Must a Function be Bounded for the Antiderivative to Exist over a Given Interval?

Question

In my Calculus class, we were given the following definition of antiderivative:

Let $f$ be a function defined on an interval.

An antiderivative of $f$ is any function $F$ such that $F' = f$.

The collection of all antiderivatives of $f$ is denoted $\displaystyle \int f(x) dx$.

My question is, don't we have to say that $f$ should be bounded in the definition?

If not, then $f$ is not integrable by definition, so we can't say anything about $\displaystyle \int f(x) dx$, right?

I'm not sure if I'm thinking about this the right way.

Answer 1

The existence of an antiderivative and being integrable are distinct (although related) concepts.

Take$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\mathbb R\\&&x\mapsto&\begin{cases}x^2\sin\left(\frac1{x^2}\right)&\text{ if }x>0\\0&\text{ if }x=0.\end{cases}\end{array}$$Then $f$ is differentiable, but $f'$ is unbounded. But, in particular, $f$ is an antiderivative of $f'$.

Answer 2

No, the function does not have to be bounded to have an integral.

Consider $$ \int _0^1 \frac {dx}{\sqrt x}$$ which is an improper integral because the integrand is not bounded on $(0,1)$.

However the anti derivative is $2\sqrt x$ which results in a bounded value for

$$ \int _0^1 \frac {dx}{\sqrt x} =2 $$

Answer 3

If $f$ is continuous on the interval, no additional condition is needed for it to have an antiderivative. Pick any point $c$ in the interval and $\int_c^x f(x')dx'$ will be an antiderivative, since $f(x)$ is bounded on $[c,x]$.

If $f$ is not continuous, then things get trickier. For the normal sort of functions you encounter in calculus class, it is sufficient for the function to be bounded on every finite closed subinterval of the domain.

Answer 4

Continuity over the given interval is all that is required for the integral (term synonymous with antiderivative) to exist over that interval.

When our interval is $[a, b]$ we may invoke the Fundamental Theorem of Calculus which provides the guarantee.

On the other hand, if we have say, continuity over an unbounded interval, say $[a, \infty)$, we may restructure the (improper) integral so that we take the limit as the variable, say $b$, approaches infinity of the comparable definite integral over $[a,b]$. This permits us the use of the FTC to find an antiderivative before taking the limit.