Equivalence of Definitions of Real Natural Logarithm/Proof 1

Definition 1 implies Definition 2
Let $\map F x$ be $\ds \int_1^x \frac {\d t} t$.

Let $\map f t$ be $\ds \int \frac {\d t} t$.

Then:
 * $\dfrac {\d t} t = \dfrac 1 t$

Or:
 * $\dfrac {\d x} x = \dfrac 1 x$

Also:
 * $\map F x = \map f x - \map f 1$

Therefore:

Furthermore:
 * $\map F 1 = \map f 1 - \map f 1 = 0$

The result follows from the fifth definition of the exponential function:
 * $\map F x \equiv e^x$

Definition 2 implies Definition 1
Let $\map f t$ be $\ds \int \frac 1 t \rd t$.

Then:
 * $\map F x = \map f x + C$

When $\map F x = 0$:
 * $x = e^{\map F x} = 1$
 * $\map F 1 = \map f 1 + C = 0 \implies \map f 1 = - C$

Therefore:
 * $\map F x = \map f x - \map f 1$

Therefore:
 * $\ds \map F x = \int_1^x \frac {\d t} t$

Therefore: