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Let $f$ be continuous on $\mathbb{R}$. Due to FTC I, we know that a function of the form∗ $F(x) = \int_a^xf(t)\operatorname dt$ is always an antiderivative of $f(x)$. In this question you will investigate whether all antiderivatives of $f(x)$ can be expressed in this form∗. For simplicity, let us further assume $f$ is non-negative $(i.e. ∀x ∈ \mathbb{R}, f(x) ≥ 0)$.

(a) Suppose$\lim\limits _{A\rightarrow\infty}\int_0^Af(t)\operatorname dt$ or $\lim\limits _{A\rightarrow-\infty}\int_A^0f(t)\operatorname dt$ is finite, show there is an antiderivative $G(x)$ of $f(x)$ which does not equal $\int_a^xf(t)\operatorname dt$ for any a $\in \mathbb{R}$

(b)Suppose $\lim\limits _{A\rightarrow\infty}\int_0^Af(t)\operatorname dt=\infty$ and $\lim\limits _{A\rightarrow-\infty}\int_A^0f(t)\operatorname dt=\infty$, show for any antiderivative $G(x)$ of $f(x), ∃a ∈ \mathbb{R} \text{ s.t.} G(x) = \int_a^xf(t)\operatorname dt$

Hint: Think about whether antiderivatives of f(x) need to have zeroes.


What I have tried so far:

Look thorugh (a) and (b), it's saying if $f$ is continuous on $\mathbb{R}$ we have:

$(\lim\limits _{A\rightarrow\infty}\int_0^Af(t)\operatorname dt=\pm\infty \wedge \lim\limits _{A\rightarrow-\infty}\int_A^0f(t)\operatorname dt=\pm\infty )\leftrightarrow \forall G(x), ∃a ∈ R \text{ s.t.} G(x) = \int_a^xf(t)\operatorname dt$

(This is a stronger version of the question, since negation of finite also include $-\infty$, I'm not sure if this is still true, but this should implies what the question is asking to prove)

By assumption, $f$ is non-negative, then we don't need to consider the $-\infty$ cases, just show the following would be sufficient:

$(\lim\limits _{A\rightarrow\infty}\int_0^Af(t)\operatorname dt=\infty \wedge \lim\limits _{A\rightarrow-\infty}\int_A^0f(t)\operatorname dt=\infty )\leftrightarrow \forall G(x), ∃a ∈ R \text{ s.t.} G(x) = \int_a^xf(t)\operatorname dt$

I don't have the Intuition of why this is true, at least it's not very trivial to me..

So, first I tried to break it into definitions:

1.$\lim\limits _{A\rightarrow\infty}\int_0^Af(t)\operatorname dt=\pm\infty$

$\Leftrightarrow \forall N\in \mathbb{R},\exists M\in \mathbb{R} s.t. A>M\rightarrow(\int_0^Af(t)\operatorname dt>N\vee \int_0^Af(t)\operatorname dt<N)$

2.$\lim\limits _{A\rightarrow-\infty}\int_A^0f(t)\operatorname dt=\pm\infty$

$\Leftrightarrow \forall N\in \mathbb{R},\exists M\in \mathbb{R} s.t. A<M\rightarrow(\int_0^Af(t)\operatorname dt>N\vee \int_0^Af(t)\operatorname dt<N)$

3.$\forall G(x), ∃a ∈ R \text{ s.t.}G(x) = \int_a^xf(t)\operatorname dt$ (not sure about this one)

$\Leftrightarrow\forall G(x), ∃a ∈ R \text{ s.t.}\forall n \in \mathbb{R}, \forall \varepsilon>0, \exists \delta>0\text{ s.t. } \exists P\in \mathbb{P}$ s.t.

$( \text{$P$ is a partition of [a,n]} \wedge l(P)<\delta)\rightarrow|S(f(t),P)-G(n)|<\varepsilon$

But those doesn't looks like very useful...where should I start?

Any help or hint or suggestion would be appreciated.

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  • $\begingroup$ Hint: Consider when $a = \pm\infty$. $\endgroup$ – Jakobian Jul 13 '19 at 21:29
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Well, at first, let $f:\mathbb{R}\to\mathbb{R}$ be some comntinuous function. Then, as you stated, every function of the form: $$G(x)=\int_a^xf(t)dt,$$ is an antiderivative of $f$. Let $F(x)$ be some antiderivative of $f$. Then, we have $F'(x)=f(x)$ for every $x\in\mathbb{R}$. Thus, there exists some constant $c_a\in\mathbb{R}$ such that: $$F(x)=\int_a^xf(t)dt+c_a.$$ Inversely, it is evident that a function of the form $\int_a^xf(t)dt+c$ is an anti-derivative of $f$. So, we have proved the following:

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. Then the set: $$\mathcal{A}:=\left\{\int_a^xf(t)dt+c:a,c\in\mathbb{R}\right\}$$ contains exactly all the anti-derivatives of $f$.

In general, since $f(x)\geq0$, we have that: $$\int_0^xf(t)dt\geq0,\ \forall\ x>0,$$ and,similarly: $$\int_x^0f(t)dt\geq0,\ \forall\ x<0.$$

Also, since $f(x)\geq0$we get that any anti-derivative of $f$ is increasing.

For the first question, let, W.L.o.G. $$\lim_{x\to+\infty}\int_0^xf(t)dt=L<+\infty.$$ Also, let $$F(x):=\int_0^xf(t)dt.$$ Since $F$ is an anti-derivative of $f$, $F$ is increasing and: $$F(x)\leq L.$$

From the above, we also have that $L\geq0$. Then, the function: $$G(x)=\int_0^xf(t)dt-L-1$$ is an anti-derivative of $f$ with $G(x)\leq-1<0$ for each $x>0$. So, $G$ has no roots, thus, cannot be of the form: $$\int_a^xf(t)dt,$$ since any such function has at least one root ($a$ is always a root).

For the second question, let $G$ be an antiderivative of $f$. Then, $G$ can be written in the form: $$G(x)=\int_a^xf(t)dt+c.$$ We can now do the following trick: $$G(x)=\int_a^xf(t)dt+c=\int_0^xf(t)dt+\underbrace{\int_a^0f(t)dt+c}_{C}=\int_0^xf(t)dt+C.$$

Now, the two given assumptions imply that: $$\lim_{x\to+\infty}G(x)=+\infty\text{ and }\lim_{x\to-\infty}G(x)=-\infty,$$ and, since $G$ is continuous, we get that $G(\mathbb{R})=\mathbb{R}.$ Particularly, this implies that there exists some $x_0\in\mathbb{R}$ such that $G(x_0)=0$, or, equivalently: $$\int_0^{x_0}f(t)dt+C=0\Leftrightarrow C=\int_{x_0}^0f(t)dt.$$ Thus, we have: $$G(x)=\int_0^xf(t)dt+\int_{x_0}^0f(t)dt=\int_{x_0}^xf(t)dt,$$ which was our goal.

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