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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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∎

Depends on

Used by

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