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Math Help - Why is the integral of Re(f(x)) the real part of the integral of f(x)?

  1. #1
    Junior Member Hardwork's Avatar
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    Why is the integral of Re(f(x)) the real part of the integral of f(x)?

    Why is \int\Re\left\{f(x)\right\right}}\;{dx} = \Re\int{f(x)\;{dx}}?
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  2. #2
    Junior Member Hardwork's Avatar
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    Perhaps a little context might help. I was trying to follow the following integration technique on Wikibooks:

    Spoiler:
    This technique requires an understanding and recognition of complex numbers.

    Specifically Euler's formula:

    [LaTeX ERROR: Convert failed]

    Recognize, for example, that the real portion:

    [LaTeX ERROR: Convert failed]

    Given an integral of the general form:

    [LaTeX ERROR: Convert failed]

    We can complexify it:

    [LaTeX ERROR: Convert failed]

    [LaTeX ERROR: Convert failed]

    With basic rules of exponents:

    [LaTeX ERROR: Convert failed]

    It can be proven that the "real portion" operator can be moved outside the integral:

    [LaTeX ERROR: Convert failed]

    The integral easily evaluates:

    [LaTeX ERROR: Convert failed]

    Multiplying and dividing by (1-2i):

    [LaTeX ERROR: Convert failed]

    Which can be rewritten as:

    [LaTeX ERROR: Convert failed]

    Applying Euler's forumula:

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    Expanding:

    [LaTeX ERROR: Convert failed]

    Taking the Real part of this expression:

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    So:

    [LaTeX ERROR: Convert failed]
    I was just curios on the highlighted part where it says: 'it can be proven that the "real portion" operator can be moved outside the integral:'.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Hardwork View Post
    Why is \int\Re\left\{f(x)\right\right}}\;{dx} = \Re\int{f(x)\;{dx}}?
    Because, in essence, that is how we defined it.
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