 1}(1 - F(x))dx[/tex]

    ?
     
  4. Apr 9, 2005 #3

    xanthym

    User Avatar
    Science Advisor

    From the problem statement, we have:

    [tex] 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)[/tex]

    Define the following:

    [tex] 2: \ \ \ \ g\left(x\right) \ \, = \, \ x^{\displaystyle \alpha} F\left(x\right) [/tex]

    Thus:

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

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

    [tex] 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) [/tex]

    [tex] 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 [/tex]

    [tex] 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 [/tex]

    [tex] 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 [/tex]


    [tex] 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 [/tex]

    [tex] 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 ) [/tex]

    [tex] 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.}} [/tex]

    ~~
     
    Last edited: Apr 9, 2005
  5. Apr 9, 2005 #4

    quasar987

    User Avatar
    Science Advisor
    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

    [tex]\int_{a}^{\infty} = \lim_{M\rightarrow \infty}\int_{a}^{M}[/tex]

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

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

    with u = x^alpha and dv = f(x)dx
     
  6. Apr 9, 2005 #5
    Oh sorry, it is in fact,
    [tex]F(x) = \int_{0}^{x}f(s)ds[/tex]

    xanthym, I have no idea what you did. Though I do understand that
    [tex] \ \ \ \ g\left(x\right) \ \, = \, \ x^{\displaystyle \alpha} F\left(x\right) [/tex].
    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?
     
  7. Apr 9, 2005 #6

    Imo

    User Avatar

    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.
     
  8. Apr 9, 2005 #7

    xanthym

    User Avatar
    Science Advisor

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


    ~~
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Determine divergence/convergence of an integral?
  1. Converges or diverges (Replies: 8)

Loading...