# Why is the integral of Re(f(x)) the real part of the integral of f(x)?

• Nov 2nd 2010, 05:45 PM
Hardwork
Why is the integral of Re(f(x)) the real part of the integral of f(x)?
Why is $\int\Re\left\{f(x)\right\right}}\;{dx} = \Re\int{f(x)\;{dx}}$?
• Nov 4th 2010, 03:34 PM
Hardwork
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:

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Recognize, for example, that the real portion:

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Given an integral of the general form:

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We can complexify it:

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With basic rules of exponents:

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

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The integral easily evaluates:

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Multiplying and dividing by (1-2i):

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Which can be rewritten as:

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Applying Euler's forumula:

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Expanding:

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Taking the Real part of this expression:

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So:

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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:'.
• Nov 4th 2010, 05:29 PM
Drexel28
Quote:

Originally Posted by Hardwork
Why is $\int\Re\left\{f(x)\right\right}}\;{dx} = \Re\int{f(x)\;{dx}}$?

Because, in essence, that is how we defined it.