# Show that ${\partial^2F \over \partial{x}\partial{y}}$ vanishing implies $F$ is sum of arbitrary functions of $x$ and $y$

Can someone please provide a snappy demonstration of why ${\partial^2F \over \partial{x}\partial{y}} \equiv 0 \rightarrow F = f(x) + g(y)$ over some region? I can see it's "obviously" true but I'm struggling to find an elegant way to show it.

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Let $G(x,y)=\frac{\partial F}{\partial y}$. Fix $y$. Then $$\frac{\partial G}{\partial x}=0\implies G=c$$ where $c$ is a constant. Index it by the $y$ we had fixed. Applying definitions, $$\frac{\partial F}{\partial y}=c_y=c(y)$$ Now fix $x$ and allow $y$ to vary. By the FTC, $$F = g(y)+d$$ where $g$ is an antiderivative of $c$ and $d$ is a constant. Index $d$ by the $x$ we had fixed. Putting it all together, $$F(x,y)=g(y)+d_x=f(x)+g(y)$$ after suitable renaming.

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I marked this correct but I'm still not that happy. Can't we do both directions at once via circles or rectangles or something. It's completely symmetrical after all. And does $${\partial G \over \partial x} = 0 \rightarrow G = c$$ Why couldn't $${\partial F \over \partial y} = y ?$$ Do we need to explicitly consider the single-variable function along the line we fix? I'm still waiting for the proof of this "from The Book". –  TheMathemagician Jan 14 '13 at 16:42
Pay more attention to my solution... $$\frac{\partial F}{\partial y}=y$$ is allowed. In fact, notice I wrote $$\frac{\partial F}{\partial y}=c(y)$$ Take $c(y)=y$ to obtain your special case. –  pre-kidney Jan 17 '13 at 5:39

Hint. Use the Leibniz integral rule: if we have an integral of the form

$$\int_{y_0}^{y_1} f(x, y) \,dy$$

then for $x\in [x_0,x_1]$ the derivative of this integral is thus expressible

$${d\over dx} \left ( \int_{y_0}^{y_1} f(x, y) \,dy \right )= \int_{y_0}^{y_1} f_x(x,y)\,dy$$

provided that $f$ and its partial derivative $f_x$ are both continuous over a region in the form $[x_o,x_1]\times[y_0,y_1]$.

And use the Fundamental Theorem of Calculus and Leibniz Integral Rule on the functions $f_x=\frac{\partial f}{\partial x}$, $f_y=\frac{\partial f}{\partial y}$, $\frac{\partial f}{\partial x\partial y}$ and $\frac{\partial f}{\partial y\partial x}$. Explore a simetry $$\frac{\partial f}{\partial x\partial y}= \frac{\partial f}{\partial y\partial x}$$

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