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

Statement

For every $x > 0$ ,

$L^{'} (x) = \frac{1}{x},$

and $L (1) = 0$ .

Facts & Assumptions

Given: $x > 0$ and $L$ as defined.

[F1]

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

[L1]

If an integrand is Riemann integrable on a compact interval and continuous at $c$ , then its integral function with fixed lower endpoint has derivative equal to the integrand at $c$ (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive).

[L2]

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 $a, b, c$ ).

[F2]

An oriented integral reverses sign when its endpoints are reversed and is $0$ when the endpoints agree (The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

Proof

technique · direct

1.1

Choose $a, b$ with $0 < a < x < b$ . For every $z \in (a, b)$ , additivity gives $L (z) = \int_{1}^{a} \frac{d t}{t} + \int_{a}^{z} \frac{d t}{t} .$

F1L2

1.2

By [F1] and the equal-endpoint convention [F2], $L (1) = \int_{1}^{1} d t / t = 0$ .

F1F2

2.1

The first term in step 1.1 is constant in $z$ . Since $1 / t$ is continuous at $x$ , [L1] gives $L^{'} (x) = 1 / x$ .

step 1.1L1algebra

3.1

Since $x > 0$ was arbitrary, steps 2.1 and 1.2 prove both claims.

step 2.1step 1.2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 74 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

The integral logarithm satisfies L′(x)=1/xL′(x)=1/x for x>0x>0 and L(1)=0L(1)=0