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The Integral Logarithm and the Equivalence of Its Characterisations: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

$\log 2$ from four characterisations: hyperbolic area, the alternating harmonic series, Landau iterates and $\exp^{- 1} (2)$

Example

The number $\log 2$ has the four equal descriptions

$\int_{1}^{2} \frac{d t}{t} = \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n} = \lim_{n \to \infty} 2^{n} (2^{1 / 2^{n}} - 1) = \exp^{- 1} (2) .$

Facts & Assumptions

Given: The positive input $2$ .

[L1]

The inverse-exponential, integral, continued Mercator, Landau-limit, and normalised functional-equation characterisations all define $\log$ (Five characterisations of the natural logarithm are equivalent: inverse exponential, integral, continued Mercator series, Landau root limit and the normalised functional equation).

[L2]

At $u = 1$ , the Mercator theorem gives $\log 2 = \sum_{n = 1}^{\infty} (- 1)^{n + 1} / n$ ([The power series for log(1+x) on (-1,1], including the Abel endpoint](/item/thm-log-one-plus-x-power-series)).

[L3]

For $x > 0$ , $\log x = \lim_{n \to \infty} 2^{n} (x^{1 / 2^{n}} - 1)$ (Landau's root limit: log x is the limit of 2^n times (x^(1/2^n) minus 1)).

[F1]

The natural logarithm is the inverse of exponential (The natural logarithm as the inverse of the exponential function).

Verification

technique · direct

1.1

Apply the integral characterisation in [L1] at $x = 2$ to obtain $\log 2 = \int_{1}^{2} d t / t$ .

1.2

Apply [L2] to obtain the alternating harmonic series value.

1.3

Apply [L3] at $x = 2$ to obtain the Landau limit.

1.4

By [F1], $\log 2$ is the unique real $y$ with $\exp (y) = 2$ , namely $\exp^{- 1} (2)$ .

2.1

Steps 1.1 through 1.4 identify all four displayed quantities with the same value, as asserted by the equivalence theorem [L1].

step 1.1step 1.2step 1.3step 1.4L1∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

Dropping $f (e) = 1$ leaves the whole family $c \log$ , including logarithms to other bases and the zero function

Example

Without the normalisation $f (e) = 1$ , the continuous product-to-sum functions are exactly

$f_{c} (x) = c \log x (c \in R) .$

They include the zero function and, for suitable nonzero $c$ , logarithms to other bases.

Facts & Assumptions

Given: $c \in R$ .

[L1]

Every continuous product-to-sum function is uniquely $c \log x$ , including $c = 0$ (Every continuous $f$ with $f (x y) = f (x) + f (y)$ is $f (x) = c \log x$ for a unique $c$ , including $c = 0$ ).

[L2]

The natural logarithm is continuous and satisfies $\log (x y) = \log x + \log y$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L3]

The natural logarithm satisfies $\log (e) = 1$ ( $\log$ is the unique continuous $f : (0, \infty) \to R$ with $f (x y) = f (x) + f (y)$ and $f (e) = 1$ ).

[F1]

For $b > 0$ , $b \neq 1$ , $\log_{b} x = \log x / \log b$ (The logarithm to a positive base other than one).

Verification

technique · direct

1.1

By [L2], $f_{c} (x y) = c \log (x y) = c \log x + c \log y = f_{c} (x) + f_{c} (y)$ , and $f_{c}$ is continuous.

L2algebra

1.2

Also $f_{c} (e) = c \log (e) = c$ , so only $c = 1$ meets the normalisation $f (e) = 1$ .

L3algebra

1.3

If $b > 0$ , $b \neq 1$ , then [F1] identifies $\log_{b}$ with the member $c = 1 / \log b$ . The case $c = 0$ is the zero function and cannot equal $\log_{b}$ , because $\log_{b} b = 1$ .

F1algebra

2.1

The classification theorem [L1] shows that steps 1.1 through 1.3 exhaust all continuous product-to-sum functions, not merely a subfamily.

step 1.1step 1.2step 1.3L1∎

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Assuming choice, a Hamel-basis additive map transported through $\exp$ gives a discontinuous logarithmic function that is not $c \log$

Statement refuted

Continuity cannot be omitted from the multiplicative-to-additive characterisation. Assuming the Axiom of Choice, there is a function $f : (0, \infty) \to R$ satisfying $f (x y) = f (x) + f (y)$ that is discontinuous and is not $c \log$ for any scalar $c$ .

