Does the complex conjugate of an integral equal the integral of the conjugate? 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 $L^1$?
 A: Yet another variation: let be $\overline{\gamma}(t) = \overline{\gamma(t)}, t\in[a,b]$. Then,
$$
\overline{\int_{\gamma}f(z)\,dz} =
\overline{\int_a^b f(\gamma(t))\gamma'(t)\,dt} =
\int_a^b\overline{f(\gamma(t))}\,\overline{\gamma'(t)}\,dt =
\int_a^b\overline{f(\overline{\overline{\gamma}(t)})}\,\overline{\gamma}'(t)\,dt =
\int_{\overline{\gamma}}\overline{f(\overline{z})}\,dz.
$$
A: 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
$$ \overline{\int_C f(z) \; dz} = \overline{\int_I f(\gamma(t)) \gamma'(t) \; dt} = \int_I \overline{f(\gamma(t)) \gamma'(t)} \; dt= \int_C \overline{f(z)} \; \overline{dz}. $$
A: 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, 
$$\int\limits_{\gamma} \overline{\xi} \; d\xi = 2\pi i.$$
A: In general, answer is "no", because
$$\overline{ \int f(z) dz} = \overline{\int \left( \text{Re}f(z) + i\text{Im}f(z)\right)dz}=\\ \int \overline{ \left( \text{Re}f(z) + i\text{Im}f(z)\right)(dx+i dy)}=\int\overline{{\left(\text{Re}f(z)dx - \text{Im}f(z)dy\right)+ i(\text{Re}f(z)dy+\text{Im}f(z)dx)}}=\\ \int{\left(\text{Re}f(z)dx - \text{Im}f(z)dy\right)}-i \cdot\int{\left(\text{Re}f(z)dy+\text{Im}f(z)dx\right)}$$
