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Thread: 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 $\displaystyle \int\Re\left\{f(x)\right\right}}\;{dx} = \Re\int{f(x)\;{dx}}$?
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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:

    $\displaystyle \cos \theta + i\cdot \sin \theta = e^{i \cdot \theta}$

    Recognize, for example, that the real portion:

    $\displaystyle \mathrm{Re}\{ e^{i \cdot \theta} \} = \cos \theta$

    Given an integral of the general form:

    $\displaystyle \int e^{x} \cos {2x} \; dx$

    We can complexify it:

    $\displaystyle \int \mathrm{Re}\{ e^{x} (\cos {2x} + i\cdot \sin {2x}) \} \; dx$

    $\displaystyle \int \mathrm{Re}\{ e^{x} (e^{i 2x}) \} \; dx$

    With basic rules of exponents:

    $\displaystyle \int \mathrm{Re}\{ e^{x + i2x} \} \; dx$

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

    $\displaystyle \mathrm{Re}\{ \int e^{x(1 + 2i)} \; dx \}$

    The integral easily evaluates:

    $\displaystyle \mathrm{Re}\{ \frac{e^{x(1 + 2i)}}{1 + 2i} \}$

    Multiplying and dividing by (1-2i):

    $\displaystyle \mathrm{Re} \{ \frac{1 - 2i}{5} e^{x(1 + 2i)} \}$

    Which can be rewritten as:

    $\displaystyle \mathrm{Re} \{ \frac{1 - 2i}{5} e^{x} e^{i2x} \}$

    Applying Euler's forumula:

    $\displaystyle \mathrm{Re} \{ \frac{1 - 2i}{5} e^{x} (\cos 2x + i\cdot \sin 2x) \}$

    Expanding:

    $\displaystyle \mathrm{Re} \{ \frac{e^{x}}{5} (\cos 2x +2 \sin 2x) + i\cdot \frac{e^{x}}{5} (\sin 2x -2 \cos 2x) \}$

    Taking the Real part of this expression:

    $\displaystyle \frac{e^{x}}{5} (\cos 2x +2 \sin 2x)$

    So:

    $\displaystyle \int e^{x} \cos {2x} \; dx = \frac{e^{x}}{5} (\cos 2x +2 \sin 2x)+C$
    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 $\displaystyle \int\Re\left\{f(x)\right\right}}\;{dx} = \Re\int{f(x)\;{dx}}$?
    Because, in essence, that is how we defined it.
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