Properties of the Integral and the Working FTC: Examples and Counterexamples

Let $m \in \mathbb{N}$ and let $p_m : [0,1] \to \mathbb{R}$ be $p_m(x) = x^{m}$ (Integer powers $a^m$). Then $p_m$ is integrable on $[0,1]$ and

$\int_0^1 x^{m}\,\mathrm{d}x \;=\; \frac{1}{\iota(m+1)} ,$

where $\iota(m+1)$ is the canonical natural of $m+1$ in $\mathbb{R}$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field) and is positive because $m+1 \ge 1$ (Canonical naturals are positive and strictly increasing).

The $\iota$ is not decoration. A natural number is a von Neumann natural, that is a set, so $m+1$ is not an element of $\mathbb{R}$ and $1/(m+1)$ is not an expression of the field; what the display says is $1/\iota(m+1)$, and that is why the reader meets $\iota$ here at all.

Two independent checks are carried out below: the value at $m = 0$ against the published formula for the integral of a constant (If $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ for every partition $P$; in particular every constant function is integrable, with $\int_a^b c = c(b-a)$), and the monotonicity of the answer in $m$ against the pointwise inequality $x^{m+1} \le x^{m}$ on $[0,1]$.

• The primitive is the only input, and the index range inside it is the trap. For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$; for $n = 0$ it is the constant $1$, with derivative $0$; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term states its power rule for $n \ge 1$ only, because at $n = 0$ the formula $\iota(0)c^{-1}$ mentions $c^{-1}$, which is undefined at $c = 0$. The exponent used above is $m+1$, which is $\ge 1$ for every $m \in \mathbb{N}$ including $m = 0$, so claim 2 of that lemma applies with no case split and the example is correct at $m = 0$ as well.

• The same computation from the definition is a different exercise. The previous page's companion evaluates $\int_0^1 x^{2}$ from Darboux sums on uniform partitions and a closed form for $\sum_{k<n}\iota(k)^{2}$; that is the value $1/3$, which is $1/\iota(3)$ and agrees with the formula above at $m = 2$. The point of computing by the fundamental theorem is that no closed form for a power sum is needed at any $m$.

• Linearity extends this to every polynomial. For $P(x) = \sum_{j<n}a_jx^{j}$ one gets $\int_0^1 P = \sum_{j<n}a_j/\iota(j+1)$ from Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and $\int_a^b(\lambda f+\mu g) = \lambda\int_a^b f + \mu\int_a^b g$; that is a routine consequence and is not stated as a separate claim.

Let $f : [0,2] \to \mathbb{R}$ be

$f(x) \;:=\; \begin{cases} 1 & 0 \le x < 1, \\ 3 & 1 \le x \le 2. \end{cases}$

Then $f$ is integrable on $[0,2]$ and

$\int_0^2 f \;=\; 1 + 3 \;=\; 4 .$

Two routes are carried out below. The first splits at the jump with For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then $\int_a^b f = \int_a^c f + \int_c^b f$; with the oriented form for arbitrary $a,b,c$ and evaluates each piece with Changing an integrable function at finitely many points changes neither its integrability nor its integral and the constant case of If $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ for every partition $P$; in particular every constant function is integrable, with $\int_a^b c = c(b-a)$. The second computes lower and upper Darboux sums directly on a three-point partition and squeezes.

Why this is the smallest interesting example. The jump sits at the interior point $1$, which is a partition point of both routes, and a partition point lies in two subintervals (Partition of $[a,b]$ as a finite strictly increasing list $a = t_0 < t_1 < \dots < t_n = b$, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions). That is exactly the case singled out in the proof of Changing an integrable function at finitely many points changes neither its integrability nor its integral, and it is why the value of $f$ at the single point $1$ is irrelevant to both the integrability and the value while the upper sums still see it.

• The two sums of $P_{\eta}$ are not symmetric, and that is a statement about these partitions. For each $\eta \in (0,1)$ the lower sum $L(f,P_{\eta})$ is exactly $4$ while the upper sum is $4 + 2\eta > 4$, so this family already locates the integral exactly from below while its upper sums only approach it. Nothing is claimed here about an arbitrary partition: a partition whose points do not include $1$ has a subinterval on which $f$ takes both values, and its lower sum is then strictly below $4$.

• The value at the jump is irrelevant to the integral, but not to the sums. Replacing $f(1) = 3$ by any real leaves $f$ integrable with the same integral, by Changing an integrable function at finitely many points changes neither its integrability nor its integral. What it changes is which partitions realise the extremes: with $f(1) := 1$ the subinterval $[1-\eta,1]$ carries $m_1 = M_1 = 1$, while $[1,2]$ now carries $m_2 = 1$ and $M_2 = 3$, so $L(f,P_{\eta}) = 2$ and $U(f,P_{\eta}) = 4$ and the asymmetry moves to the other side. The integral is $4$ either way.

