Results 1 to 3 of 3

Math Help - Definite integral

  1. #1
    Member roshanhero's Avatar
    Joined
    Aug 2008
    Posts
    176

    Definite integral

    Integrate:
    \int_{-\infty }^{\infty}\frac{x^2}{1+x^4}
    Please give me the hints only
    Last edited by roshanhero; June 9th 2010 at 09:04 PM. Reason: My mistake
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    ....

    First show

     \int_0^{\infty} \frac{x^2}{1 + x^4}~dx converges

    Then we can write

     \int_{-\infty}^{\infty} \frac{x^2}{1 + x^4}~dx = 2 \int_0^{\infty}\frac{x^2}{1+x^4}~dx

    For the integral  2 \int_0^{\infty}\frac{x^2}{1+x^4}~dx

    Write the numerator as  \frac{1}{2} ( (x^2+1)+(x^2-1))

    Then it equals to :

     \int_0^{\infty}\frac{(x^2+1)+(x^2-1)}{1+x^4}~dx

     =\int_0^{\infty}\frac{x^2+1}{1+x^4}~dx  + \int_0^{\infty}\frac{x^2-1}{1+x^4}~dx


    For the integral  \int_0^{\infty} \frac{x^2+1}{1+x^4}~dx

     =  \int_0^{\infty} \frac{1+1/x^2}{1/x^2+x^2}~dx

     = \int_0^{\infty} \frac{d(x-1/x)}{ \left( x - \frac{1}{x} \right)^2 + 2 } then use substitution :  x-1/x = t ...

    But the second integral should be zero :  I = \int_0^{\infty}\frac{x^2-1}{1+x^4}~dx

    After substituting  x = 1/t it becomes

     I = \int_0^{\infty} \frac{1- t^2}{1+t^4}~dt = - \int_0^{\infty} \frac{ t^2-1}{1+t^4}~dt = -I ~~~\implies I = 0
    Last edited by simplependulum; June 9th 2010 at 09:20 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The complex analysis offers a faster way to arrive to the result. If we apply the residue theorem we have...

    \int_{-\infty}^{+\infty} \frac{x^{2}}{1+x^{4}}\cdot dx = 2 \pi i \sum_{k} r_{k} (1)

    ... where the r_{k} are the residue of the poles of f(z) = \frac{z^{2}}{1+z^{4}} with positive imaginary part. If z_{0} is a pole of f(z) is [applying l'Hopital rule] ...

    r_{z=z_{0}} = \lim_{z \rightarrow z_{0}} (z-z_{0}) f(z) = \frac{1}{4 z_{0}} (2)

    Now the poles of f(z) with positive imaginary part are z= \frac{i \pm 1}{\sqrt{2}} so that is...

    \int_{-\infty}^{+\infty} \frac{x^{2}}{1+x^{4}}\cdot dx = 2 \pi i \frac{\sqrt{2}}{4} (\frac{1}{i-1} + \frac{1}{i+1}) = \frac{\pi}{\sqrt{2}} (3)

    Kind regards

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

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: December 5th 2011, 05:21 PM
  2. Replies: 4
    Last Post: April 13th 2011, 02:08 AM
  3. definite integral/ limit of integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2010, 04:00 AM
  4. Definite Integral 1/(x*(1-x)^0.5)
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 20th 2010, 01:39 PM
  5. Definite Integral help please
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 22nd 2008, 12:28 PM

Search Tags


/mathhelpforum @mathhelpforum