佐里奇《数学分析II》P136问题：求含参变量积分函数的偏导数

Alright, let's work through this problem step by step—we're going to derive the partial derivatives of ( F(x) ) straight from the definition, which will help us see exactly why the formula works, especially since ( f ) is continuous. This is a core result in multivariable calculus, and breaking it down from first principles is a great way to solidify your understanding.

We'll start by formalizing the problem setup, then derive the partial derivative for an arbitrary coordinate ( x_k ), and note the special case where ( n=1 ) (which matches the initial claim ( \frac{\partial F}{\partial x_1} = f(x) )).

Problem Setup #

First, let's fix the notation clearly to avoid confusion:

• ( I_{a,b} = { x = (x_1, x_2, \dots, x_n) \in \mathbb{R}^n \mid a_i \leq x_i \leq b_i \text{ for each } i=1,\dots,n } ): an n-dimensional closed rectangular box (interval) in ( \mathbb{R}^n ).
• ( f: I_{a,b} \to \mathbb{R} ): a continuous function on ( I_{a,b} ) (continuity is critical here—it ensures integrability and lets us use uniform continuity later).
• ( F(x) = \int_{I_{a,x}} f(t) dt ): where ( I_{a,x} = { t = (t_1, \dots, t_n) \in \mathbb{R}^n \mid a_i \leq t_i \leq x_i \text{ for each } i=1,\dots,n } ), the sub-box from the corner ( a = (a_1, \dots, a_n) ) to ( x ).

Deriving ( \frac{\partial F}{\partial x_k}(x) ) from First Principles #

We'll use the definition of the partial derivative for an arbitrary ( k \in {1, \dots, n} ):
$$
\frac{\partial F}{\partial x_k}(x) = \lim_{\Delta x_k \to 0} \frac{F(x + \Delta x_k e_k) - F(x)}{\Delta x_k}
$$
where ( e_k = (0, \dots, 1, \dots, 0) ) is the k-th standard basis vector (1 in the k-th position, 0 elsewhere).

Step 1: Rewrite the Integral Difference #

The function ( F(x + \Delta x_k e_k) ) is the integral over the box ( I_{a, x + \Delta x_k e_k} ), which is exactly ( I_{a,x} ) plus a "slab" where ( t_k ) ranges from ( x_k ) to ( x_k + \Delta x_k ), and all other ( t_i ) stay within ( [a_i, x_i] ). Let's call this slab ( J_k(\Delta x_k) ). So:
$$
F(x + \Delta x_k e_k) - F(x) = \int_{J_k(\Delta x_k)} f(t) dt
$$
Formally, ( J_k(\Delta x_k) = { t \in \mathbb{R}^n \mid a_i \leq t_i \leq x_i \text{ for } i \neq k, , x_k \leq t_k \leq x_k + \Delta x_k } ).

Step 2: Iterate the Integral with Fubini's Theorem #

Since ( f ) is continuous (hence integrable) on ( I_{a,b} ), Fubini's theorem lets us rewrite the integral over ( J_k(\Delta x_k) ) as an iterated integral, with the inner integral over ( t_k ):
$$
\int_{J_k(\Delta x_k)} f(t) dt = \int_{I_{a,x}^{(k)}} \left( \int_{x_k}^{x_k + \Delta x_k} f(t_1, \dots, t_k, \dots, t_n) dt_k \right) dt_1 \dots dt_{k-1} dt_{k+1} \dots dt_n
$$
Here, ( I_{a,x}^{(k)} ) is the (n-1)-dimensional box: ( { (t_1, \dots, t_{k-1}, t_{k+1}, \dots, t_n) \mid a_i \leq t_i \leq x_i \text{ for } i \neq k } ).

Step 3: Apply the Mean Value Theorem for Integrals #

For each fixed ( (t_1, \dots, t_{k-1}, t_{k+1}, \dots, t_n) \in I_{a,x}^{(k)} ), the function ( f(\cdot, t_1, \dots, t_{k-1}, t_{k+1}, \dots, t_n) ) is continuous in ( t_k ) (since ( f ) is continuous on ( I_{a,b} )). The Mean Value Theorem for Integrals tells us there exists some ( \xi_k \in [x_k, x_k + \Delta x_k] ) (depending on the fixed coordinates) such that:
$$
\int_{x_k}^{x_k + \Delta x_k} f(t_1, \dots, t_k, \dots, t_n) dt_k = f(t_1, \dots, \xi_k, \dots, t_n) \cdot \Delta x_k
$$

Step 4: Simplify and Take the Limit #

Substitute this back into the iterated integral:
$$
F(x + \Delta x_k e_k) - F(x) = \Delta x_k \cdot \int_{I_{a,x}^{(k)}} f(t_1, \dots, \xi_k, \dots, t_n) dt_1 \dots dt_{k-1} dt_{k+1} \dots dt_n
$$
Divide both sides by ( \Delta x_k ):
$$
\frac{F(x + \Delta x_k e_k) - F(x)}{\Delta x_k} = \int_{I_{a,x}^{(k)}} f(t_1, \dots, \xi_k, \dots, t_n) dt_1 \dots dt_{k-1} dt_{k+1} \dots dt_n
$$

Now take the limit as ( \Delta x_k \to 0 ). As ( \Delta x_k \to 0 ), ( \xi_k \to x_k ) (since ( \xi_k ) is between ( x_k ) and ( x_k + \Delta x_k )). Because ( f ) is uniformly continuous on the compact set ( I_{a,b} ), the function ( f(t_1, \dots, \xi_k, \dots, t_n) ) converges uniformly to ( f(t_1, \dots, x_k, \dots, t_n) ) as ( \Delta x_k \to 0 ). Uniform convergence allows us to interchange the limit and the integral:
$$
\lim_{\Delta x_k \to 0} \frac{F(x + \Delta x_k e_k) - F(x)}{\Delta x_k} = \int_{I_{a,x}^{(k)}} f(t_1, \dots, x_k, \dots, t_n) dt_1 \dots dt_{k-1} dt_{k+1} \dots dt_n
$$

Special Case: ( n=1 ) #

When ( n=1 ), the (n-1)-dimensional integral ( \int_{I_{a,x}^{(1)}} \dots ) collapses to a single point (since there are no other variables to integrate over), so the result simplifies to ( f(x_1) ). This matches the initial claim ( \frac{\partial F}{\partial x_1} = f(x) ).

Final Result #

For general ( n \geq 1 ) and each ( k=1,\dots,n ), the partial derivative of ( F(x) ) with respect to ( x_k ) is:
$$
\boxed{\frac{\partial F}{\partial x_k}(x) = \int_{I_{a,x}^{(k)}} f(x_1, \dots, x_k, \dots, x_n) dt_1 \dots dt_{k-1} dt_{k+1} \dots dt_n}
$$
Where ( I_{a,x}^{(k)} ) is the (n-1)-dimensional box obtained by fixing ( t_k = x_k ) and integrating over all other coordinates from ( a_i ) to ( x_i ).

内容的提问来源于stack exchange，提问作者Our