Logarithm as an integral

Intuition [Beginner]

There is a curve in mathematics that does something remarkable: the area under $y=1/x$ from $x=1$ to $x=a$ grows exactly like a logarithm. The bigger $a$ gets, the more area you accumulate, and the rate of accumulation slows down in a precise, predictable way.

This matters because it gives us a second, independent way to define logarithms. Instead of starting with "the inverse of an exponential," we start with a geometric quantity — area — and build the entire theory of logarithms and exponentials from scratch. No prior knowledge of powers or roots required.

Why does this concept exist? Because defining the natural logarithm as an area under a curve lets us prove every property of logarithms and exponentials from a single geometric starting point, without circular reasoning.

Visual [Beginner]

The picture shows the hyperbola $y=1/x$ drawn on a coordinate plane. A shaded region fills the area between the curve and the horizontal axis, starting at $x=1$ and stretching to $x=4$. The area of this shaded region is a specific number: the natural logarithm of $4$.

The key observation is that the area grows slowly. Doubling the right endpoint does not double the area — it adds the same amount each time you multiply by a fixed factor.

Worked example [Beginner]

Compute the area under $y=1/x$ from $x=1$ to $x=3$ numerically by estimating with rectangles.

Step 1. Divide the interval $[1,3]$ into four equal strips of width $0.5$. The left endpoints are $x=1.0$, $1.5$, $2.0$, $2.5$.

Step 2. Evaluate $1/x$ at each left endpoint: $1/1.0=1.0$, $1/1.5=0.667$, $1/2.0=0.5$, $1/2.5=0.4$.

Step 3. Multiply each height by the width $0.5$ and add: $0.5\times 1.0+0.5\times 0.667+0.5\times 0.5+0.5\times 0.4=0.5+0.333+0.25+0.2=1.283$.

What this tells us: the area under $1/x$ from $1$ to $3$ is approximately $1.283$, and the exact value (the natural logarithm of $3$) is about $1.099$. Our left-endpoint estimate overshoots because the curve is decreasing — each rectangle extends above the curve.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Definition (The natural logarithm). Define $L:(0,\infty )\to \mathbb{R}$ by

$$L(x)={\int }_1^x\frac{dt}{t}.$$

For $x>1$, $L(x)$ is the area under $1/t$ from $1$ to $x$. For $0<x<1$, $L(x)$ is the negative of the area from $x$ to $1$. The function $L$ is well-defined because $1/t$ is continuous on $(0,\infty )$, hence integrable on every closed subinterval.

By the fundamental theorem of calculus 02.04.04, $L$ is differentiable on $(0,\infty )$ with

$$L^{\prime }(x)=\frac{1}{x}.$$

Counterexamples to common slips

• $L(x)$ is not the area from $0$ to $x$. The function $1/t$ blows up as $t\to 0^{+}$, so the integral from $0$ to any positive number diverges. The lower limit must be a fixed positive number; the choice $1$ is conventional.

• $L$ is not the area under an arbitrary function. The integrand must be $1/t$ specifically. Choosing $1/t^2$ or $1/\sqrt{t}$ gives a completely different function.

• $L(x)$ can be negative. For $0<x<1$, the integral ${\int }_1^{x}dt/t$ runs "backwards" (the upper limit is smaller than the lower limit), yielding a negative value.

Key theorem with proof [Intermediate+]

Theorem (Functional equations for $L$). The function $L(x)={\int }_1^{x}dt/t$ satisfies:

1. $L(xy)=L(x)+L(y)$ for all $x,y>0$.
2. $L(x^p)=p\cdot L(x)$ for all $x>0$ and all $p\in \mathbb{R}$.
3. $L$ is strictly increasing, with range $(-\infty ,\infty )$.

Consequently, there exists a unique number $e>1$ with $L(e)=1$.

Proof of (1). Fix $x>0$. Define $f(y)=L(xy)={\int }_1^{xy}dt/t$. By the chain rule and the fundamental theorem of calculus,

$$f^{\prime }(y)=\frac{1}{xy}\cdot x=\frac{1}{y}.$$

But $L^{\prime }(y)=1/y$ as well, so $f^{\prime }(y)=L^{\prime }(y)$ for all $y>0$. Therefore $f(y)-L(y)$ has derivative zero everywhere on $(0,\infty )$, hence $f(y)-L(y)=C$ for some constant $C$. Evaluating at $y=1$: $f(1)=L(x\cdot 1)=L(x)$, while $L(1)={\int }_1^{1}dt/t=0$. So $C=L(x)-0=L(x)$. Thus

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

$\square$

Proof of (2). For integer $n\ge 1$, property (2) follows from (1) by induction: $L(x^n)=L(x\cdot x^{n-1})=L(x)+L(x^{n-1})$, and the base case $L(x^1)=1\cdot L(x)$ is immediate. For $n=0$: $L(x^0)=L(1)=0=0\cdot L(x)$. For negative integers $n=-m$ with $m>0$: from (1), $0=L(1)=L(x^m\cdot x^{-m})=L(x^m)+L(x^{-m})$, so $L(x^{-m})=-L(x^m)=-m\cdot L(x)$.

