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Show the integral diverges

  1. Mar 19, 2005 #1

    quasar987

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    Here's the problem:

    Consider a > 0. Show that the integral

    [tex]\int_{a+1}^{\infty} \frac{1}{1/a-1/x}dx [/tex]

    diverges. Do so without using the fact that the primitive of 1/x is ln x. :eek:

    I have no clue what to do!
     
  2. jcsd
  3. Mar 19, 2005 #2
    What's [tex] \lim_{x\rightarrow \infty} \frac{1}{\frac{1}{a} - \frac{1}{x}}[/tex]?
     
    Last edited: Mar 19, 2005
  4. Mar 19, 2005 #3
    Evaluate the integral as
    [tex]\lim_{b\rightarrow \infty}\int_{a+1}^{b}\frac{1}{\frac{1}{a}-\frac{1}{x}}\ {dx}[/tex]
    [tex]=\lim_{b\rightarrow \infty}\int_{a+1}^{b}a-\frac{a^2}{a-x}\ dx[/tex]
    ...
    ... you can do the intermediary steps here... you'll work out that you get something like the following:
    [tex]\lim_{b\rightarrow \infty}ab-a^2\log{a-b}-a^2-a+\log{1}[/tex]

    as you know, [itex]ab[/itex] goes to infinity as b goes to infinity. ln(a-b) also becomes undefined as you subtract infinity from any number, since log is only defined for positive numbers.
     
  5. Mar 19, 2005 #4
    Answering my query is a whole lot easier, and, unlike scholzie's method, does not require you to break the rules of the question!

     
  6. Mar 20, 2005 #5

    xanthym

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    Note that since {a > 0}:

    [tex] :(1): \ \ \ \ \ 0 \ \ < \ \ a \ \ = \ \ \frac {1} {1/a} \ \ < \ \ \frac{1} {1/a-1/x} \ \ \ \ \color{blue} \mathbb \forall \ \ x \ \ge \ (a + 1) [/tex]

    [tex] :(2): \ \ \ \ \ \ \Longrightarrow \ \ \ \ 0 \ \ < \ \ \int_{a+1}^{\infty} a \ dx \ \ < \ \ \int_{a+1}^{\infty} \frac {1} {1/a-1/x} \ dx [/tex]

    [tex] :(3): \ \ \ \ \ \ \Longrightarrow \ \ \ \ 0 \ \ < \ \ \left [ ax \right ]_{a + 1}^{\infty} \ \ < \ \ \int_{a+1}^{\infty} \frac {1} {1/a-1/x} \ dx [/tex]

    [tex] :(4): \ \ \ \ \color{red} \Longrightarrow \ \ (\infty) \ < \ \int_{a+1}^{\infty} \frac {1} {1/a-1/x} \ dx [/tex]

    Thus, subject integral diverges.


    ~~
     
    Last edited: Mar 20, 2005
  7. Mar 20, 2005 #6
    Didnt see the bit about not using the antiderivative ln(x)... I was only trying to help...
     
  8. Mar 20, 2005 #7

    xanthym

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    No problem ... several techniques are now available for comparison.


    ~~
     
    Last edited: Mar 20, 2005
  9. Mar 20, 2005 #8
    Yep, nothing wrong with posting alternate methods of solution, I was just trying to avoid getting him confused.
     
  10. Mar 20, 2005 #9

    quasar987

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    Ok, thanks everyone!
     
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