Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $L^1$?

share|cite|improve this question
2  
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
up vote 12 down vote accepted

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

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

share|cite|improve this answer
    
And why does $ \overline{\int f(z) \; dz} = \int \overline{f(z) \; dz} $ hold? – Karlo Jun 15 at 11:44
    
@Karlo, The reason is explained above. – Sangchul Lee Jun 16 at 9:55
    
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 at 9:58
1  
@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 at 10:01

In general, answer is "no", because $$\overline{ \int f(z) dz} = \overline{\int \left( \text{Re}f(z) + i\text{Im}f(z)\right)dz}=\\ \int \overline{ \left( \text{Re}f(z) + i\text{Im}f(z)\right)(dx+i dy)}=\int\overline{{\left(\text{Re}f(z)dx - \text{Im}f(z)dy\right)+ i(\text{Re}f(z)dy+\text{Im}f(z)dx)}}=\\ \int{\left(\text{Re}f(z)dx - \text{Im}f(z)dy\right)}-i \cdot\int{\left(\text{Re}f(z)dy+\text{Im}f(z)dx\right)}$$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.