Results 1 to 4 of 4

Math Help - Taking down an integral

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Taking down an integral

    I was wondering which method you'd use to take down the following integral: \int _0^{\infty} \frac{dx}{(d^2+x^2)^2}.
    I need it to be "killed", so if any of you, inte-killers, are interested, please have fun.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,911
    Thanks
    774
    Hello, arbolis!

    \int _0^{\infty} \frac{dx}{(a^2+x^2)^2}.

    Let: . x \:=\:a\tan\theta \quad\Rightarrow\quad dx \:=\:a\sec^2\!\theta\,d\theta
    . . Then: . a^2+x^2 \:=\:a^2\sec^2\!\theta


    Substitute: . \int \frac{a\sec^2\!\theta\,d\theta}{(a^2\sec^2\!\theta  )^2} \;=\; \frac{1}{a^3}\int\frac{dx}{\sec^2\!\theta} \;=\;\frac{1}{a^3}\int\cos^2\!\theta\,dx \;=\;\frac{1}{2a^3}\int(1 + \cos2\theta)\,dx

    . . . . . . . . =\;\frac{1}{2a^3}\bigg(\theta + \frac{1}{2}\sin 2\theta \bigg) + C \;=\;\frac{1}{2a^3}\bigg(\theta + \sin\theta\cos\theta\bigg) + C


    Back-substitute: . \tan\theta \,=\,\frac{x}{a} \quad\Rightarrow\quad \sin\theta \,=\,\frac{x}{\sqrt{a^2+x^2}} \qquad\Rightarrow\quad \cos\theta \,=\,\frac{a}{\sqrt{a^2+x^2}}

    . . and we have: . \frac{1}{2a^3}\bigg(\arctan\frac{x}{a} + \frac{ax}{a^2+x^2}\bigg) + C




    Now we evaluate from 0 to \infty

    When we substitute x = 0, we get 0.


    We don't substitute \infty. we have to sneak up on it.

    We substitute x = b, then take limit as b gets infinite.

    We have: . \lim_{b\to\infty}\frac{1}{2a^3}\bigg(\arctan\frac{  b}{a} + \frac{ab}{a^2+b^2}\bigg)


    . . The first limit is: . \lim_{b\to\infty}\,\arctan\frac{b}{a} \;=\;\arctan\infty \;=\;\frac{\pi}{2}

    . . The second limit is: . \lim_{b\to\infty}\frac{ab}{a^2+b^2}

    . . Divide top and bottom by b^2\!: . \lim_{b\to\infty}\,\frac{\frac{a}{b}}{\frac{a^2}{b  ^2} + 1} \;=\;\frac{0}{0+1} \:=\:0


    Therefore, we have: . \frac{1}{2a^3}\bigg(\frac{\pi}{2} + 0\bigg) \;=\;\frac{\pi}{4a^3}


    Am I close?


    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    okay then, let's kill it.

    \int_{0}^{\infty }{\frac{dx}{\left( d^{2}+x^{2} \right)^{2}}}=\frac{1}{d^{2}}\left( \int_{0}^{\infty }{\frac{dx}{d^{2}+x^{2}}}-\int_{0}^{\infty }{\frac{x^{2}}{\left( d^{2}+x^{2} \right)^{2}}} \right)\,dx, well actually, i'll let you to kill it, 'cause you just need to integrate by parts the last integral and you're done.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Thanks to both! That was helpful.
    Having the result, I realize I made an error setting up this integral. It came from an attempt to find the electric field of a semi-infinite charged line. I should have reached \frac{1}{d^2}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Taking the Integral - did I do it right?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 22nd 2010, 01:20 AM
  2. Replies: 5
    Last Post: December 2nd 2009, 11:15 PM
  3. Taking The Derivative
    Posted in the Calculus Forum
    Replies: 9
    Last Post: November 18th 2009, 01:49 PM
  4. taking integral of incomplete Beta function
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: October 30th 2009, 08:42 AM
  5. Taking integrals with e
    Posted in the Calculus Forum
    Replies: 8
    Last Post: April 18th 2008, 06:44 PM

Search Tags


/mathhelpforum @mathhelpforum