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Let be $f:\mathbb{R}\rightarrow \mathbb{C}$ Why $|\int_{-\infty}^{\infty}f(x)dx|\leq\int_{-\infty}^{\infty}|f(x)|dx$?

pdta:$|\cdot|$ is module of complex numbers

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    A related question: [Proving two integral inequalities](http://math.stackexchange.com/questions/216213/proving-two-integral-inequalities)2012-10-19

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Note that $$\left| \int_a^b f(x)dx \right| = e^{-i\theta}\int_a^b f(x)dx = \int_a^b e^{-i\theta}f(x)dx$$ where $\theta = \operatorname{Arg}\int_a^b f(x)dx$. But $\left| \int_a^b f(x)dx \right|$ is real, so $$\left| \int_a^b f(x)dx \right| = \operatorname{Re}\int_a^b e^{-i\theta}f(x)dx = \int_a^b \operatorname{Re}(e^{-i\theta}f(x))dx \leq \int_a^b \left| e^{-i\theta}f(x) \right|dx = \int_a^b \left| f(x) \right|dx.$$

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    Why $\int_a^b {\left| {{e^{ - i\theta }}f(x)} \right|dx} = \int_a^b {\left| {f(x)} \right|dx}$?2012-10-19
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    What does Euler formula tell us about $\left| {{e^{ - i\theta }}} \right|$?2012-10-19
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    mmm I think that $|f(x)|e^{i\theta} = f(x)$2012-10-19
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    I am not sure why you have $\left| {f(x)} \right|{e^{i\theta}}$. Euler formula gives: $\left| {{e^{ - i\theta }}} \right| = \sqrt {{{\cos }^2}(\theta ) + {{\sin }^2}(\theta )} = \ldots $ so $\left| {{e^{ - i\theta }}f(x)} \right| = \left| {{e^{ - i\theta }}} \right|\left| {f(x)} \right| = \ldots $2012-10-20
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    mmmm and $|e^{-i\theta}|$ is one?, Why?2012-10-28
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    and another question Why $\operatorname{Re} \int_a^b {{e^{ - i\theta }}f(x)dx = } \int_a^b {\operatorname{Re} ({e^{ - i\theta }}f(x))dx}$2012-10-28
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    OK, here is the [definition](http://mathworld.wolfram.com/ComplexModulus.html) of the mod function and you MUST know that ${\cos ^2}(\theta ) + {\sin ^2}(\theta ) = 1$ and that $\sqrt 1 = 1$. See [Argand diagram](http://mathworld.wolfram.com/ArgandDiagram.html) if you are not convinced.2012-10-29
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    Regarding Re, note that $\int_a^b {{e^{ - i\theta }}f(x)dx}$ is a number. So it make not difference if you evaluate the integral and take the real part of if you take the real part and evaluate.2012-10-29
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    $\int_a^b {{e^{ - i\theta }}f(x)dx}$ is a number? real?2012-10-29
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    It is a number, it could be complex or it could be real. It depends on $f(x)$.2012-10-29
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    but I don't understand yet Why "... not difference if you evaluate the integral and take the real part of if you take the real part and evaluate.", This is a property between integral and RE function?2012-10-29
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    This is exactly the same as $\operatorname{Re} ( - i) = - \operatorname{Re} i$, for example. If you are not convinced, find a proof in a complex analysis book.2012-10-29
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    @glebovg thanks for the answer does this inequality has a name to remember and to quote back when using it thanks2013-12-26
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    I do not think this inequality has a name. Sorry.2014-01-05