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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 $L^1$?

Thomas Russell
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4 Answers4

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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

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

Sangchul Lee
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    And why does $ \overline{\int f(z) ; dz} = \int \overline{f(z) ; dz} $ hold? – Karlo Jun 15 '16 at 11:44
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    @Karlo, The reason is explained above. – Sangchul Lee Jun 16 '16 at 09:55
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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? – Karlo Jun 16 '16 at 09:58
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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. – Sangchul Lee Jun 16 '16 at 10:01
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    Hello, This is some time ago, but still: What happens if the integral $${\int_I f(\gamma(t)) \gamma'(t) ; dt}$$ is not defined in terms of a Riemann integral? – Diger May 21 '19 at 14:06
  • @Diger In most complex analysis textbooks, the integral of a complex-valued function is defined to split across the real and imaginary parts, regardless of whether the underlying integration is Riemann's or not. See, for instance, the discussion in this answer, – Mark S. Feb 13 '25 at 03:04
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In general, answer is "no", because $$\overline{ \int f(z) dz} = \overline{\int \left( \text{Re}f(z) + i\text{Im}f(z)\right)dz}=\\ \int \overline{ \left( \text{Re}f(z) + i\text{Im}f(z)\right)(dx+i dy)}=\int\overline{{\left(\text{Re}f(z)dx - \text{Im}f(z)dy\right)+ i(\text{Re}f(z)dy+\text{Im}f(z)dx)}}=\\ \int{\left(\text{Re}f(z)dx - \text{Im}f(z)dy\right)}-i \cdot\int{\left(\text{Re}f(z)dy+\text{Im}f(z)dx\right)}$$

M. Strochyk
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4

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

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

Martín-Blas Pérez Pinilla
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  • if this holds then we can go further as $dz=\overline{d\bar z}$ so $$\int_{\gamma }f(z)dz=\int_{\bar \gamma }f(\bar z)d\bar z$$ –  Jun 14 '22 at 05:53