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice).

[L1]

Under choice, $R$ has a Hamel basis over $Q$ ; a chosen basis element has an additive coefficient map $g : R \to R$ , and there is a nonzero complementary vector on which that coefficient map vanishes (Assuming the Axiom of Choice, $R$ has a Hamel basis over $Q$ : there is $B \subseteq R$ such that every real is a finite $Q$ -linear combination of elements of $B$ in exactly one way, and each basis vector carries a well-defined $Q$ -linear coefficient map).

[L2]

An additive real function that is continuous at one point is scalar multiplication (Six regularity conditions each force an additive $f : R \to R$ to be $x \mapsto f (1) x$ : continuity at a single point, monotonicity on a nondegenerate interval, boundedness above on one, boundedness below on one, constancy of sign on one, and a graph that is not dense in $R^{2}$ ).

[L3]

$\exp (s + t) = \exp (s) \exp (t)$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ ).

[L4]

Exponential is continuous and strictly increasing (The exponential function is strictly increasing) and is a bijection from $R$ onto $(0, \infty)$ (The exponential is a continuous bijection from $R$ onto $(0, \infty)$ ).

[L5]

A composite of continuous functions is continuous (A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs).

[F1]

$\log$ is the inverse of $\exp$ (The natural logarithm as the inverse of the exponential function).

Counterexample

technique · direct

1.1

Choose a Hamel basis element $b_{*}$ , its coefficient map $g$ , and a nonzero vector $w$ in the complementary span. Then $g (b_{*}) = 1$ and $g (w) = 0$ .

L1given

1.2

For $x > 0$ , let $t$ be the unique real with $x = \exp (t)$ , and define $f (x) : = g (t)$ . This is well defined by bijectivity in [L4].

L4construct

2.1

The map $g$ is not scalar multiplication. If $g (t) = c t$ , then $0 = g (w) = c w$ and $w \neq 0$ force $c = 0$ , contradicting $g (b_{*}) = 1$ .

step 1.1algebra

2.2

If $x = \exp (s)$ and $y = \exp (t)$ , then [L3] gives $x y = \exp (s + t)$ , so $f (x y) = g (s + t) = g (s) + g (t) = f (x) + f (y)$ .

step 1.2L3L1

3.1

If $f = c \log$ , then composing with exponential and using [F1] gives $g (t) = f (\exp t) = c t$ , contradicting step 2.1.

step 1.2step 2.1F1

3.2

If $f$ were continuous, then $g = f \circ \exp$ would be continuous by [L4] and [L5]. The regularity theorem [L2] would make $g$ scalar multiplication, again contradicting step 2.1.

step 1.2step 2.1L4L5L2

4.1

Thus the constructed $f$ satisfies the functional equation but is discontinuous and is not a scalar multiple of $\log$ .

step 2.2step 3.1step 3.2∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

For every $x > 2$ the uncontinued Mercator series at $u = x - 1$ diverges

Statement refuted

The Mercator expression $\sum_{n = 1}^{\infty} (- 1)^{n + 1} (x - 1)^{n} / n$ does not define the natural logarithm for every $x > 0$ . For every $x > 2$ , this uncontinued series diverges.

Facts & Assumptions

Given: $x > 2$ and $u : = x - 1 > 1$ .

[L1]

The ratio test says that a series diverges if the lower limit of the absolute ratios of successive nonzero terms is greater than $1$ (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence).

[L2]

If a series converges, then its terms tend to $0$ (If a series converges then its terms tend to $0$ ).

[L3]

The Mercator series gives local data on $- 1 < u \leq 1$ ; a separate product law continues those data uniquely to all positive inputs (The Mercator series, its value at $1$ and the product law determine $\log$ on all positive reals, while the series alone is only local).

Counterexample

technique · direct

1.1

Let $a_{n} = (- 1)^{n + 1} u^{n} / n$ for $n \geq 1$ . Every $a_{n}$ is nonzero, and $∣ \frac{a_{n + 1}}{a_{n}} ∣ = u \frac{n}{n + 1} ⟶ u > 1.$

givenalgebra

2.1

The ratio test [L1] therefore makes $\sum_{n} a_{n}$ divergent.

step 1.1L1

2.2

More explicitly, choose $q$ with $1 < q < u$ . The ratios in step 1.1 are at least $q$ for all sufficiently large $n$ , so $∣ a_{n} ∣$ then grows by a factor at least $q$ and cannot tend to $0$ ; [L2] again rules out convergence.

step 1.1L2choosealgebra

3.1

This does not conflict with [L3]: for $x > 2$ , the direct substitution $u = x - 1$ lies outside the local interval, and the value at $x$ is obtained by the product-law continuation instead.

step 2.1L3∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

Integral bounds alone give $2 < e < 4$ ; the sharper published bound is $2 < e < 3$

Example

Elementary integral bounds give $2 < e < 4$ . The sharper published estimate is $2 < e < 3$ .

