Can the interval defined in the fundamental theorem of calculus be $[-\infty,\infty]$? With regards to the fundamental theorem of calculus, the statement defines a continuous function $f$ inside a closed interval $[a,b]$. Most examples I can find online uses finite numbers for $a$ and $b$. However, in problem 3 c) starting at the bottom of page 12 of this problem set, the solution involves using the theorem where the one of the limits is $\infty$ or $-\infty$. Is this valid because $(-\infty,\infty)$ is considered both open and closed per this answer?
 A: Riemann integrals are defined on bounded intervals. Functions can be Riemann integrable on these intervals, a term which I won't define, but is weaker than continuity but stronger than boundedness. That is, a Riemann integrable function on a bounded interval is always bounded (i.e. there exists some $M$ such that $|f(x)| \le M$ for all $x$ in the interval), and every continuous function is always integrable (a consequence of the fundamental theorem of calculus).
We can extend this definition of Riemann integration to unbounded intervals, or intervals for which the function is unbounded (e.g. when asymptotes are involved), using limits. These are known as improper integrals. I won't go into the asymptotes; let's examine the integrals on unbounded intervals.
The improper integral $\int_{-\infty}^a f(t) \, \mathrm{d}t$ is defined to be
$$\int_{-\infty}^a f(t) \, \mathrm{d}t = \lim_{x\to -\infty}\int_x^a f(t) \, \mathrm{d}t.$$
If the definite integrals inside the limit don't exist, or the limit of this function of $x$ doesn't exist, then the improper integral doesn't exist. Similarly,
$$\int_a^\infty f(t) \, \mathrm{d}t = \lim_{x\to \infty}\int_a^x f(t) \, \mathrm{d}t.$$
On the other hand, we define
$$\int_{-\infty}^\infty f(t) \, \mathrm{d}t = \int_{-\infty}^a f(t) \, \mathrm{d}t + \int_a^\infty f(t) \, \mathrm{d}t,$$
where $a$ is some arbitrary element of $\Bbb{R}$ (it doesn't matter which; changing it will not change the result). Note that this requires both the one-sided improper integrals to exist independently of each other; if one or both of them do not exist, then the full integral does not exist.
So, let's examine this with respect to the fundamental theorem of calculus. Note that, for a continuous function on $\Bbb{R}$, it is not possible to substitute in $\infty$ or $-\infty$, as such points simply don't exist in the domain. It is sometimes possible, however, to compute limits to $\infty$ or $-\infty$.
Let's say we have a continuous function $f : \Bbb{R} \to \Bbb{R}$ and a corresponding antiderivative $F : \Bbb{R} \to \Bbb{R}$. Then,
$$\int_{-\infty}^a f(t) \, \mathrm{d}t = \lim_{x \to -\infty} \int_x^a f(t) \, \mathrm{d}t = \lim_{x \to -\infty} [F(a) - F(x)].$$
Since $a$ is constant with respect to $x$, this limit will exist if and only if $\lim_{x \to -\infty} F(x)$ exists, and so
$$\int_{-\infty}^a f(t) \, \mathrm{d}t = F(a) - \lim_{x \to -\infty} F(x),$$
which is as close to the FTC applying to unbounded intervals as you're going to get. Similarly,
$$\int_a^\infty f(t) \, \mathrm{d}t = \lim_{x \to \infty} F(x) - F(a).$$
We finally get,
$$\int_{-\infty}^\infty f(t) \, \mathrm{d}t = \int_{-\infty}^a f(t) \, \mathrm{d}t + \int_a^\infty f(t) \, \mathrm{d}t = \lim_{x \to \infty} F(x) - \lim_{x \to -\infty} F(x),$$
which again exists if and only if both limits exist independently.
So, short answer is, yes, the FTC applies with unbounded intervals, provided you know to take limits rather than "substitute" in infinity.