• Additivity is what makes the first route short. Without For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then $\int_a^b f = \int_a^c f + \int_c^b f$; with the oriented form for arbitrary $a,b,c$ the only route is the second one, which needs the Darboux sums of a function that is not constant on any partition containing the jump in the interior of a subinterval.

Let $p \in \mathbb{Q}$ with $p > 0$ (Order on the rationals) and define

$f_p : [0,\infty) \to \mathbb{R}, \qquad f_p(t) \;:=\; (t+1)^{-p} ,$

the rational power of the positive base $t+1 \ge 1$ (Rational powers $a^r$ of a positive base). Then $f_p$ is nonnegative and nonincreasing, so The integral test: for $f \ge 0$ nonincreasing on $[0,\infty)$, $\sum_k f(k)$ converges if and only if the sequence $\bigl(\int_0^N f\bigr)_N$ is bounded, with $\int_0^N f \le \sum_{k<N} f(k) \le f(0) + \int_0^N f$ applies, and its terms are

$f_p(k) \;=\; \frac{1}{\iota(k+1)^{p}} \qquad (k \in \mathbb{N}) .$

The series $\sum_k f_p(k)$ is exactly the $p$-series $\sum_{k \ge 1} 1/\iota(k)^{p}$ in the sense of Series, partial sums, convergence and the sum, divergence, and the tail series, which converges if and only if $p > 1$ (For rational $p > 0$, $\sum 1/k^p$ converges iff $p > 1$). The integral test therefore delivers, with no primitive computed anywhere:

$\Bigl(\textstyle\int_0^N f_p\Bigr)_{N \in \mathbb{N}} \ \text{ is bounded above} \qquad \Longleftrightarrow \qquad p > 1 .$

The cross-check. At $p = 2$ the integral can also be computed directly: the primitive $G(t) = -(t+1)^{-1}$ gives

$\int_0^N (t+1)^{-2}\,\mathrm{d}t \;=\; 1 - \frac{1}{\iota(N+1)} \;<\; 1 ,$

so the sequence is bounded by $1$, in agreement with the verdict above at $p = 2 > 1$. At $p = 1$ the verdict is that $\bigl(\int_0^N (t+1)^{-1}\bigr)_N$ is unbounded, since the harmonic series diverges. No named logarithmic primitive is available from the current dependency vocabulary, and none is needed for this conclusion.

The exponent must be rational. Real exponents do not exist in this library at this point in the reading order (Why real exponents are deferred on the rational-powers page), so "for $p \in [1,\infty)$" is not a statement that can be made here.

• The test is run backwards here, and that is the point. The usual textbook order computes $\int_1^N t^{-p}\,\mathrm{d}t$ with a primitive and reads off the convergence of the series. That route is unavailable at $p = 1$ in this library, because the primitive is the logarithm and the logarithm is built on a later page. Running the equivalence in the other direction costs nothing: the published For rational $p > 0$, $\sum 1/k^p$ converges iff $p > 1$ settles the series for every rational $p>0$, and The integral test: for $f \ge 0$ nonincreasing on $[0,\infty)$, $\sum_k f(k)$ converges if and only if the sequence $\bigl(\int_0^N f\bigr)_N$ is bounded, with $\int_0^N f \le \sum_{k<N} f(k) \le f(0) + \int_0^N f$ transfers the verdict to the integrals.

• The index shift is real and is checked in step 2.2. The $p$-series starts at $k = 1$ because $0^{p} = 0$ has no reciprocal, while a sequence in this library is a function on $\mathbb{N}$, which contains $0$; the integrand $(t+1)^{-p}$ is shifted by exactly one for that reason, and $f_p(k) = 1/\iota(k+1)^{p}$. Substituting $t^{-p}$ for $(t+1)^{-p}$ would put an undefined value at $t = 0$ and make $\int_0^N$ improper.

• An independent elementary route to the series verdict. For a nonincreasing nonnegative sequence, $\sum a_k$ converges iff $\sum 2^k a_{2^k}$ converges applies to the nonnegative nonincreasing family $(1/\iota(k)^{p})_{k\ge1}$ and turns it into $\sum_{j\ge0}\iota(2)^{j}/\iota(2^{j})^{p}$, a geometric series; that is the standard elementary route to the same verdict, and it is noted here for orientation only. No claim is made about how For rational $p > 0$, $\sum 1/k^p$ converges iff $p > 1$ is itself proved, and nothing above depends on this remark.

Write $\sqrt{x} := x^{1/2}$ for the unique nonnegative square root of $x \ge 0$ (Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$, Rational powers $a^r$ of a positive base) and put

$H : [0,1] \to \mathbb{R}, \qquad H(x) \;:=\; 2\sqrt{x} .$