A proof involving derivatives of improper integrals of the form $\int_c^\infty f(x,t)dt$

Question

Given $F(x) = \int_c^\infty f(x,t)dt$ exists for each $x\in[a,b]$ and is continuous, $f(x,t), f_x(x,t)$ continuous on $D = [a, b] \times [c, \infty)$, show that

$$F'(x) = \int_c^\infty f_x(x,t)dt\text{,}$$

provided this integral converges uniformly.

This problem is very similar to the one here except that this integral is improper. I've been able to show the the proper case using a different approach from the one in the other post but my technique required using an $\epsilon$/(width of the interval), but this approach doesn't apply here.

Answer 1

Consider the sequence of functions $F_n:[a,b] \to \mathbb{R}$, where

$$F_n(x) = \int_c^{c+n} f(x,t) \, dt$$

Since $f,f_x$ are continuous on the compact set $[a,b] \times [c,c+n]$, it follows (from the linked result) that $F_n$ is differentiable on $[a,b]$ and

$$F_n'(x) = \int_c^{c+n} f_x(x,t) \, dt$$

By hypothesis, $F(x) = \int_c^\infty f(x,t) \, dt$ exists for all $x \in [a,b]$ and, thus, $F_n(x) \to F(x)$ as $n \to \infty$. We are also give that the improper integral $\int_c^\infty f_x(x,t) \, dt$ is uniformly convergent for $x \in [a,b]$, which implies that $F_n'(x) \to \int_c^\infty f_x(x,t) \, dt$ uniformly as $n \to \infty$ for $x \in [a,b]$.

Now we can apply a well-known theorem (Theorem 7.17 in Priciples of Mathematical Analyis by Rudin)

Consider a sequence of differentiable functions $F_n: [a,b] \to \mathbb{R}$. Suppose there is a point $x_0 \in [a,b]$ such that the sequence of real numbers $(F_n(x_0))$ converges, and the sequence of derivatives $F_n'$ converges uniformly to a function $G$ on $[a,b]$. Then there exists a differentiable function $F: [a,b] \to \mathbb{R}$ such that $F_n(x) \to F(x)$ and $F_n'(x) \to G(x)= F'(x)$ as $n \to \infty$ for all $x \in [a,b]$

Applying this theorem, we get

$$F'(x) = \lim_{n \to \infty} F_n'(x) =\lim_{n \to \infty} \int_c^{c+n} f_x(x,t) \, dt= \int_c^\infty f_x(x,t) \, dt$$