Antiderivative definition

Question

Recently, I was trying to derive the antiderivative of $\frac{1}{x}$ by taking

$$\lim_{p\to-1}\;\;\int{x^p}$$

but quickly realized that it would not work as it gives

$$\frac{x^0}{0}$$

however, if I instead took

$$\lim_{p\to-1}\;\;\int_a^x{x^p}$$

This gives

$$\lim_{p\to-1}\;\;\frac{x^{p+1}-a^{p+1}}{p+1}$$

Which evaluated with L'Hopital's rule returned

$$\ln(x)-\ln(a)$$

where $\ln(a)$ is just some constant and returns the expected result.

This leads me to ask, is the antiderivative defined as a definite integral with variable upper bound, or is there something else I'm not understanding here? I looked at the fundamental theorem of calculus, for hints and this seems to be the case, but there was nothing definitive that said this was true.

Answer 1

You're close to a good answer.

The logarithm function

Formally, the logarithm function is defined as a function that is a solution to the logarithmic differential equation

$$ \begin{cases}F'(x) = \frac{1}{x}, \text{ for all } x>0 \\ F(1) = 0 \end{cases} $$

rather than as an antiderivative of the derivative of $\frac{1}{x}$.

Such function satisfies the three properties of a logarithmic operation which can be proved through properties of the differential equation. The logarithm is an antiderivative only in the sense that it is the solution to the differential equation that is being integrated.

To illustrate, consider this property of the logarithm.

i) $F(ab) = F(a) +F(b)$ (Fitzpatrick, Theorem 5.1.i)

Define a function $g:(0, \infty) \to \mathbb{R}$ as $g(x) = F(ax) - F(a) - F(x)$. It follows that

$$ g'(x) = \frac{\mathrm{d}}{\mathrm{d}x}\left( F(ax) - F(a) - F(x) \right) = \frac{a}{ax} - 0 - \frac{1}{x} = \frac{1}{x} - \frac{1}{x} = 0$$

This satisfies the criteria for the identity criterion, so we can then conclude $F(ax) = F(a) + F(x)$.

Indefinite Integration

Indefinite integration is a special case of the first part of the First Fundamental Theorem of Calculus (integrating derivatives).

The theorem states that for a function $F:[a,b] \to \mathbb{R}$ that is continuous over $[a,b]$ (a closed interval) that has a continuous and bounded derivative over the open interval $(a,b)$, the function $F$ satisfies $$\int_a^b F'(x) \mathrm{d}x = F(b) - F(a) $$

We'd like to pick values of $a$ and $b$ such that $$ \int_a^b F'(x) \mathrm{d}x = F(t) $$ Which is obtained by selecting $a = t$ and $b$ such that $F(b) = 0$. (Note that $b = 1$ satisfies $ln(1) = 0$, the initial condition of the logarithm differential equation.)

This gives $$ \int_a^t F'(x) \mathrm{d}x = F(t) $$ Which can then be differentiated pursuant to the Second Fundamental Theorem of Calculus (differentiating integrals) at $t = x$ to obtain $F'(x)$.

Answer 2

This is a version of the fundamental theorem of calculus. The rate of change of the definite integral with respect to the upper limit is the integrand.

But your reasoning here is suspiciously circular since to use L'Hopital you must already know a lot about the natural logarithm - probably something equivalent to what you are trying to prove.

Answer 3

$$\int\frac{1}{x}dx=\ln|x|+C$$ if this is true, then: $$\frac{1}{x}=\frac{d}{dx}\left[\ln|x|\right]$$ let $$y=\ln(x)$$ then $$e^y=x$$ $$e^y\frac{dy}{dx}=1$$ $$\frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}$$