TITLE: Does the complex conjugate of an integral equal the integral of the conjugate?
QUESTION [47 upvotes]: 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 $L^1$?

REPLY [3 votes]: Yet another variation: let be $\overline{\gamma}(t) = \overline{\gamma(t)}, t\in[a,b]$. Then,
$$
\overline{\int_{\gamma}f(z)\,dz} =
\overline{\int_a^b f(\gamma(t))\gamma'(t)\,dt} =
\int_a^b\overline{f(\gamma(t))}\,\overline{\gamma'(t)}\,dt =
\int_a^b\overline{f(\overline{\overline{\gamma}(t)})}\,\overline{\gamma}'(t)\,dt =
\int_{\overline{\gamma}}\overline{f(\overline{z})}\,dz.
$$