# A Proof Related to the Fundamental Theorem of Calculus

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

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. AN\vee \int_0^Af(t)\operatorname dt

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.

• Hint: Consider when $a = \pm\infty$. Jul 13 '19 at 21:29

Well, at first, let $$f:\mathbb{R}\to\mathbb{R}$$ be some continuous 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.