# Change of variable formula when integrand involves partial derivatives

A well known change of variable theorem of integral calculus says the following.

Let $$X$$ and $$Y$$ be two open sets in $$\mathbb{R}^2$$ and $$\phi: X \rightarrow Y$$ be a diffeomorphism. Then $$\begin{eqnarray}\tag{1} \int\limits_{Y} f(y_1,y_2) dy_1 dy_2=\int\limits_{X} f(\phi(x_1,x_2)) \left|J_{\phi}(x_1,x_2)\right| dx_1 dx_2, \end{eqnarray}$$ where $$J_{\phi}$$ denotes the Jacobian matrix of $$\phi.$$

What is the corresponding change of variable formula if the integrand involves derivatives. More precisely, for $$f \in C^1$$ what is the change of variable formula if the integrand is $$\begin{eqnarray} \int\limits_{Y} \left(\partial_1f(y_1,y_2)+\partial_2f(y_1,y_2)\right) dy_1 dy_2=\int\limits_{X} \cdot\cdot\cdot\cdot dx_1 dx_2. \end{eqnarray}$$ How to prove the change of variable formula for such integrand using the change of formula (1) ?

• Try to compute the partial derivative of y_1 or y_2 in terms of x_1 and x_2 applying the chain rule. That would give you the change of variable. Finally you should multiply by |J|. Commented Apr 7, 2021 at 18:16
• Good question. I always get confused with this kind of computations. Commented Apr 7, 2021 at 19:32
• Unless there is something I am missing, the answer is straight forward: With $F(y_1,y_2)=\partial_1f(y_1,y_2)+\partial_2f(y_1,y_2)$, $$\int_{\phi(X)}F(\mathbf{y})\,d\mathbf{y}=\int_X F(\phi(\mathbf{x})|J_\phi(\mathbf{x})|\,d\mathbf{x}=\int_X \Big((\partial_1f)(\phi(\mathbf{x})) +(\partial_2f)(\phi(\mathbf{x}))\Big)|J_\phi(\mathbf{x})|\,d\mathbf{x}$$ The $\phi$ has nothing to do with $f$. Commented Apr 7, 2021 at 20:27
• @Oliver Diaz as far as I know, it does not work that way Commented Apr 7, 2021 at 20:41
• @Sameera: You may be confusing the change of variables through diffeomorphisms with the Stokes theorem: $\int_S \operatorname{div}(f_1,f_2)\,d\mathbf{x}=\int_{\partial S}(f_1,f_2)\cdot \mathbf{n}\,d\sigma$ Commented Apr 7, 2021 at 20:47

One element of confusion is coming from a somewhat imprecise notation. Writing $$\partial_1 f$$ is asking for trouble when you invoke a change of variables. If instead you write $$\frac{\partial f}{\partial y_1}$$ or $$\partial_{y_1} f$$ this removes (in principle) any ambiguity. For example if $$f(y_1,y_2)=y_1^2 y_2$$ then (omitting domains) $$\int \partial_{y_1} f (y_1,y_2) dy_1\, dy_2 =\int 2 y_1 y_2 \, dy_1\, dy_2$$ when changing variables gives $$\int \partial_{y_1}f\circ \phi(x_1,x_2) \; J(\phi)\; dx_1 dx_2 = \int 2\phi_1(x_1,x_2)\phi_2(x_1,x_2)\; J(\phi)\;dx_1 dx_2$$

I think that you just has a confusion with the notation, by example we will have that

$$\int_A \partial _1 f(x,y)d(x,y)=\int_{\phi ^{-1}(A)}(\partial _1f\circ \phi)(x,y)|\det [\partial \phi(x,y)]|d(x,y) \tag1$$

for some diffeomorphism $$\phi$$. Here $$\partial_1 f$$ is a function as any other (and above I used the symbol $$\partial$$ alone without subscript to denote the Fréchet derivative of the function $$\phi$$). Note that $$\partial _1 f\circ \phi$$ and $$\partial _1(f\circ \phi )$$ are two different things, the first one is the composition of the functions $$\partial _1 f$$ and $$\phi$$, but the second is the derivative respect to the first argument of the function $$f\circ \phi$$. Maybe something like that was your source of confusion.

Also observe that I used the notation $$(h\circ l)(x)$$ instead of $$h(l(x))$$, the first is preferably in almost all situations by many reasons, one of them is the enhanced clarity in it meaning because there is a separation between the notation for functions and it arguments.

Also note that the notation $$\partial_1$$ is more clear and simple than something like $$\partial_x$$: just observe that in (1) I didn't needed to denote in the second integral the arguments using different letters, however if I would used $$\partial_x$$ instead then I would need to change the name of the argument $$x$$ to something else because otherwise will be confusing, but it will be confusing anyway because then there is no argument $$x$$ to know what is the partial derivative that you took!