For $p=1/n$ with $n$ a positive integer: set $y=x^{1/n}$, so $y^n=x$. Then $L(x)=L(y^n)=n\cdot L(y)$, giving $L(y)=L(x)/n$, i.e., $L(x^{1/n})=(1/n)L(x)$.

For rational $p=m/n$: $L(x^{m/n})=L((x^{1/n}{)}^m)=m\cdot L(x^{1/n})=m\cdot (1/n)\cdot L(x)=(m/n)L(x)$.

For real $p$: since $L$ is continuous (being differentiable) and the rationals are dense in $\mathbb{R}$, the identity $L(x^p)=p\cdot L(x)$ for all rational $p$ extends by continuity to all real $p$.

$\square$

Proof of (3). Since $L^{\prime }(x)=1/x>0$ for all $x>0$, $L$ is strictly increasing. For $x>1$, $L(x)={\int }_1^{x}dt/t>0$, and as $x\to \infty$, $L(x)\to \infty$ (since $1/t\ge 1/x$ on $[1,x]$ gives $L(x)\ge (x-1)/x$, though more precisely $L(2^n)=n\cdot L(2)$ grows without bound). For $0<x<1$, $L(x)=-{\int }_x^{1}dt/t<0$, and as $x\to 0^{+}$, $L(x)\to -\infty$. By the intermediate value theorem, $L$ is surjective onto $\mathbb{R}$.

Since $L$ is strictly increasing and continuous with range $(-\infty ,\infty )$, it has a continuous strictly increasing inverse $L^{-1}:\mathbb{R}\to (0,\infty )$. Define $e=L^{-1}(1)$, the unique positive number with $L(e)=1$.

$\square$

Bridge. This theorem is the foundational reason that the logarithm defined by area satisfies every algebraic property one expects from a logarithm. The central insight is that the functional equation $L(xy)=L(x)+L(y)$ follows from the derivative $L^{\prime }=1/x$ alone, and this is exactly the bridge that identifies the area-based logarithm with the inverse of the exponential function. The result builds toward the power series for $e^x$ 02.03.03 by providing an independent, non-circular definition of $e$ and $\ln$, and appears again in 02.04.06 when the improper integral ${\int }_1^{\infty }dt/t$ diverges — a consequence of $L(x)\to \infty$.

Exercises [Intermediate+]

Advanced results [Master]

Theorem 1 (The exponential as the inverse of $L$). Define $\exp :\mathbb{R}\to (0,\infty )$ by $\exp =L^{-1}$. Then $\exp$ is the unique function satisfying ${\exp }^{\prime }(x)=\exp (x)$ for all $x\in \mathbb{R}$ with $\exp (0)=1$.

The proof follows from the inverse function theorem: ${\exp }^{\prime }(x)=1/L^{\prime }(\exp (x))=1/(1/\exp (x))=\exp (x)$.

Theorem 2 (The identity $e^x=\exp (x)$). For rational $x=p/q$, $\exp (x)=e^{p/q}$ (the positive $q$-th root of $e^p$). By continuity of $\exp$ and density of $\mathbb{Q}$ in $\mathbb{R}$, $\exp (x)=e^x$ for all real $x$, where $e^x$ is defined by continuity from rational exponents.

Theorem 3 (The power series for $e^x$). The exponential function has the Taylor expansion

$$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

convergent for all $x\in \mathbb{R}$. The proof uses the Taylor remainder theorem: the $n$-th derivative of $\exp$ is $\exp$ itself, so the Lagrange remainder at order $n$ is $|R_n(x)|=|e^c{x}^{n+1}/(n+1)!|$ for some $c$ between $0$ and $x$. Since $e^c\le e^{|x|}$ and $|x{|}^{n+1}/(n+1)!\to 0$ as $n\to \infty$, the remainder vanishes.

Theorem 4 (The series for $\ln (1+x)$). For $|x|<1$,

$$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots =\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}{x}^n}{n}.$$

This is obtained by integrating the geometric series $1/(1+t)=1-t+t^2-t^3+\cdots$ term by term from $0$ to $x$. Convergence at $x=1$ (giving $\ln 2=1-1/2+1/3-\cdots$) follows from the alternating series test.

