Results 1 to 6 of 6

Math Help - Integral Calculation - Urgent

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    3

    Integral Calculation - Urgent

    Hello,

    I need to show that an integral from minus infinity to infinity is smaller than infinity. All the expressions that are integrated are positive. What should I do? Can you recommend me some theorems that I can use? Thank you in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by joreto_co View Post
    Hello,

    I need to show that an integral from minus infinity to infinity is smaller than infinity. All the expressions that are integrated are positive. What should I do? Can you recommend me some theorems that I can use? Thank you in advance!
    My best recommendation (for the moment at least) is that you post the integral here.
    Last edited by mr fantastic; May 29th 2008 at 05:08 AM. Reason: Misunderstood the question
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Please try to post the image again, it isn't uploaded.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2008
    Posts
    3
    Here in the attachment is what I have to prove
    Attached Thumbnails Attached Thumbnails Integral Calculation - Urgent-help.jpg  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    \int_{-\infty}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx

    As |x| and x^2 are positive, we can write this as,

    2\int_{0}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx

    So checking the convergence of \int_{0}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx is enough.

    \int_{0}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx = \left [ \int_{0}^{1} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx \right ]  ~+~ \left [ \int_{1}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx \right ]

    We can easily prove that \int_{1}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx converges.

    For 1\leq x \leq \infty,
    0\leq \frac{\ln x \cdot a \cdot e^{-\frac{a}{x^2}}}{x^3} \leq \frac{\ln x \cdot a}{x^3}

    \int_1^{\infty} \frac{\ln x \cdot a}{x^3}~dx = \frac{a}{4}

    So \int_{1}^{\infty} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx converges too.

    Now prove the convergence of \int_{0}^{1} \frac{|\ln |x|| \cdot a \cdot e^{-\frac{a}{x^2}}}{|x|^3}~dx and it'll be over.
    (It looks a little long and I can't think of anything easy at the moment..)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2008
    Posts
    3
    And does someone have an ideal for the second equation?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Calculation of the integral
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 30th 2011, 10:03 AM
  2. Integral calculation
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 27th 2010, 02:41 PM
  3. Integral calculation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 17th 2009, 12:22 AM
  4. Integral calculation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 16th 2009, 02:01 PM
  5. Today's calculation of integral #15
    Posted in the Calculus Forum
    Replies: 10
    Last Post: May 10th 2008, 07:15 AM

Search Tags


/mathhelpforum @mathhelpforum