Results 1 to 4 of 4

Math Help - Improper integrals:

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    Improper integrals:

    \int_{\frac{1}{2}}^{2}\frac{dz}{z(ln{z})^{\frac{1}  {5}}}

    so:

    \lim_{t \to 1^{-}} \int_{\frac{1}{2}}^{t} \frac{dz}{z(ln{z})^{\frac{1}{5}}} + \lim_{t\to 1^{+}} \int_{t}^{2}\frac{dz}{z(ln{z})^{\frac{1}{5}}}


    \lim_{t\to{1^{-}}} \frac{5}{4} (ln{z})^{\frac{4}{5}} \big|_{\frac{1}{2}}^{t} + \lim_{t\to{1^{+}}} \frac{5}{4} (ln{z})^{\frac{4}{5}}\big|_{t}^{2}

    when i evaluate this.... the answer doesnt appear to be the sam as the back of book:
    Answer at back of book : 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    \int_{\frac{1}{2}}^{2}\frac{dz}{z(ln{z})^{\frac{1}  {5}}}

    so:

    \lim_{t \to 1^{-}} \int_{\frac{1}{2}}^{t} \frac{dz}{z(ln{z})^{\frac{1}{5}}} + \lim_{t\to 1^{+}} \int_{t}^{2}\frac{dz}{z(ln{z})^{\frac{1}{5}}}


    \lim_{t\to{1^{-}}} \frac{5}{4} (ln{z})^{\frac{4}{5}} \big|_{\frac{1}{2}}^{t} + \lim_{t\to{1^{+}}} \frac{5}{4} (ln{z})^{\frac{4}{5}}\big|_{t}^{2}

    when i evaluate this.... the answer doesnt appear to be the sam as the back of book:
    Answer at back of book : 0
    \lim_{t\to{1^{-}}} \frac{5}{4} (ln{z})^{\frac{4}{5}} \big|_{\frac{1}{2}}^{t} + \lim_{t\to{1^{+}}} \frac{5}{4} (ln{z})^{\frac{4}{5}}\big|_{t}^{2} = \lim_{t \to 1} \left[ \frac {5}{4} (\ln t )^{4/5} - \frac {5}{4}( - \ln 2)^{4/5} \right] + \lim_{t \to 1} \left[ \frac {5}{4} (\ln 2)^{4/5} - \frac {5}{4}( \ln t)^{4/5} \right]

    = - \frac {5}{4} (-\ln 2)^{4/5} + \frac {5}{4}(\ln 2)^{4/5}

    = - \frac {5}{4} \left[ (\ln 2)^4 \right]^{1/5} + \frac {5}{4} \left[ (\ln 2)^4 \right]^{1/5}

    = 0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381
    thanks:
    i got it now so \ln{\frac{1}{2}} = \ln{1} - \ln{2} = -\ln{2}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    thanks:
    i got it now so \ln{\frac{1}{2}} = \ln{1} - \ln{2} = -\ln{2}
    Or ln \left ( \frac{1}{2} \right ) = ln \left ( 2^{-1} \right ) = -ln(2)
    Either way.

    (I can't believe I forgot to take power of the ln(). This was driving me crazy earlier. )

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Improper Integrals
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: February 15th 2011, 12:09 AM
  2. improper integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 7th 2009, 05:06 AM
  3. Improper Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 17th 2009, 08:29 PM
  4. Improper Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 5th 2008, 11:57 AM
  5. Improper Integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 24th 2007, 05:05 PM

Search Tags


/mathhelpforum @mathhelpforum