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Complexifying an integral.

  1. Jul 3, 2011 #1
    1. The problem statement

    1.1. Is it possible to do [itex] \int\ sin{x}\, \ cos{x}\, \ e^x\, \ dx\ [/itex] by complexifying the integral? (Note: not by integration by parts.)

    Complexifying the Integral (Arthur Mattuck, MIT) [9:23]


    1.2. When is it appropriate to complexify an integral, beside the condition that the integrand can be expressed as [itex] Re (\ e^{\alpha x})\, \ [/itex]?

    2. The attempt at a solution

    2.1.

    [tex] \begin{equation*}
    \begin{split}
    \int\ sin{x}\, \ cos{x}\, \ e^{x}\, \ dx\ =\\
    \int\ Re(\ e^{i(\frac{\pi}{2}\ -\ x)}\ )\, \ Re(\ e^{ix})\, \ e^{x}\ dx\ =\\
    Re\int\ e^{i(\frac{\pi}{2}\ -\ x)}\, \ e^{ix}\, \ e^{x}\ dx\ =\\
    Re\int\ i\ e^x\, \ dx\ =\\
    - Im( e^x )\ + \ C\, \, (?)
    \end{split}
    \end{equation*} [/tex]
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Jul 3, 2011 #2

    micromass

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    Hi HotMintea! :smile:

    You've made a mistake in your solution. You seem to think that Re(z)Re(z')=Re(zz'), but this is not the case. The real part doesn't behave that way. Thus you cannot say

    [tex]Re(e^{i(\pi/2-x)})Re(e^{ix})=Re(e^{i(\pi/2-x)+ix})[/tex]

    Instead, it would be better to use some trigonometric formula's in the beginning:

    [tex]\sin(x)\cos(x)=\frac{\sin(2x)}{2}[/tex]
     
  4. Jul 3, 2011 #3
    Hi micromass! :smile:

    Thank you for pointing out my mistake and also for the suggestion.

    [tex]
    \begin{equation*}
    \begin{split}
    \int\ sin{x}\, \ cos{x}\, \ e^x\ dx\, = \, \frac{1}{2}\int\ sin{2x}\, \ e^x\ dx\, \ = \\ \frac{1}{2}\, \ Re\int\ e^{i(\frac{\pi}{2} - 2x)}\, \ e^x\, \ dx\, = \, \ \frac{1}{2}\ Re(\frac{i}{1-2i}\ e^{(1-2i)x})\, \ + \, \ C\, \ = \\ \frac{1}{10}\ e^x\ Re((i-2)(cos{2x}-isin{2x}))\, +\, \ C\, \ = \, \frac{1}{10}\ e^x\, \ (sin{2x}-2cos{2x})\, \ + \, \ C
    \end{split}
    \end{equation*}
    [/tex]

    It now seems to me that complexification works and is effective if an integrand can be rewritten as [itex] Re(e^{\alpha x}[/itex]). Am I correct?
     
  5. Jul 3, 2011 #4

    micromass

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  6. Jul 3, 2011 #5
     
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