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

Statement

The function $L : (0, \infty) \to R$ is continuous and strictly increasing.

Facts & Assumptions

Given: $L$ on $(0, \infty)$ .

[L1]

$L$ is differentiable and $L^{'} (x) = 1 / x$ for $x > 0$ (The integral logarithm satisfies $L^{'} (x) = 1 / x$ for $x > 0$ and $L (1) = 0$ ).

[L2]

Differentiability at a point implies continuity there (A function differentiable at $c$ is continuous at $c$ ).

[L3]

If a function is continuous on $[a, b]$ and differentiable on $(a, b)$ , then $f (b) - f (a) = f^{'} (c) (b - a)$ for some $c \in (a, b)$ (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

Proof

technique · direct

1.1

By [L1] and [L2], $L$ is continuous at every point of $(0, \infty)$ .

L1L2

1.2

Let $0 < a < b$ . Applying [L3] gives a $c \in (a, b)$ such that $L (b) - L (a) = L^{'} (c) (b - a) = \frac{b - a}{c} > 0.$

L1L3algebra

2.1

Hence $L (a) < L (b)$ whenever $0 < a < b$ , so $L$ is strictly increasing; step 1.1 supplies continuity.

step 1.1step 1.2∎

Depends on

Used by

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Sources

OpenStax, Calculus Volume 1, Section 6.7 (standard reference, not scraped)

The integral logarithm is continuous and strictly increasing on (0,∞)(0,∞)