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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.

$$

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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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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

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

In general, answer is “no”, because

$$\overline{ \int f(z) dz} = \overline{\int \left( \text{Re}f(z) + i\text{Im}f(z)\right)dz}=\\ \int \overline{ \left( \text{Re}f(z) + i\text{Im}f(z)\right)(dx+i dy)}=\int\overline{{\left(\text{Re}f(z)dx – \text{Im}f(z)dy\right)+ i(\text{Re}f(z)dy+\text{Im}f(z)dx)}}=\\ \int{\left(\text{Re}f(z)dx – \text{Im}f(z)dy\right)}-i \cdot\int{\left(\text{Re}f(z)dy+\text{Im}f(z)dx\right)}$$

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,

$$\int\limits_{\gamma} \overline{\xi} \; d\xi = 2\pi i.$$

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