TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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

Statement

For all $x, y > 0$ ,

$L (x y) = L (x) + L (y) .$

Facts & Assumptions

Given: $y > 0$ , with $x$ ranging over $(0, \infty)$ .

[L1]

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

[L2]

If $f$ is differentiable at $g (x)$ and $g$ at $x$ , then $(f \circ g)^{'} (x) = f^{'} (g (x)) g^{'} (x)$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L3]

A continuous function on an order-convex interval whose derivative vanishes throughout the interior is constant (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

[L4]

A differentiable function is continuous (A function differentiable at $c$ is continuous at $c$ ).

[L5]

Sums, differences and scalar multiples of functions differentiable at a point are differentiable there, with the corresponding derivatives (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Proof

technique · direct

1.1

Define $h (x) : = L (x y) - L (x)$ on $(0, \infty)$ . By [L1] and [L2] each term is differentiable, so [L5] makes $h$ differentiable with $h^{'} (x) = \frac{y}{x y} - \frac{1}{x} = 0.$

L1L2L5algebra

2.1

By [L4], $h$ is continuous, so [L3] makes it constant on $(0, \infty)$ .

step 1.1L4L3

3.1

Evaluating at $x = 1$ gives $h (x) = h (1) = L (y) - L (1) = L (y)$ . Therefore $L (x y) - L (x) = L (y)$ , which is the claimed product law.

step 2.1L1algebra∎

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Sources

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

The integral logarithm satisfies L(xy)=L(x)+L(y)L(xy)=L(x)+L(y) for all positive xx and yy