Let $\psi(x,\xi)$ take inputs $x = (x_1,...,x_n),\ \xi = (\xi_1,...,\xi_n)\in\mathbb{R}^n$ and let $\psi$ have continuous first partial derivatives. I wish to show that

$$ \frac{\partial}{\partial x_j}\int_{B_r(x)} \psi(x,\xi)\ d\xi = \int_{B_r(x)} \psi_{x_j}(x,\xi) + \psi_{\xi_j}(x,\xi)\ d\xi $$

I've made the variable substitution $\eta = \xi - x$ which has yielded

$$ \frac{\partial}{\partial x_j}\int_{B_r(x)} \psi(\xi-\eta,\xi)\ d\xi $$

This feels like the correct direction but how would I differentiate $\psi$ with this variable substitution?

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The problem here is that $x$ occurs in the region of integration. As long as that is the case, we cannot take the derivative without using the (rather complicated) Leibniz integral rule in higher dimensions. Simply renaming the variables and using the chain rule does not really help here.

Instead we need to understand $\eta = \xi - x$ as a change of the integration variable. It has unit Jacobian and it shifts the region of integration from $B_r(x)$ to $B_r(0)$. This yields $$\int_{B_r(x)} \psi(x,\xi) \, \mathrm{d} \xi = \int_{B_r(0)} \psi(x,\eta + x) \, \mathrm{d} \eta \, .$$ Let $\partial_k \psi$ denote the partial derivative of $\psi$ with respect to the $k$-th variable for $1 \leq k \leq 2n$ . Then by the chain rule we have $$ \frac{\partial}{\partial x_j} \psi(x,\eta + x) = \partial_j \psi(x,\eta+x) + \partial_{n+j} \psi(x,\eta+x) \, , \, 1 \leq j \leq n \, .$$ We can now differentiate under the integral sign and return to the original variable of integration afterwards to obtain \begin{align} \frac{\partial}{\partial x_j} \int_{B_r(x)} \psi(x,\xi) \, \mathrm{d} \xi &= \int_{B_r(0)} [\partial_j \psi(x,\eta+x) + \partial_{n+j} \psi(x,\eta+x)] \, \mathrm{d} \eta \\ &= \int_{B_r(x)} [\partial_j \psi(x,\xi) + \partial_{n+j} \psi(x,\xi)] \, \mathrm{d} \xi \\ &= \int_{B_r(x)} \left[\frac{\partial}{\partial x_j} \psi(x,\xi) + \frac{\partial}{\partial \xi_j} \psi(x,\xi)\right] \, \mathrm{d} \xi \end{align} for $1 \leq j \leq n$ as desired.

  • Wonderfully explained, thank you. – user512116 Aug 10 at 10:46
  • @Bailey You're welcome :) – ComplexYetTrivial Aug 10 at 10:47

Making the change to $\eta = \xi - x$, then $\partial/\partial x_j$ is taken via the chain rule, since now $\eta$ is a function of $x$.

  • Thanks. I've written $\frac{\partial \psi}{\partial x_j} = \frac{\partial \psi}{\partial \eta}\frac{\partial \eta}{\partial x}\frac{\partial x}{\partial x_j} + \frac{\partial \psi}{\partial \eta}\frac{\partial \eta}{\partial \xi}\frac{\partial \xi}{\partial x_j}$. Is this the correct form of the chain rule to apply? – user512116 Aug 10 at 6:48

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