Theorem 5 (The limit definition of $e$). The number $e$ satisfies

$$e=\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n.$$

To see this, observe that $L\left((1+1/n{)}^n\right)=n\cdot L(1+1/n)=n\cdot {\int }_1^{1+1/n}dt/t$. Since $1/t$ is continuous at $t=1$, for large $n$ we have ${\int }_1^{1+1/n}dt/t\approx (1/n)\cdot (1/1)=1/n$. Thus $L((1+1/n{)}^n)\to 1$, and applying $\exp$ gives $(1+1/n{)}^n\to e$.

Theorem 6 ($e$ is irrational). If $e=p/q$ for positive integers $p,q$, then for any $n>q$, the number $n!(e-s_n)$ (where $s_n={\sum }_{k=0}^{n}1/k!$) is a positive integer (since $q\mid n!$). But $0<n!(e-s_n)=n!{\sum }_{k=n+1}^{\infty }1/k!<n!\cdot 1/(n\cdot n!)=1/n$, which is less than $1$ for $n\ge 2$. A positive integer less than $1$ is impossible.

Theorem 7 (The inequality $(1+1/n{)}^n<e<(1+1/n{)}^{n+1}$). From the proof of Theorem 5, $L((1+1/n{)}^n)=n\cdot L(1+1/n)$. The integral ${\int }_1^{1+1/n}dt/t$ satisfies $1/(1+1/n)\cdot (1/n)<{\int }_1^{1+1/n}dt/t<1\cdot (1/n)$ by bounding $1/t$ between its minimum and maximum on $[1,1+1/n]$. This gives $n/(n+1)<L((1+1/n{)}^n)<1$, so $(1+1/n{)}^n<e$. Similarly, $L((1+1/n{)}^{n+1})=(n+1)L(1+1/n)$, and the upper bound gives $(n+1)\cdot 1/(n+1)=1$, so $L((1+1/n{)}^{n+1})>1$, hence $(1+1/n{)}^{n+1}>e$.

Synthesis. The foundational reason this unit works is that a single geometric definition — the area under $1/x$ — generates every algebraic, analytic, and series property of logarithms and exponentials. The central insight is that $L^{\prime }(x)=1/x$ forces the functional equation $L(xy)=L(x)+L(y)$, and this is exactly the bridge from geometry to algebra. Putting these together with the inverse function theorem, the exponential $\exp =L^{-1}$ satisfies ${\exp }^{\prime }=\exp$, and the pattern generalises to the full Taylor expansion $e^x=\sum x^n/n!$. The bridge is that the integral definition builds toward the power series for $e^x$ 02.03.03 without circular reasoning, appears again in the divergence of ${\int }_1^{\infty }dt/t$ 02.04.06, and identifies $L(x)$ with the precalculus logarithm 00.05.02 via the functional equations.

Full proof set [Master]

Proposition 1 (Derivative of the exponential). The function $\exp =L^{-1}$ satisfies ${\exp }^{\prime }(x)=\exp (x)$ for all $x\in \mathbb{R}$.

Proof. Let $y=\exp (x)$, so $x=L(y)$. By the inverse function theorem,

$${\exp }^{\prime }(x)=\frac{1}{L^{\prime }(\exp (x))}=\frac{1}{1/\exp (x)}=\exp (x).$$

The inverse function theorem applies because $L$ is $C^1$ with $L^{\prime }(y)=1/y\ne 0$ for all $y>0$. $\square$

Proposition 2 (Power series for $e^x$). The Taylor series of $\exp$ at $x=0$ converges to $\exp (x)$ for every $x\in \mathbb{R}$.

Proof. Since ${\exp }^{(n)}(x)=\exp (x)$ for every $n$, the Taylor polynomial of degree $n$ at $0$ is $T_n(x)={\sum }_{k=0}^n{x}^k/k!$. By the Lagrange form of the remainder, for each $x$ there exists $c$ between $0$ and $x$ such that

$$|R_n(x)|=\left|\frac{{\exp }^{(n+1)}(c)}{(n+1)!}{x}^{n+1}\right|=\frac{e^c|x{|}^{n+1}}{(n+1)!}.$$

For $|x|\le M$, $e^c\le e^M$, so $|R_n(x)|\le e^M|x{|}^{n+1}/(n+1)!\to 0$ as $n\to \infty$ (since $a^n/n!\to 0$ for any fixed $a$). Hence $\exp (x)={\lim }_{n\to \infty }{T}_n(x)={\sum }_{k=0}^{\infty }{x}^k/k!$. $\square$

