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$f:[a,b]\mapsto \mathbb{C}$ is Riemann integrable. It is given also that $a,b \in\ $ and that $a<b $. How do I show that:

$\int_{a}^{b} \! \overline{f(x)} \, dx=\overline{\int_{a}^{b} \! f(x) \, dx } $

$\int f(x) dx = \int \operatorname{Re} f(x) dx + i\int \operatorname{Im} f(x) dx$

$\int_{a}^{b} \! \overline{ \operatorname{Re} f(x) + i\int \operatorname{Im} } \, dx=\overline{\int_{a}^{b} \! \operatorname{Re} f(x) + i\int \operatorname{Im} \, dx } $

$\int_{a}^{b} \!{\operatorname{Re} f(x) dx - i\int \operatorname{Im} f(x) } \, dx={\int_{a}^{b} \! \operatorname{Re} f(x) dx - i\int \operatorname{Im} f(x) dx } $

Is that all that I need for proof?

galaxy--
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Maica
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  • Use the summation definition and apply the conjugation, you'll see that it converges to the conjugation of the sum – Chickenmancer Apr 16 '17 at 16:29
  • @Justin Assuming that you already have linearity, you need not go all the way back to Riemann/Darboux sums. If you don't have linearity already proven (and don't want to prove it) then that is a good approach. – Hayden Apr 16 '17 at 16:32
  • Good point @hayden – Chickenmancer Apr 16 '17 at 18:29

1 Answers1

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While your presentation of the proof isn't super clear, you have the right idea. A bit cleaner would be: \begin{align} \overline{\int_a^b{f(x) d x}} & = \overline{\int_a^b{[\mathrm{Re}f(x)+i\mathrm{Im}f(x)] dx}} \\ & = \overline{\int_a^b{\mathrm{Re}f(x) dx} + i\int_a^b{\mathrm{Im}f(x) dx}} \\ & = \int_a^b{\mathrm{Re}f(x) dx} - i\int_a^b{\mathrm{Im}f(x) dx} \\ & = \int_a^b{[ \mathrm{Re}f(x) - i\mathrm{Im}f(x)] dx} \\ & = \int_a^b{ \overline{f(x)} dx} \end{align}

Hayden
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