Let $f:\Omega\subset\mathbb{C}^n\rightarrow\mathbb{C}$ a holomorphic function. For any $m\leq n$ I would like to find an $m-primitive$ of $f$, which is still holomorphic in each variable.
If we assume that for any $z=(z_1,...,z_n)\in\Omega$, the set $\Omega^m(z):=\{w\in\mathbb{C}\mid (z_1,...,z_{m-1},w,z_{m+1},...,z_n\in\Omega\}$ is simply connected, we could define
$$F^m(z):=\int_\gamma f(z_1,...,z_{m-1},w,z_{m+1},...,z_n)dw,
$$
where $\gamma$ is any path from a chosen $z^0\in\Omega$, with $z_m^0=z_m$ to $z$ entirely contained in $\Omega^m(z)$.
It is easy to see that $F^m$ is a $m-$primitive of $f$. But, what about the other variables? Is $F^m$ still holomorphic w.r.t. $z_i$, for $i=1,...,n$?
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If $\Omega\subset \mathbb{C}^n$, and $f:\Omega \rightarrow \mathbb{C}$ is a let's say a smooth function, then you can look at $f$ as a function of $2n$ variables mapping to $\mathbb{R}^2$. First suppose that $n=1$, then it is a classical theorem of complex calculus that $f:\Omega \rightarrow \mathbb{C}$ is holomorphic, if and only it satisfies the Cauchy-Riemann equations, i.e. if $f(x,y)=(f_1(x,y),f_2(x,y))$, then $\partial_{x}f_1=\partial_y f_2$, and $\partial_{y}f_1=-\partial{x}f_2$. (Satisfiyng the Cauchy-Riemann equation means that the tangent map of $f$ in the basis $\partial_x$ and $\partial_y$ is a Complex linear map.) Now we can encode this system in one clever equation. Let $\partial_{\overline{z}}=1/2(\partial_x+i\partial_y)$. This operator acts on complex valued function as follows, if $f(x,y)=f_1(x,y)+f_2(x,y)$, then $$\partial_{\overline{z}}f(x,y)=1/2(\partial_xf_1(x,y)-\partial_yf_2(x,y)+i(\partial_xf_2(x,y)+\partial_yf_1(x,y)).$$ It follows that $f$ satisfies the Cauchy-Riemann eqations if and only if $\partial_{\overline{z}}f=0$. Now suppose that $n>1$, and suppose that $z_j=x_j+iy_j$, then we can define the operators $\partial_{z_j}=1/2(\partial_{x_j}+i\partial_{y_j})$, and this acts on complex valued functions just like before. Now we see, that a function $f$ is partially homolorphic, if and only if $\partial_{\overline{z}_j}f=0$ for all $j$. One can show using Cauchy's integral formula, that a smooth function is holomorphic if and only if it is partially holomorphic. Now getting back to your question, you only have to check that $\partial_{\overline{z}_j}F^m(Z)=0$, for all $j$, and this follows from the fact that $f$ is smooth , so you can swap the derivation and integration. (For the variable $m$ to show that $\partial_{\overline{z}_m}F^m(z)=0$ you just have to copy the proof of the one variable case when we make the primitive function.)
