Results 1 to 6 of 6
Like Tree2Thanks
  • 1 Post By HallsofIvy
  • 1 Post By Prove It

Math Help - integration by complex exponentials

  1. #1
    Member
    Joined
    Feb 2010
    From
    in the 4th dimension....
    Posts
    122
    Thanks
    9

    integration by complex exponentials

    I have seen that one can use the euler's formula to complexify the integration, and integrate functions like e^{nx} \cos mx  dx easily.(The method has a wikipedia page:Integration using Euler's formula - Wikipedia, the free encyclopedia)

    How can i do \int \frac{dx}{13+3\cos x+4\sin x} using complex exponentials.

    PS:I know this can be easily done with standard trig substitutions and might turn out to be longer using complex exponentials, but I want to know if and how can it be done with this method ...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,567
    Thanks
    1409

    Re: integration by complex exponentials

    Quote Originally Posted by earthboy View Post
    I have seen that one can use the euler's formula to complexify the integration, and integrate functions like e^{nx} \cos mx  dx easily.(The method has a wikipedia page:Integration using Euler's formula - Wikipedia, the free encyclopedia)

    How can i do \int \frac{dx}{13+3\cos x+4\sin x} using complex exponentials.

    PS:I know this can be easily done with standard trig substitutions and might turn out to be longer using complex exponentials, but I want to know if and how can it be done with this method ...
    cos(x)= \frac{e^{ix}+ e^{-ix}}{2} and sin(x)= \frac{e^{ix}- e^{-ix}}{2} so that integral, in complex exponentials, is \int\frac{dx}{13+ \frac{3e^{ix}+ 3e^{-ix}}{2}+ \frac{4e^{ix}- 4e^{-ix}}{2}}
    = 2\int \frac{dx}{26+ 7e^{ix}- e^{ix}}
    Multiply both numerator and denominator by e^{ix} to get
    2\int \frac{e^{ix}dx}{26e^{ix}+ 7e^{2ix}- 1}

    Now let u= e^{ix} so that du= ie^{ix}dx, -i du= e^{ix}dx and that integral becomes
    -2i\int \frac{du}{7u^2+ 26u- 1}.

    That denominator can be factored (though not easily- use the quadratic formula to find the roots). Then it will be of the form
    -\frac{2i}{7}\int \frac{du}{(u- a)(u- b)}du
    where a and b are the roots. That can be integrated by "partial fracations".
    Last edited by HallsofIvy; February 5th 2013 at 09:02 AM.
    Thanks from earthboy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    From
    in the 4th dimension....
    Posts
    122
    Thanks
    9

    Re: integration by complex exponentials

    Quote Originally Posted by HallsofIvy View Post
    cos(x)= \frac{e^{ix}+ e^{-ix}}{2} and sin(x)= \frac{e^{ix}- e^{-ix}}{2}
    small doubt:
    is \sin x= \frac{e^{ix}- e^{-ix}}{2} or \sin x= \frac{e^{ix}- e^{-ix}}{2i}
    both are seem to be used in books, but what is the difference?
    Last edited by earthboy; February 5th 2013 at 05:42 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,504
    Thanks
    1400

    Re: integration by complex exponentials

    Quote Originally Posted by earthboy View Post
    small doubt:
    is \sin x= \frac{e^{ix}- e^{-ix}}{2} or \sin x= \frac{e^{ix}- e^{-ix}}{2i}
    both are seem to be used in books, but what is the difference?
    It's \displaystyle \begin{align*} \sin{(x)} = \frac{e^{i\,x} - e^{-i\,x}}{2i} \end{align*}.
    Thanks from earthboy
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2010
    From
    in the 4th dimension....
    Posts
    122
    Thanks
    9

    Re: integration by complex exponentials

    Quote Originally Posted by HallsofIvy View Post

    That denominator can be factored (though not easily- use the quadratic formula to find the roots). Then it will be of the form
    -\frac{2i}{7}\int \frac{du}{(u- a)(u- b)}du
    where a and b are the roots. That can be integrated by "partial fracations".
    -\frac{2i}{7}\int \frac{du}{(u- a)(u- b)}du = -\frac{2i}{7} \frac{\log(u-a)-\log(u-b)}{a-b}
    where a=-13-4\sqrt{11} and b= 4\sqrt{11}-13

    how do I separate the real and imaginary part here to get the answer,especially what to do with the \frac{2i}{7} ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2010
    From
    in the 4th dimension....
    Posts
    122
    Thanks
    9

    Re: integration by complex exponentials

    Quote Originally Posted by earthboy View Post
    -\frac{2i}{7}\int \frac{du}{(u- a)(u- b)}du = -\frac{2i}{7} \frac{\log(u-a)-\log(u-b)}{a-b}
    where a=-13-4\sqrt{11} and b= 4\sqrt{11}-13

    how do I separate the real and imaginary part here to get the answer,especially what to do with the \frac{2i}{7} ?
    Any hint on how to separate the real and the imaginary part and find the answer from here?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. complex exponentials
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: January 26th 2013, 02:40 AM
  2. Quick question regarding complex exponentials
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: April 30th 2011, 07:09 PM
  3. Complex exponentials
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 28th 2010, 01:46 PM
  4. Converting Sinosoids to complex exponentials
    Posted in the Advanced Applied Math Forum
    Replies: 8
    Last Post: April 17th 2009, 12:59 PM
  5. complex number exponentials
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: September 10th 2007, 06:59 PM

Search Tags


/mathhelpforum @mathhelpforum