Facts & Assumptions

Given: The integral function $L$ and the number $e$ .

[L1]

$e$ is the unique positive number with $L (e) = \int_{1}^{e} d t / t = 1$ (The number $e$ is the unique $x > 0$ satisfying $\int_{1}^{x} \frac{d t}{t} = 1$ ).

[L2]

$L (x y) = L (x) + L (y)$ (The integral logarithm satisfies $L (x y) = L (x) + L (y)$ for all positive $x$ and $y$ ).

[L3]

$L$ is strictly increasing (The integral logarithm is continuous and strictly increasing on $(0, \infty)$ ).

[L4]

If $m \leq f \leq M$ on $[a, b]$ , then $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L5]

Oriented integrals satisfy $\int_{u}^{v} f + \int_{v}^{w} f = \int_{u}^{w} f$ (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 $a, b, c$ ).

[L6]

The published sharper bound is $2 < e < 3$ (The elementary numerical bound $2 < e < 3$ ).

Verification

technique · direct

1.1

On $[1, 3 / 2]$ , one has $2 / 3 \leq 1 / t \leq 1$ , so [L4] gives $\frac{1}{3} \leq \int_{1}^{3 / 2} \frac{d t}{t} \leq \frac{1}{2} .$

L4algebra

1.2

On $[3 / 2, 2]$ , one has $1 / 2 \leq 1 / t \leq 2 / 3$ , so [L4] gives $\frac{1}{4} \leq \int_{3 / 2}^{2} \frac{d t}{t} \leq \frac{1}{3} .$

L4algebra

1.3

The stronger estimate $2 < e < 3$ is the published result [L6]; it is cited here, not reproved.

2.1

By additivity [L5], steps 1.1 and 1.2 yield $7 / 12 \leq L (2) \leq 5 / 6$ . Thus $L (2) < 1$ and $2 L (2) \geq 7 / 6 > 1$ .

step 1.1step 1.2L5algebra

3.1

The product law gives $L (4) = L (2 \cdot 2) = 2 L (2) > 1$ .

step 2.1L2

4.1

Since $L (e) = 1$ by [L1] and $L$ is strictly increasing by [L3], $L (2) < L (e) < L (4)$ implies $2 < e < 4$ .

step 2.1step 3.1L1L3

5.1

Steps 4.1 and 1.3 establish both stated brackets.

step 4.1step 1.3∎

Sources

Standard references

Recommended treatments; not extraction sources.

The Integral Logarithm and the Equivalence of Its Characterisations: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

log⁡2log2 from four characterisations: hyperbolic area, the alternating harmonic series, Landau iterates and exp⁡−1(2)exp−1(2)

Example

Facts & Assumptions

Verification

Dropping f(e)=1f(e)=1 leaves the whole family clog⁡clog, including logarithms to other bases and the zero function

Example

Facts & Assumptions

Verification

Assuming choice, a Hamel-basis additive map transported through exp⁡exp gives a discontinuous logarithmic function that is not clog⁡clog

Statement refuted

Facts & Assumptions

Counterexample

For every x>2x>2 the uncontinued Mercator series at u=x−1u=x−1 diverges

Statement refuted

Facts & Assumptions

Counterexample

Integral bounds alone give 2<e<42<e<4; the sharper published bound is 2<e<32<e<3

Example

Facts & Assumptions

Verification

Sources

Standard references

$\log 2$ from four characterisations: hyperbolic area, the alternating harmonic series, Landau iterates and $\exp^{- 1} (2)$

Dropping $f (e) = 1$ leaves the whole family $c \log$ , including logarithms to other bases and the zero function

Assuming choice, a Hamel-basis additive map transported through $\exp$ gives a discontinuous logarithmic function that is not $c \log$

For every $x > 2$ the uncontinued Mercator series at $u = x - 1$ diverges

Integral bounds alone give $2 < e < 4$ ; the sharper published bound is $2 < e < 3$