Proposition 3 (Mercator's series for $\ln (1+x)$). For $x\in (-1,1]$,

$$\ln (1+x)=\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}{x}^n}{n}.$$

Proof. For $|t|<1$, the geometric series gives $1/(1+t)={\sum }_{n=0}^{\infty }(-t{)}^n={\sum }_{n=0}^{\infty }(-1{)}^n{t}^n$. Integrating term by term from $0$ to $x$ (justified by uniform convergence on $[-r,r]$ for any $r<1$):

$$\ln (1+x)=L(1+x)-L(1)={\int }_0^x\frac{dt}{1+t}=\sum _{n=0}^{\infty }(-1{)}^n{\int }_0^x{t}^{n}dt=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{n+1}}{n+1}=\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}{x}^n}{n}.$$

At $x=1$, the series becomes $1-1/2+1/3-1/4+\cdots$, which converges by the alternating series test. Abel's theorem for power series guarantees that the sum at $x=1$ equals ${\lim }_{x\to 1^{-}}\ln (1+x)=\ln 2$. $\square$

Connections [Master]

• Fundamental theorems of calculus 02.04.04. The entire development rests on FTC1: differentiating $L(x)={\int }_1^{x}dt/t$ gives $L^{\prime }(x)=1/x$, and the functional equations for $L$ follow from this derivative. The inverse function theorem application that yields ${\exp }^{\prime }=\exp$ is itself a consequence of FTC1 applied through the inverse.

• Improper integrals 02.04.06. The integral ${\int }_1^{\infty }dt/t$ diverges, which is the statement $L(x)\to \infty$ as $x\to \infty$. This divergence is the reason the natural logarithm grows without bound, and the comparison with $1/t^p$ for $p>1$ (which converges) is what makes $1/t$ the boundary case in the $p$-test for improper integrals.

• Infinite series and convergence tests 02.03.03. The power series $e^x=\sum x^n/n!$ and $\ln (1+x)=\sum (-1{)}^{n+1}{x}^n/n$ are foundational examples in the theory of Taylor series. The ratio test applied to $x^n/n!$ gives convergence for all $x$, and the alternating series test justifies convergence of the Mercator series at $x=1$.

Historical & philosophical context [Master]

Saint-Vincent 1647, in his Opus Geometricum ^{[SaintVincent1647]}, discovered that the area under the hyperbola $y=1/x$ satisfies a logarithmic law: equal areas correspond to equal ratios of the bounding abscissae. This was the first recognition that the quadrature of the hyperbola produces a logarithmic function. Mercator 1668, in Logarithmotechnia ^{[Mercator1668]}, derived the power series $\ln (1+x)=x-x^2/2+x^3/3-\cdots$ by integrating the geometric series term by term, providing the first series expansion of the natural logarithm. Newton 1665, in De Analysi ^[Newton1665], had already obtained the general binomial series and the expansion of $\ln (1+x)$ independently, recognising the connection between areas under curves and infinite series.

Euler 1748, in the Introductio in analysin infinitorum ^[Euler1748], systematised the entire theory: he defined $e$ as the number whose hyperbolic logarithm is $1$, proved $e^x=\sum x^n/n!$, and showed that every property of logarithms and exponentials follows from these definitions. Euler's treatment is essentially the modern one, replacing the ad hoc constructions of the seventeenth century with a clean analytical framework.

Bibliography [Master]

@book{SaintVincent1647,
  author = {Saint-Vincent, Gr\'egoire de},
  title = {Opus Geometricum Quadraturae Circuli et Sectionum Coni},
  publisher = {Ioannis et Iacobi Meursiorum},
  year = {1647},
}

@book{Mercator1668,
  author = {Mercator, Nicolaus},
  title = {Logarithmotechnia: Sive Methodus Construendi Logarithmos},
  publisher = {M. Pitt},
  year = {1668},
}

@article{Newton1665,
  author = {Newton, Isaac},
  title = {De Analysi per Aequationes Numero Terminorum Infinitas},
  journal = {Manuscript, circulated 1669, published 1711 by William Jones},
  year = {1665},
}

@book{Euler1748,
  author = {Euler, Leonhard},
  title = {Introductio in Analysin Infinitorum},
  publisher = {Marcus-Michaelis Bousquet},
  year = {1748},
}

@book{Rudin1976,
  author = {Rudin, Walter},
  title = {Principles of Mathematical Analysis},
  publisher = {McGraw-Hill},
  year = {1976},
  edition = {3rd},
}

@book{Apostol1967,
  author = {Apostol, Tom M.},
  title = {Calculus, Volume 1},
  publisher = {John Wiley \& Sons},
  year = {1967},
}