Real part of complex integral Let $x:\Omega \subset 
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\mathbb{R}^2
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\rightarrow 
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\mathbb{R}
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$ harmonic (I do not know if you need) and $\frac{\partial x}{\partial u_{1}},\frac{\partial x}{%
\partial u_{2}}:\Omega \subset 
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\mathbb{R}^2
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\rightarrow 
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\mathbb{R}
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$ their partial derivatives, with $\Omega $ simply connected open. Show that 
$$ 
x\left( z\right) =\Re \int_{0}^{z}\left( \frac{\partial x}{%
\partial u_{1}}\left( z\right) -i\frac{\partial x}{\partial u_{2}}\left(
z\right) \right) dz. $$
I have looked in books of complex analysis, but I can not find the property that allows me this equality.
 A: I don't think this has much to do with harmonic functions or complex analysis. Below I use different notation, otherwise I'll get confused. 
Assume $\Omega$ is open and connected (I don't think we need simple connectivity.) I'll be writing $w=x+iy$ for $w\in \mathbb C.$ Let $u:\Omega \to \mathbb R$ be continuously differentiable. Suppose $0,z\in \Omega$ and $\gamma:[a,b]\to \Omega$ is a contour with $\gamma(a)=0,\gamma (b)=z.$ Then
$$\tag 1\text { Re }\int_\gamma (u_x(w)-iu_y(w))\, dw = u(z)-u(0).$$
Proof:  Write $\gamma(t)= x(t)+iy(t).$ Then the left side of $(1)$ equals
$$\text { Re } \left [\int_a^b (u_x(\gamma(t))-iu_y(\gamma(t)))(x'(t)+iy'(t))\,dt \right ]$$ $$ = \int_a^b [u_x(\gamma(t))x'(t) + u_y(\gamma(t))y'(t)]\,dt $$ $$ = \int_a^b [u(\gamma(t))]'\, dt = u(\gamma(b)) - u(\gamma(a)) = u(z)-u(0).$$
The only things we have used are the chain rule and the FTC.
A: Ok, some comments: 


*

*The notation in the integral is a little bit confusing; Let us reinterpret it as being


$$ \text{Re}\left( \int_0^z \left(\frac{\partial x}{\partial u_1}(w) - i \frac{\partial x}{\partial u_2}(w)\right) \mathrm{d}w\right).$$


*

*Also, we shall assume that $0 \in \Omega,$ as otherwise the integration does not make sense. 

*Define $y:\Omega \subset \mathbb{R}^2 \to \mathbb{R}$ to be $x'$s harmonic conjugate. This exists, as $\Omega$ is open and simply connected. Therefore, letting $f(z) = x(z) + iy(z),$ we obtain, from the Cauchy-Riemann equation, that the integral (inside the real part) above is 
$$ \int_0^z \frac{\partial f}{\partial u_1}(w) \,\mathrm{d}w.$$
(Just use $\partial x/\partial u_2 = - \partial y/\partial u_1$). 


*

*The last step was only possible because, from the definition, $f$ is holomorphic. Therefore, it does not matter in which direction we take the partial derivative: it shall always yield the (complex) derivative $f'(z)$ of $f.$ Therefore, we rewrite the integral one more time as 


$$ \int_0^z f'(w) \, \mathrm{d}w. $$ 


*

*This last is a complex antiderivative to $f'$ on $\Omega,$ and differs from $f$ by only a constant factor. Explicitly, we get to the conclusion that 


$$ \int_0^z f'(w) \, \mathrm{d}w = f(z) - f(0).$$ 


*

*We finish by noticing that the harmonic function $x(z) - x(0)$ is the real part of $f(z) - f(0),$ which then yields 


$$ \text{Re}\left( \int_0^z \left(\frac{\partial x}{\partial u_1}(w) - i \frac{\partial x}{\partial u_2}(w)\right) \mathrm{d}w\right) = $$ 
$$ \text{Re} \left(\int_0^z f'(w) \, \mathrm{d}w \right) = \text{Re}(f(z) - f(0)) = x(z) - x(0).$$ 
(This means: you need a $x(0)$ to be added to the right hand side of your formula, but, aside from that, the result follows.) 
