Prob. 9, Chap. 6, in Baby Rudin: Integration by parts for improper integrals

Here is Prob. 9, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Show that integration by parts can sometimes be applied to the "improper" integrals defined in Exercises 7 and 8. (State appropriate hypotheses, formulate a theorem, and prove it.) For instance show that $$\int_0^\infty \frac{\cos x}{1+x} \ \mathrm{d} x = \int_0^\infty \frac{\sin x}{(1+x)^2} \ \mathrm{d} x.$$ Show that one of these integrals converges absolutely, but that the other does not.

Here are the links to my Math SE posts on Probs. 7 and 8, Chap. 6 in Baby Rudin, 3rd edition:

Prob. 7 (a), Chap. 6, in Baby Rudin: If $f$ is integrable on $[c, 1]$ for every $c>0$, then $\int_0^1 f(x) \ \mathrm{d}x =$ . . .

Prob. 7 (b), Chap. 6, in Baby Rudin: Example of a function such that $\lim_{c \to 0+} \int_c^1 f(x) \ \mathrm{d}x$ exists but . . .

Prob. 8, Chap. 6, in Baby Rudin: The Integral Test for Convergence of Series

And, here is Theorem 6.22 (integration by parts) in Baby Rudin:

Suppose $F$ and $G$ are differentiable functions on $[a, b]$, $F^\prime = f \in \mathscr{R}$, and $G^\prime = g \in \mathscr{R}$. Then $$\int_a^b F(x) g(x) \ \mathrm{d} x = F(b)G(b) - F(a) G(a) - \int_a^b f(x) G(x) \ \mathrm{d} x.$$

My Attempt:

Here I will only be formulating the analog of the result in Prob. 7 (a).

Suppose $F$ and $G$ are real differentiable functions on $(0, 1]$ such that $F^\prime = f \in \mathscr{R}$ and $G^\prime = g \in \mathscr{R}$ on $[c, 1]$ for every $c > 0$. Suppose that $\lim_{x \to 0+} \left[ F(x) G(x) \right]$ exists and is finite. If $f \in \mathscr{R}$ and $g \in \mathscr{R}$ on $[0, 1]$, then we have $$\int_0^1 F(x) g(x) \ \mathrm{d} x = F(1) G(1) - \lim_{c \to 0+} F(c) G(c) \ - \ \int_0^1 f(x) G(x) \ \mathrm{d} x.$$

Is this result correct? If so, is this the result required by Rudin?

Now here is my proof:

By Theorem 6.22 in Baby Rudin, we see that for any $c$ such that $0 < c < 1$, we have $$\int_c^1 F(x) g(x) \ \mathrm{d} x = F(1)G(1) - F(c) G(c) - \int_c^1 f(x) G(x) \ \mathrm{d} x.$$ Now applying the conclusion of Prob. 7 (a) to this identity we obtain \begin{align} \int_0^1 F(x) g(x) \ \mathrm{d} x &= \lim_{c \to 0+} \int_c^1 F(x) g(x) \ \mathrm{d} x \\ &= \lim_{c \to 0+} \left( F(1)G(1) - F(c) G(c) - \int_c^1 f(x) G(x) \ \mathrm{d} x \right) \\ &= F(1) G(1) - \lim_{c \to 0+} F(c) G(c) - \lim_{c \to 0+} \int_c^1 f(x) G(x) \ \mathrm{d} x \\ &= F(1) G(1) - \lim_{c \to 0+} F(c) G(c) \ - \ \int_0^1 f(x) G(x) \ \mathrm{d} x, \end{align} as required.

Is my proof correct and rigorous enough for Rudin?