Let $k\in\mathbb{N}$ and $f$ such that $f^{(i)}$ for $i=0,...,k-1$ is continuously differentiable on the bounded interval $\mathbb{I}\subset\mathbb{R}$ with $f^{(k)}$ absolute continuous on $\mathbb{I}$. Then, by the Taylor's theorem with the integral form of the remainder we have $$ f(x)=f(0)+f'(0)x+...+\frac{1}{k!}f^{(k)}(0)x^k+ R_k(x) $$ with $$ R_k(x) = \frac{1}{k!} \int_0^x (x-t)^k f^{(k+1)}(t) \, dt = \frac{x^k}{k!} \int_0^1 (1-s)^k f^{(k+1)}(xs) \, ds. $$ This can be proven by doing integration by parts a sufficient number of times. Now, I am wondering: what if $\mathbb{I}=\mathbb{R}$?
When I look at $R_k(x)$, then the inequality $$ |R_k(x)| = |\frac{1}{k!} \int_0^x (x-t)^k f^{(k+1)}(t)\, dt| \leq \frac{1}{k!}\int_0^x |x-t|^k|f^{(k+1)}(t)|\, dt $$ suggests to me that we need something like "$f$ must satisfy $\int_0^x |x-t|^k|f^{(k+1)}(t)|\, dt<\infty$ for all $x\in\mathbb{R}$" needs to hold. When $x$ is bounded, absolute continuity covers this. This can be seen from $$ \int_0^x |x-t|^k|f^{(k+1)}(t)|\, dt \leq \int_0^c |c-t|^k|f^{(k+1)}(t)|\, dt \leq|c|^k\int_0^c |f^{(k+1)}(t)|\, dt<\infty, $$ where we assumed $0 \leq x\leq c < \infty$. We can't do this for $x$ that gets arbitrarily large. So I think we need something extra for that. "$f$ must satisfy $\int_0^x |x-t|^k|f^{(k+1)}(t)|\, dt<\infty$ for all $x\in\mathbb{R}$" combined with the absolute continuity reduces to "$f$ must satisfy $\lim_{x\to\pm\infty}sgn(x)\int_0^x |x-t|^k|f^{(k+1)}(t)|\, dt<\infty$". This is not an expression I have been able to find elsewhere.
Question: Which additional explicit conditions do we have to place on $f$, next to $f^{(i)}$ for $i=0,...,k-1$ being continuously differentiable on $\mathbb{R}$ with $f^{(k+1)}$ existing in weak sense and being in $L^1(\mathbb{R})$, such that the integral form of the Taylor remainder theorem holds for all $x\in\mathbb{R}$?
Hunch: My guess is that I am missing something about the decay of $f^{(i)}$. Probably these need to go to zero at infinity instead of just being bounded.
The integral form of the remainder is valid even if $I = \mathbf R$. Note the integral form of the remainder doesn't even involve endpoints of $I$. Did you actually find a place in the derivation of the integral form of the remainder where boundedness of $I$ is ever used? I doubt it.
A reason you are seeing books present the integral form of the remainder only for bounded intervals is because if you want to take an additional step and get a nice bound on the remainder that doesn't involve integrals, then it is practical to do this when $I$ is bounded, since on bounded intervals the higher derivatives of $f$ will be bounded (continuous functions on closed bounded intervals are bounded).
The point is that power series often converge uniformly only on bounded intervals, not on the whole real line, even if the radius of convergence is infinite. For example, the power series $\sum x^n/n!$ converges for each $x \in \mathbf R$, but it doesn't converge uniformly on the whole real line, only on each bounded interval such as $[-c,c]$.