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Let be $f:\mathbb{R}\rightarrow \mathbb{C}$. This $$\operatorname{Re} \int_a^b {{e^{ - i\theta }}f(x)dx = } \int_a^b {\operatorname{Re} ({e^{ - i\theta }}f(x))dx}$$

is True? Why?

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What is your definition of $\int_a^b f(x)\,dx$ for a complex-valued $f$? –  wj32 Oct 29 '12 at 0:03
    
$\int_a^b f(x)\,dx$ is real number –  juaninf Oct 29 '12 at 0:13
    
That can't be right - what is $\int_0^1 i\,dx$ then? –  wj32 Oct 29 '12 at 0:14
    
mmmm understand –  juaninf Oct 29 '12 at 0:16
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1 Answer

Let $f:X\rightarrow \mathbb{C}$ be any complex valued function on a domain where integration makes sense. We can break up $f$ into $f_r$ which is real valued and $f_i$ which is real valued, so that $f=f_r+if_i$. Then $$\int f dx = \int (f_r+if_i) dx = \int f_r dx +i \int f_i dx.$$ $\int f_r dx$ and $\int f_i dx$ should both be real since their integrands are, and so it shouldn't matter whether you apply the $Re$ operator before or after the integration.

The only missing step is proving that if $f$ is integrable, so are $Re(f)$ and $Im(f)$. I'm afraid I don't have a proof for that, though I doubt it would be false.

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