I am new to Complex and have done Vector Calculus quite a long time back. Please explain me the details of the proof given in my book.
Theorem: If $f$ and $g$ are $C^1$ functions on the rectangle $R=\{(x,y) \in \mathbb{R}^2: |x-a|<\delta, |y-b|< \epsilon\}$ and if $\frac{\partial f}{\partial y}\equiv \frac{\partial g}{\partial x}$ on $R$, then $\exists h$ such that $h$ is $C^2$ and $\frac{\partial h}{\partial x}\equiv f$ and $\frac{\partial h}{\partial y}\equiv g$
Proof: Set $h(x,y)=\int_a^x f(t,b)dt + \int_b^y g(x,s)ds$. Then by Fundamental theorem of Calculus $\frac{\partial h}{\partial y}(x,y)= g(x,y)$. This I understand. $\int_a^x f(t,b)dt$ would be a function in $x$ and so the differentiation of this w.r.t. $y$ would be 0.
Now, to calculate $\frac{\partial h}{\partial x}$ by Fundamental theorem $\frac{\partial }{\partial x}\int_a^x f(t,b)dt=f(x,b)$.
Also, $\frac{\partial }{\partial x} \int_b^y g(x,s)ds= \int_b^y \frac{\partial }{\partial x}g(x,s)ds=\int_b^y \frac{\partial }{\partial y}f(x,s)ds=f(x,y)-f(x,b)$. Thus $\frac{\partial h}{\partial x}= f$.
My question is why can't you conclude $\frac{\partial h}{\partial x}= f$ by Fundamental theorem of Calculus by the same way that we did for $\frac{\partial h}{\partial y}(x,y)= g(x,y)$? Why are we computing $\frac{\partial h}{\partial x}= f$ by such a convoluted procdure?
Note that
\begin{eqnarray} \frac{\partial h}{\partial x} &=& \frac{\partial }{\partial x} \left\{ \int_a^x f(t,b)dt + \int_b^y g(x,s)ds\right\} \\ &=& \underbrace{\color{red}{\frac{\partial }{\partial x} \int_a^x f(t,b)dt}}_{\text{apply first fundamental th.}} + \underbrace{\color{blue}{\frac{\partial }{\partial x} \int_b^y g(x,s)ds}}_{\text{move the deriv. inside}} \\ &=& \color{red}{f(x, b)} + \underbrace{\color{blue}{\int_b^y \frac{\partial }{\partial x} g(x,s)ds}}_{\text{use } \partial g/\partial x = \partial f/\partial y} \\ &=& \color{red}{f(x, b)} + \underbrace{\color{blue}{\int_b^y \frac{\partial }{\partial y} f(x,s)ds}}_{\text{use second fundamental th.}} \\ &=& \color{red}{f(x, b)} + \color{blue}{f(x,y) - f(x,b)} \\ &=& f(x,y) \end{eqnarray}
You are indeed using both parts of the fundamental theorem