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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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    I do not think this inequality has a name. Sorry.2014-01-05