Results 1 to 2 of 2

Thread: Integration

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    39
    Thanks
    1

    Integration

    Could you help me with this?

    By considering the integral of $\displaystyle e^{-yx^2} $ over a suitable region of $\displaystyle \mathbb{R}^2 $, show that for $\displaystyle 0<a<b<\infty $,

    $\displaystyle \int_{0}^{\infty} \frac{e^{-ax^2} - e^{-bx^2}}{x^2} dx = \sqrt{\pi} ( \sqrt{b} - \sqrt {a} ) $
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    the hint was already given, we have that $\displaystyle \frac{e^{-ax^2}-e^{-bx^2}}{x^2}=\int_a^b e^{-tx^2}\,dt,$ now reverse integration order (justified by Tonelli's Theorem since the integrand is non negative, becuase of $\displaystyle b>a$) and use the fact that $\displaystyle \int_{x\ge0}e^{-x^2}\,dx=\frac{\sqrt\pi}2,$ hence, the result.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Nov 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: Nov 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: Sep 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: Feb 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum