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Stuck in Integrating e^(i*x)cos(x)

  1. Feb 28, 2016 #1
    I'am trying to prove

    [tex]\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}[/tex]

    Wolfram tells so http://integrals.wolfram.com/index.jsp?expr=e^(i*x)cos(x)&random=false

    But I am stuck in obtaining the first term:

    My step typically involved integration by parts:

    let [itex]u=e^{ix}cos(x)[/itex] and [itex]dv=dx[/itex]

    so:

    [tex]du=-e^{ix}sin(x)dx+icos(x)e^{ix}[/tex][tex]du=ie^{ix}(sin(x)+cos(x))dx[/tex][tex]du=ie^{2ix}dx[/tex]

    the other one is just: [itex]v=x[/itex]

    then:
    [tex]\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\int xe^{2ix}dx[/tex]

    We will again do another integration by parts for the second term, so we let [itex]u=x[/itex] and [itex]dv=e^{2ix}dx[/itex] then solving further, we obtain:
    [tex]\int xe^{2ix}dx=\frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}[/tex]
    plugging it back to the original problem, then doing simple distribution, we will obtain:
    [tex]\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\left ( \frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}\right )[/tex]
    [tex]\int e^{ix}cos(x) dx=xe^{ix}cos(x)- \frac{1}{2}xe^{2ix}-\frac{1}{4}ie^{2ix}[/tex]
    notice that we have proved the 2nd term, but the other half is badly away from what the Integration Table and Tools says:
    [tex]\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}[/tex]

    What have I gone wrong?
     
  2. jcsd
  3. Feb 28, 2016 #2

    SteamKing

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    You can side step a lot of the work above if you remember:

    $$cos\, x = \frac{e^{ix}+e^{-ix}}{2}$$
     
  4. Feb 28, 2016 #3
    I'm sorry, I couldn't follow your answer. But the integral is quite easy. Try using Euler's formula for e^x. Once you do ( and use a simple trigonometric identity) it's an easy
    integration.
    .
    .
    Edit: Or you can use the way in the post above mine, your convenience.
     
    Last edited: Feb 28, 2016
  5. Feb 28, 2016 #4

    Mark44

    Staff: Mentor

    All you really need to do is to show that ##\frac d{dx}[\frac x 2 - \frac 1 4 ie^{2ix}] = e^{ix}\cos(x)##. This isn't that hard to do.
     
  6. Feb 28, 2016 #5

    mathman

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    Use streamking's approach [itex]cos(x)=\frac{e^{ix}+e^{-ix}}{2}[/itex]. It is now a trivial problem.
     
  7. Feb 28, 2016 #6
    Darn! Applied steamking's approached, solved in 2 lines only. -_-

    Thanks everyone!
     
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