# 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 $L^1$?

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actually it depends on the domain ex. $\int_{\mathbb{R}}g(x)^{2}dx=\infty$ so $g(x)\notin L^{2}[\mathbb{R}]$. – TKM Feb 10 '14 at 1:48

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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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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