Differentiation under the integral sign in a triple integral

Question

Let $F(a,b,c) = \iiint\limits_{V}f(x,y,z,a,b,c) dx\ dy\ dz$,

V = $\{(x,y,z)| \ ... \}$ a connected, bounded and closed domain in $R^3$,

and $B((0,0,0),r))$ a ball in $R^3$ as a domain for $(a,b,c)$.

Under the assumption that $f$ and $\frac{\partial f}{\partial a}$ for instance are continuous for $(x,y,z) \in V$ and $(a,b,c) \in B$

is it possible to claim that:

$\frac{\partial}{\partial a}F(a,b,c) = \frac{\partial}{\partial a} \iiint\limits_{V}f(x,y,z,a,b,c) dx\ dy\ dz = \iiint\limits_{V} \frac{\partial}{\partial a}f(x,y,z,a,b,c) dx\ dy\ dz $ ?

Answer 1

If you assume $f$ and $\dfrac{\partial f}{\partial a}$ are continuous as "a function of $6$ variables" $(x,y,z,a,b,c) \in V \times B$, then yes, the result is true, and the proof is pretty easy; it uses the fact that continuous functions on compact sets are uniformly continuous. Below, I present a statement which is valid in higher dimensions. You could probably get away with weaker hypotheses, but the version I'm stating is valid for many applications (at least the ones I've come across), and it is pretty easy to prove; the key ingredients are the (single-variable) mean-value theorem (for derivatives) and the fact that continuous fucntions on compact sets are uniformly continuous.

Theorem: Let $f: \mathbb{R^n} \times \mathbb{R^m} \to \mathbb{R}$ be a continuous function. Let $V \subset \mathbb{R^n}$ be a compact Jordan-measurable set. Now, define the function $F:\mathbb{R^m} \to \mathbb{R}$ by \begin{equation} F(\eta) = \int_V f(\xi, \eta) \, d^n \xi \end{equation} (here $d^n \xi$ means $d \xi_1 \cdots d \xi_n$). If $\dfrac{\partial f}{\partial \eta_i} : \mathbb{R^n}\times \mathbb{R^m} \to \mathbb{R}$ is continuous ($1 \leq i \leq m$), then, for every $\eta \in \mathbb{R^m}$, we have that \begin{equation} \dfrac{\partial F}{\partial \eta_i}(\eta) = \int_V \dfrac{\partial f}{\partial \eta_i}(\xi, \eta) \, d^n \xi. \end{equation}

A few points to note are the following:

• if you don't know/understand what Jordan-measurable means, don't worry about it; it's just a condition to ensure the set $V$ is nice enough so that the integrals appearing actually exist. For example, if $n = 3$, then $V$ being a cube/ellipsoid/pumpkin are all examples of compact Jordan-measurable sets
• Of course, there is no need to have the domain of $f$ be all of $\mathbb{R^n} \times \mathbb{R^m}$. I simply didn't bother with smaller domains to make the statement more readable.
• By writing $\xi = (x,y,z)$ and $\eta = (a,b,c)$ hopefully you can see how to recover the special case you mentioned