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Math Help - Convergence Problem... Problems

  1. #1
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    Convergence Problem... Problems

    Currently taking integral calculus and I have a problem that I have no idea how to approach.

    Given the integral from 1 to infinity of dx/(x^p *(ln(x))^a)

    - For what values of p and a does this converge

    Even if you cant get the full solution, any thoughts on how to approach it would be greatly appreciated.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by scg780 View Post
    Currently taking integral calculus and I have a problem that I have no idea how to approach.

    Given the integral from 1 to infinity of dx/(x^p *(ln(x))^a)

    - For what values of p and a does this converge

    Even if you cant get the full solution, any thoughts on how to approach it would be greatly appreciated.
    For this integral to converge requires that the limit:

    \lim_{u\to \infty} \int_1^u\frac{1}{x^p (\ln(x))^a}\;dx

    exists. Now as \ln(x) increases more slowly than any power of x this is equivalent to asking for what values of p does:

    \lim_{u\to \infty} \int_1^u\frac{1}{x^p}\;dx

    converge (with p=1 possibly being a special case).

    CB
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  3. #3
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    ok. Thank you. but i'm still confused on what the value of "a" would be
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  4. #4
    MHF Contributor chisigma's Avatar
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    Integrating by parts we obtain...

    \int \frac{dx}{x^{p}\cdot \ln^{a} x} = \frac{1}{1-p} \cdot \frac{1}{x^{p-1}\cdot \ln^{a} x} + \frac{a}{1-p} \int\frac{dx}{x^{p}\cdot \ln ^{a+1} x} (1)

    ... and then with another integration by parts...

    \int \frac{dx}{x^{p}\cdot \ln^{a} x} = \frac{1}{1-p} \cdot \frac{1}{x^{p-1}\cdot \ln^{a} x} + \frac{a}{(1-p)^{2}} \cdot \frac{1}{x^{p-1}\cdot \ln^{a+1} x} +

    + \frac{a\cdot (a+1)}{(1-p)^{2}}\cdot \int \frac{dx}{x^{p}\cdot \ln^{a+2} x} (2)

    If now we proceed indefinitely we obtain...

    \int \frac{dx}{x^{p}\cdot \ln^{a} x} = \frac{1}{1-p} \cdot \frac{1}{x^{p-1}\cdot \ln^{a} x} + \frac{a}{(1-p)^{2}} \cdot \frac{1}{x^{p-1}\cdot \ln^{a+1} x} + \dots

    \dots + \frac{a\cdot (a+1) \dots (a+n-1)}{(1-p)^{n}}\cdot \frac{1}{x^{p-1}\cdot \ln^{a+n} x} + \dots (3)

    At this point a careful investigation of (3) permits us to conclude that the [improper] integral...

    \int_{1}^{\infty} \frac{dx}{x^{p}\cdot \ln^{a} x} (4)

    ... converges if...

    a)  p>1

    b) a is a non positive integer...

    If a= -k, k\ge 0 is ...

    \int_{1}^{\infty}\frac{\ln^{k} x}{x^{p}} = \frac{k!}{(p-1)^{k+1}} (5)

    The convergence in the case a<0 but non integer can be easily verified by comparison...

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 22nd 2010 at 01:21 PM.
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