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Let $f$ be a complex valued function of a complex variable. Does $$ \overline{\int f(z) dz} = \int \overline{f(z)}dz \text{ ?} $$

If $f$ is a function of a real variable, the answer is yes as $$ \int f(t) dt = \int \text{Re}(f(t))dt + i\text{Im}(f(t))dt. $$

If $f$ is a complex valued function of a complex variable and belong to $L^2$, the answer is also yes as $L^2$ is a Hilbert space and, by conjugate symmetry of the inner product, $$ \overline{\langle f,g\rangle}=\langle g,f\rangle $$ where $g(z)=1$ is the identity function.

Apart from these two cases, is it otherwise true?
Is it true in $L^1$?

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    actually it depends on the domain ex. $\int_{\mathbb{R}}g(x)^{2}dx=\infty$ so $g(x)\notin L^{2}[\mathbb{R}]$.2014-02-10

3 Answers 3

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If $\int dz$ denotes a contour integral, then the answer is generally no. A correct formula is as follows:

$$ \overline{\int f(z) \; dz} = \int \overline{f(z)} \; \overline{dz}. $$

Indeed, let $\gamma : I \to \Bbb{C}$ be a nice curve parametrizing the contour $C$, then

$$ \overline{\int_C f(z) \; dz} = \overline{\int_I f(\gamma(t)) \gamma'(t) \; dt} = \int_I \overline{f(\gamma(t)) \gamma'(t)} \; dt= \int_C \overline{f(z)} \; \overline{dz}. $$

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    And why does $ \overline{\int f(z) \; dz} = \int \overline{f(z) \; dz} $ hold?2016-06-15
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    @Karlo, The reason is explained above.2016-06-16
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    More precisely, why does $\overline{\int_I f(\gamma(t)) \gamma'(t) \; dt} = \int_I \overline{f(\gamma(t)) \gamma'(t)}$ hold?2016-06-16
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    @Karlo, Essentially that is because integral is 'sum of infinitesimals' so that we can distribute conjugate to each summand. Of course, the precise justification depends on how we define integral. In this case, $\int_I \cdots \mathrm{d}t$ is simply a Riemann integral over the interval $I \subset \Bbb{R}$, and this perfectly makes sense.2016-06-16
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In general, answer is "no", because $$\overline{ \int f(z) dz} = \overline{\int \left( \text{Re}f(z) + i\text{Im}f(z)\right)dz}=\\ \int \overline{ \left( \text{Re}f(z) + i\text{Im}f(z)\right)(dx+i dy)}=\int\overline{{\left(\text{Re}f(z)dx - \text{Im}f(z)dy\right)+ i(\text{Re}f(z)dy+\text{Im}f(z)dx)}}=\\ \int{\left(\text{Re}f(z)dx - \text{Im}f(z)dy\right)}-i \cdot\int{\left(\text{Re}f(z)dy+\text{Im}f(z)dx\right)}$$

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Sangchul Lee provides a nice answer for the explicit computation of $$\overline{\int\limits_{\gamma} f(\xi) \; d\xi}.$$

An easy example demonstrating that, in general, $$\overline{\int\limits_{\gamma} f(\xi) \; d\xi} \neq \int\limits_{\gamma} \overline{f(\xi)} \; d\xi$$ is the following:

Let $\gamma$ be the curve wrapping once around the unit circle, then it is clear from elementary complex analysis that $$\overline{\int\limits_{\gamma} \xi \; d\xi} = 0;$$ whereas, $$\int\limits_{\gamma} \overline{\xi} \; d\xi = 2\pi i.$$