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I am trying to see why (in respect to a measure $\mu$), this equality is true for some $u$: $$\int e^{iu}fd\mu=\int \operatorname{Re}(e^{iu}f)d\mu$$

Is it because they have the same measure or is it generally true for a complex valued function $f$? ($\int e^{iu}fdz=\int \operatorname{Re}(e^{iu}f)dz$).

Context I am trying to understand that $\left|\int fd\mu\right|\leq\int|f|d\mu$. So there is a $u$, $u\in[0,2\pi)$ such that $\left|\int fd\mu\right|=e^{iu}\int fd\mu=\int e^{iu}fd\mu=\int \operatorname{Re}(e^{iu}f)d\mu\leq\int |e^{iu}f|d\mu=\int|f|d\mu$.

Iniciador
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  • Why would you think that the real part of a complex number is equal to the complex number? – Mark Viola Oct 29 '15 at 19:19
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    What is the source of this identity? I feel like the context that might make it true is missing. What's $u$, for example? Does $e^{iu}$ have any specific relationship to $f$, is it the complex conjugate of $f$ perhaps? – Aaron Golden Oct 29 '15 at 19:22
  • I edited,explaining a little more. – Iniciador Oct 29 '15 at 19:34
  • Is it maybe because the Im contains the sin function which is odd? – Iniciador Oct 29 '15 at 19:40
  • Something is still weird here. It's possible to choose $u$ so that $e^{iu}\int{f} = \left|{\int{f}}\right|$. $\int{f}$ is just some complex number $r e^{iv}$, so choose $u = -v$ and $e^{iu}\int{f} = e^{iu} r e^{iv} = e^{i(-v)} e^{iv} r = e^{i(v - v)}r = e^0 r = r = \left|{\int{f}}\right|$. And you can move the constant $e^{iu}$ inside the integral in the second step of your new note, but unless $f$ makes a constant ($-u$) angle with the positive real axis, then $e^{iu}f = Re(e^{iu}f)$ is not an identity. – Aaron Golden Oct 29 '15 at 20:02
  • @AaronGolden I think your explanation is what the OP needed (it should be the answer); the "$e^{iu}f = \operatorname{Re}(e^{iu}f)$" was their idea of why the integral inequality is true. It's an instance of the XY problem: one wants to know why Y is true, mistakenly thinks that it's because of X, and asks about X (instead of Y). –  Oct 30 '15 at 05:47
  • Do you mean $\left|\int{f}\right|\leq\int{\left|f\right|}$, less than or equal instead of equal? Is this a duplicate of http://math.stackexchange.com/questions/429220/integral-inequality-absolute-value? – Aaron Golden Oct 31 '15 at 19:03
  • Less than or equal. I am sorry for the confusion. I appreciate the effort helping me. So far I agree with what you say, I can find a u such that what I ask is not true or a measure that is not true. But maybe is what Normal Human says... – Iniciador Oct 31 '15 at 23:50

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