Sum of Logarithms/Natural Logarithm/Proof 2

Theorem

Then:

$\ln x + \ln y = \map \ln {x y}$

where $\ln$ denotes the natural logarithm.

$\ds \ln x + \ln y$

$=$

$\ds \int_1^x \dfrac {\d t} t + \int_1^y \dfrac {\d s} s$

$\ds $

$=$

$\ds \int_1^x \dfrac {\d t} t + \int_x^{x y} \dfrac {\d t / x} {t / x}$

Integration by Substitution: $s \mapsto t / x$, $\d s \mapsto \d t / x$, $1 \mapsto x$, $y \mapsto x y$

$\ds $

$=$

$\ds \int_1^x \dfrac {\d t} t + \int_x^{x y} \dfrac {\d t} t$

simplifying

$\ds $

$=$

$\ds \int_1^{x y} \dfrac {\d t} t$

$\ds $

$=$

$\ds \ln x y$

$\blacksquare$