Results 1 to 3 of 3

Thread: Natural Log Infinity Equivalency for Improper Integrals

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    21

    Natural Log Infinity Equivalency for Improper Integrals

    Hello.

    I do not understand how:

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

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

    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:
    $\displaystyle \int\frac{1}{z^2-4}dz, (10<z<\infty)$

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

    So,

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

    should equal:

    $\displaystyle -\frac{1}{4}ln(\frac{2}{3}$): converges to .1014
    Last edited by Jenberl; Jan 11th 2009 at 07:53 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by Jenberl View Post
    Hello.

    I do not understand how:

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

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

    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:
    $\displaystyle \int\frac{1}{z^2-4}dz, (10<z<\infty)$

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

    So,

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

    should equal:

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

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

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


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

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

    So it should tend to $\displaystyle \ln{1} = 0$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2008
    Posts
    21
    Quote Originally Posted by Prove It View Post
    OK, first off, you can NOT treat infinity as a number.

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


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

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

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

    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 11
    Last Post: Nov 14th 2010, 07:33 PM
  2. Improper Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 20th 2010, 08:38 PM
  3. Replies: 5
    Last Post: Jan 26th 2010, 04:56 AM
  4. limits at infinity of natural logs
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 2nd 2009, 06:29 PM
  5. Improper Integrals
    Posted in the Calculus Forum
    Replies: 17
    Last Post: Apr 20th 2008, 08:02 PM

Search Tags


/mathhelpforum @mathhelpforum