Differentiation Under the Integral Sign w Variable Substitution

Question

Let $f\in C^1(B_\rho(\xi))$, $\xi\in\mathbb{R}^n$ and $\rho>0$. I wish to show

$$ \frac{d}{d\rho} \int_{B_\rho(\xi)} f(x)\ dx = \frac{1}{\rho} \int_{B_\rho(\xi)} nf(x) + x\cdot Df(x)\ dx $$

I'm using the following fact: let $f\in C^1(\mathbb{R}^n)$, $\psi\in C^1(\mathbb{R}^{n+1})$ where $\psi(y,\rho) = f(y\rho)$ for $y\in\mathbb{R}^n$ and $\rho>0$. Then

$$ \frac{\partial}{\partial\rho} \psi(y,\rho) = \frac{1}{\rho} \sum_{i=1}^n y_i \frac{\partial \psi(y,\rho)}{\partial y_i} $$

I've begun with the variable substitution $x=y\rho$ which implies $f(x) = f(y\rho) = \psi(y,\rho)$. Then

$$ \frac{\partial}{\partial\rho} f(x) = \frac{\partial}{\partial \rho} \psi(y,\rho) = \frac{1}{\rho} \sum_{i=1}^n y_i \frac{\partial \psi(y,\rho)}{\partial y_i} $$

Here's the part I'm stuck on. Is the following correct?

$$ \frac{1}{\rho} \sum_{i=1}^n y_i \frac{\partial \psi(y,\rho)}{\partial y_i} = \frac{1}{\rho} \bigg[x \cdot Df(x)\bigg] $$

I think I'm forgetting an important step with the $B_\rho(\xi)$ in the bounds of integration and/or the $dx$ with the variable substitution.

Answer 1

\begin{align} \frac d{d\rho}\int_{B_\rho(\xi)}f(x) \ dx &\overset{1.}{=} \frac{d}{d\rho}\int_{B_1(0)} f(\xi + \rho y) \ \rho^{n}dy \\&\overset{2.}{=} \int_{B_1(0) }y\cdot \nabla f(\xi+\rho y)\rho^n + f(\xi+\rho y) \ n\rho^{n-1} \ dy \\&\overset{3.}{=} \frac{1}{\rho} \int_{B_\rho(\xi)} nf(x) + (x-\xi)\cdot Df(x)\ dx\end{align} where:

1. Change of variables $x-\xi= \rho y $, means that $y$ is in $B_1(0)$, and $dx = \rho^n dy$.
2. Follows from product rule, $$ \frac d{d\rho}(fg) = (\frac d{d\rho} f )g + f \frac{d}{d\rho} g $$and chain rule, $$ \frac d{d\rho} f(\vec X(\rho)) = (\nabla f)(\vec X(\rho)) \cdot\left(\frac d{d\rho} \vec X (\rho)\right).$$
3. Undoing the change of variables $x-\xi= \rho y$. Note $\rho^{n-1} dy = \frac1\rho dx$.

$Df = \nabla f$ is the gradient of $f$. Regularity is used in step 2 to allow differentiation under the integral sign.

There is an additional term which is a multiple of $$ \int_{B_\rho (\xi)} \xi \cdot \nabla f = \int_{B_\rho(\xi)} \nabla_x \cdot (\xi f(x)) \ dx = \int_{\partial B_\rho (\xi)} f \xi\cdot n $$