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Math Help - Natural Log Infinity Equivalency for Improper Integrals

  1. #1
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    Natural Log Infinity Equivalency for Improper Integrals

    Hello.

    I do not understand how:

    ln|\frac{\infty-2}{\infty+2}| is equal to ln(1)?

    Does ln(\infty) always = ln(1)?

    Thank you,
    Jen

    I am so thankful for MathMan's post on how to use LaTeX!

    *I apologize...I forgot to include the context I'm asking this for...

    I have an answer for:
    \int\frac{1}{z^2-4}dz, (10<z<\infty)

    = \frac{1}{4}ln|\frac{z-2}{z+2}| with 10<z< \infty

    So,

    \frac{1}{4}ln|\frac{\infty-2}{\infty+2}|-\frac{1}{4}ln(\frac{2}{3})

    should equal:

    -\frac{1}{4}ln(\frac{2}{3}): converges to .1014
    Last edited by Jenberl; January 11th 2009 at 07:53 PM.
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  2. #2
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    Quote Originally Posted by Jenberl View Post
    Hello.

    I do not understand how:

    ln|\frac{\infty-2}{\infty+2}| is equal to ln(1)?

    Does ln(\infty) always = ln(1)?

    Thank you,
    Jen

    I am so thankful for MathMan's post on how to use LaTeX!

    *I apologize...I forgot to include the context I'm asking this for...

    I have an answer for:
    \int\frac{1}{z^2-4}dz, (10<z<\infty)

    = \frac{1}{4}ln|\frac{z-2}{z+2}| with 10<z< \infty

    So,

    \frac{1}{4}ln|\frac{\infty-2}{\infty+2}|-\frac{1}{4}ln(\frac{2}{3})

    should equal:

    -\frac{1}{4}ln(\frac{2}{3}): converges to .1014

    OK, first off, you can NOT treat infinity as a number.

    The correct notation is \lim_{\varepsilon \to \infty} \frac{1}{4}\ln{\left |\frac{\varepsilon - 2}{\varepsilon + 2}\right | }.


    Notice that  \frac{z - 2}{z + 2} = 1 - \frac{4}{z + 2}.

    What happens as z \to \infty? The denominator gets huge, so the fraction \to 0.

    So it should tend to \ln{1} = 0.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    OK, first off, you can NOT treat infinity as a number.

    The correct notation is \lim_{\varepsilon \to \infty} \frac{1}{4}\ln{\left |\frac{\varepsilon - 2}{\varepsilon + 2}\right | }.


    Notice that  \frac{z - 2}{z + 2} = 1 - \frac{4}{z + 2}.

    What happens as z \to \infty? The denominator gets huge, so the fraction \to 0.

    So it should tend to \ln{1} = 0.

    OH!
    That makes a lot more sense now.
    Also- thank you for showing (inadvertantly) how to express absolute value notation properly with '\left' & '\right' before the '|' .

    ~Jen
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