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I'm currently studying complex analysis. My current thinking is as follows:

Let $f(t)=x(t)+iy(t)$. By definition, $$\int_{\gamma} f(t) \, \mathrm{d}t = \int_{a}^{b} f(\gamma(t)) \gamma'(t) \, \mathrm{d}t$$ and so by substitution, $$\int_{\gamma} f(t) \, \mathrm{d}t = \int_{a}^{b}x(\gamma(t))\gamma'(t) \, \mathrm{d}t + i \int_{a}^{b} y(\gamma(t))\gamma'(t) \, \mathrm{d}t.$$ Thus, $$Re\left( \int_{\gamma} f \right) =\int_{a}^{b}x(\gamma(t))\gamma'(t) \, \mathrm{d}t.$$ Now, $$\int_{\gamma} Re(f(t))\, \mathrm{d}t = \int_{\gamma}x(t) \, \mathrm{d}t = \int_{a}^{b} x(\gamma(t))\gamma'(t) \, \mathrm{d}t.$$ We see that these expressions are indeed the same, and so $Re\left( \int_{\gamma} f \right) = \int_{\gamma} Re(f)$. This seems so straightforward, and I've been trying for ages to come up with a counter example, but I haven't been able to find one. What are your thoughts?

Numbersandsoon
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    The notation $\int_\gamma f$ can be a little misleading from a formal perspective. – copper.hat Sep 21 '13 at 23:03
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    Yes, you should really avoid writing $\int_{\gamma} f$. Usually in the context of complex analysis we are working with $\int_{\gamma} f(z),dz$, but there are some situations where integrals such as $\int_\gamma f(z),|dz|$ also pop up. – mrf Sep 21 '13 at 23:08

2 Answers2

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This is not true. Take for example $\gamma$ as the unit circle and $f(z) = z$.

Then $\int_\gamma z\,dz = 0$, but \begin{align} \int_\gamma x\,dz &= \frac12\left(\int_\gamma z\,dz + \int_\gamma \bar z\,dz\right)\\ &= 0 + \frac12 \int_\gamma \frac{\bar z z}{z}\,dz \\ &= \frac12 \int_\gamma \frac{1}{z}\,dz = \pi i. \end{align}

mrf
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No: Define $f(x + iy) = iy$, and $\gamma$ is the line segment connecting $0$ and $i$. Then

$$\int_{\gamma} f = \int_0^1 (it) (i dt) = -\frac 1 2$$

On the other hand, $\Re{f} = 0$.

The flaw in the proof is that $\gamma'(t)$ need not be real.