DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-13

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The integral logarithm $L (x) : = \int_{1}^{x} \frac{d t}{t}$ for $x > 0$

Definition

For $x > 0$ , define the integral logarithm

$L (x) : = \int_{1}^{x} \frac{d t}{t},$

using the oriented integral when $x < 1$ (The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

This is well defined. The function $t \mapsto 1 / t$ is continuous wherever $t \neq 0$ by the quotient clause of Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function. For $x \neq 1$ it is therefore continuous on the nondegenerate compact interval with endpoints $1$ and $x$ , hence Riemann integrable there by A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, whose hypothesis is $a < b$ . At $x = 1$ the interval is degenerate and that theorem does not apply; there The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ stipulates $\int_{1}^{1} f = 0$ , so $L (1) = 0$ directly.

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Sources

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

The integral logarithm L(x):=∫1xdttL(x):=∫1x​tdt​ for x>0x>0