Results 1 to 5 of 5

Math Help - Improper Integral's & L'Hopital's Rule

  1. #1
    Member
    Joined
    Sep 2009
    From
    Ontario
    Posts
    162

    Improper Integral's & L'Hopital's Rule

    Hi. Im having trouble with this improper integral
    it is the integral of 5/(z^2+7z+6)dz from 5 to +infinity

    I know that you let t=+inf and set it up as the limit of the integral as t approaches +inf and solve

    I am stuck at the part after you integrate & plug +inf back in for t & I know that you're supposed to use L'hopitals rule somewhere but am having trouble doing this

    Please Help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Did you use partial fractions?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by rawkstar View Post
    Hi. Im having trouble with this improper integral
    it is the integral of 5/(z^2+7z+6)dz from 5 to +infinity

    I know that you let t=+inf and set it up as the limit of the integral as t approaches +inf and solve

    I am stuck at the part after you integrate & plug +inf back in for t & I know that you're supposed to use L'hopitals rule somewhere but am having trouble doing this

    Please Help
    You do not need L'Hopital it is a perfectly ordinary limit problem. Post what you have and we will advise you on how to proceed.

    CB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    If f(z) has only 'simple poles' \alpha_{1}, \alpha_{2}, ..., \alpha_{n} , then is...

    \displaystyle f(z)= \sum_{n=1}^{n} \frac{r_{n}}{z-\alpha_{n}} (1)

    ... where...

    \displaystyle r_{n}= \lim_{z \rightarrow \alpha_{n}} f(z)\ (z-\alpha_{n}) (2)

    Now it is evident that the limits (2) are 'ideterminate forms' of the type \frac{0}{0} so that they can be solved using l'Hopital's rule...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by chisigma View Post
    If f(z) has only 'simple poles' \alpha_{1}, \alpha_{2}, ..., \alpha_{n} , then is...

    \displaystyle f(z)= \sum_{n=1}^{n} \frac{r_{n}}{z-\alpha_{n}} (1)

    ... where...

    \displaystyle r_{n}= \lim_{z \rightarrow \alpha_{n}} f(z)\ (z-\alpha_{n}) (2)

    Now it is evident that the limits (2) are 'ideterminate forms' of the type \frac{0}{0} so that they can be solved using l'Hopital's rule...

    Kind regards

    \chi \sigma
    Now you are making things more complicated than they need be. You end up with the log of a rational function (from your equation (1)) and that can be handled perfectly well without resort to M. L'Hopital.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] L'Hopital's Rule help please
    Posted in the Calculus Forum
    Replies: 11
    Last Post: February 25th 2011, 11:49 AM
  2. Integral Involving L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 13th 2010, 12:05 PM
  3. L'Hopital's Rule and Improper Integrals
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 7th 2009, 10:13 AM
  4. Replies: 8
    Last Post: March 25th 2008, 02:35 PM
  5. l'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 10
    Last Post: October 21st 2007, 03:52 PM

Search Tags


/mathhelpforum @mathhelpforum