# The Unapologetic Mathematician

## The Natural Logarithm

Before this little break, we defined a function on the interval of integration. We proved some properties about the functions we get like this, lining them up against the Fundamental Theorem of Calculus. In particular, integrating like this can construct antiderivatives.

Now let’s consider some of the most basic functions of one variable — monomials — and their derivatives. We know that the derivative of $x^n$ is $nx^{n-1}$ whenever $n$ is an integer. Let’s try running this backwards by using Riemann integration.

First for $n\geq0$ we know that $x^n$ is defined everywhere, so we can consider the function defined for any real $x$ by

$\displaystyle f(x)=\int\limits_0^xt^ndt$

Whatever function this is will have $x^n$ as its derivative. We can see that $\frac{x^{n+1}}{n+1}$ has this derivative, and we know that any two antiderivatives differ by a constant. That is, $f(x)=\frac{x^{n+1}}{n+1}+C$ for some real constant $C$. But we can also tell that $f( 0 )=0$ because in that case we’re integrating over a degenerate interval of zero width. This tells us that $0=f(0)=\frac{0^{n+1}}{n+1}+C=C$, and we’ve determined our constant.

How about for $n\leq-2$? Now our integrand $x^n$ has an asymptote at $x=0$ so we can’t integrate across it. Let’s start at $1$ and define a function for all positive real $x$ by

$\displaystyle f(x)=\int\limits_1^xt^ndt$

Again we know that the derivative $f'(x)$ will be $x^n$, and that $\frac{x^{n+1}}{n+1}$ is such an antiderivative. We also know that $f(1)=0$, which tells us that $0=f(1)=\frac{1^{n+1}}{n+1}+C=\frac{1}{n+1}+C$ so our constant of integration is $-\frac{1}{n+1}$. That is, we’ve defined the function $f(x)=\frac{x^{n+1}-1}{n+1}$ on the interval $\left(x,\infty\right)$.

Now what happens when we take this exact same procedure and apply it to the function $\frac{1}{x}$? There is no monomial whose derivative is a scalar multiple of $\frac{1}{x}$, so the above procedure breaks down. Still, there’s some function out there. Indeed, consider the integral

$\displaystyle F(x)=\int\limits_1^x\frac{1}{t}dt$

For any positive real number $x$ the integrator $t$ is of bounded variation on $\left[1,x\right]$ (in fact it’s monotone), and $\frac{1}{t}$ is continuous for positive $t$, so the integral is indeed defined. Since the integrator is differentiable for all positive values, the integral ${F}$ must be as well, and $F'(x)=\frac{1}{x}$.

That is, this procedure has defined for us an antiderivative of $\frac{1}{x}$ on the interval $\left(0,\infty\right)$. We call this function the “natural logarithm” and denote it $\ln(x)$. Tomorrow we’ll start exploring some of its properties.

As a side note, those of you who have been paying close attention will notice that I have yet to use any function more complicated than a rational power of the variable yet. I’m following the pattern of “late transcendentals” in presenting the calculus. The alternative — “early transcendentals” — is to give a hand-waving (but not rigorous) definition of exponentials and logarithms early on to get more examples into the students’ hands. I advocate that position for college-level calculus classes for a number of reasons, but ultimately delaying the transcendentals makes for less unlearning later on.

April 7, 2008 - Posted by | Analysis, Calculus

1. [...] we defined the natural logarithm as the [...]

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2. [...] Exponential Property We’ve defined the natural logarithm and shown that it is, in fact, a logarithm. That is, it’s a homomorphism from the [...]

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3. More parse errors

Comment by Jon | April 16, 2008 | Reply

4. [...] it that makes the exponential what it is? We defined it as the inverse of the logarithm, and this is defined by integrating . But the important thing we immediately showed is that it satisfies the exponential [...]

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