The dervative of a multi-dimensional path integral

Let the dimension be arbitrary. I want to calculate

$$L_i\equiv \frac{\partial}{\partial x_i} \int_{x-\epsilon/2}^{x+\epsilon/2} d\mathbf{z} \cdot \mathbf{A}$$

There are two ways to manipulate.
Strictly speaking, in order to postulate the existence of $\phi$ such that $$d\phi(\mathbf{z}) =A_i dz_i \quad \Leftrightarrow \quad \frac{\partial \phi(\mathbf{z})}{\partial z_i} =A_i \,,$$ (where repeated indices follow the Einstein summation convention), we need the condition $$\frac{\partial A_j}{\partial z_i} = \frac{\partial A_i}{\partial z_j} .$$ But we assume it in most cases.

Method 1
\begin{alignat}{2} L_i &=&& \frac{\partial}{\partial x_i} \int_{x-\epsilon/2}^{x+\epsilon/2} d\phi \\ &=&& \frac{\partial}{\partial x_i} (\phi(x+\epsilon/2)-\phi(x-\epsilon/2)) \\ &=&& \frac{\partial}{\partial x_i} \left( \phi(x)+ \frac{\epsilon_j}{2} \frac{\partial}{\partial x_j} \phi(x)-\phi(x) -\frac{-\epsilon_j}{2} \frac{\partial}{\partial x_j} \phi(x) + {\cal O}(\epsilon^2) \right) \\ &=&& \epsilon_j \frac{\partial}{\partial x_i} A_j + {\cal O}(\epsilon^2) \end{alignat}

Method 2
\begin{alignat}{2} L_i &=&& \frac{\partial}{\partial x_i} \int_{x-\epsilon/2}^{x+\epsilon/2} d\phi \\ &=&& \frac{\partial}{\partial x_i} \phi(x+\epsilon/2) - \frac{\partial}{\partial x_i} \phi(x-\epsilon/2) \\ &=&& A_i(x+\epsilon/2) - A_i(x-\epsilon/2) \\ &=&& A_i(x) + \frac{\epsilon_j}{2} \frac{\partial}{\partial x_j} A_i -A_i(x) - \frac{-\epsilon_j}{2} \frac{\partial}{\partial x_j} A_i +{\cal O}(\epsilon^2) \\ &=&& \epsilon_j \frac{\partial}{\partial x_j} A_i +{\cal O}(\epsilon^2) \end{alignat}

It seems to me that both methods are correct. Which method is correct?
(This question is related with this post.)
Thanks.

• unclear. You should write everything explicitly in the form $\frac{\partial}{\partial h}C\int_0^1 f(x_0+hv+t a) dt=C\int_0^1 \frac{\partial}{\partial h}f(x_0+hv+t a) dt$ – reuns Nov 24 '16 at 7:53
• @user1952009 : Is it OK now? – GotchaP Nov 27 '16 at 6:05
• If you are not sure, then replace $\frac{\partial}{\partial x_i} f$ by $\frac{\partial }{\partial h} f(x+hv) = \lim_{h \to 0} \frac{f(x+hv)-f(x)}{h}$ the directional derivative in the direction $v$ (since $g(h) = f(x+hv)$ is a function of the real variable $h$, its derivative is well-defined) – reuns Nov 27 '16 at 6:11
• I'm a physicist. I used the notation of Special reativity ($x+\epsilon/2 \equiv \mathbf{x}+\mathbf{\epsilon}/2$). Is it familiar for mathematicians? – GotchaP Nov 27 '16 at 13:46
• You need to write everything explicitely : $A$ is a vector field i.e. $A(z) = [A_1(z),\ldots,A_n(z)]\in \mathbb{R}^n$ and $dz.A(z) = \sum_{m=1}^n A_m(z)dz_m$ so by definition $\int_{x -\epsilon/2}^{x +\epsilon/2} dz.A = \int_{x -\epsilon/2}^{x +\epsilon/2} \sum_{m=1}^n A_m(z)dz_m = \sum_{m=1}^n \int_0^1A_m(x -\epsilon/2+t \epsilon) d(t \epsilon)_m$ $= \sum_{m=1}^n \epsilon_m \int_0^1A_m(x -\epsilon/2+t \epsilon) dt$ and differentiating this last expression is trivial – reuns Nov 27 '16 at 14:19

The point is that in general cases we cannot postulate the existence of $\phi$ such that $$d\phi(\mathbf{z}) =A_i dz_i \quad \Leftrightarrow \quad \frac{\partial \phi(\mathbf{z})}{\partial z_i} =A_i \, .$$
Only when the condition $\frac{\partial A_j}{\partial z_i} = \frac{\partial A_i}{\partial z_j}$ is satisfied, $\phi$ exists.
If we restrice the situation to the cases where $\frac{\partial A_j}{\partial z_i} = \frac{\partial A_i}{\partial z_j}$ is satisfied, both methods are correct. Then the results will be the same : $$L_i = \sum_{j=1}^n \epsilon_j \frac{\partial}{\partial x_i} A_j + {\cal O}(|\mathbf{\epsilon}|^2)$$ (n : the dimension of the vector space)