# Primitive of Exponential of a x over x

## Theorem

$\displaystyle \int \frac {e^{a x} \rd x} x = \ln \size x + \sum_{k \mathop \ge 1} \frac {\paren {a x}^k} {k \times k!} + C$

## Proof

 $\displaystyle \int \frac {e^{a x} \rd x} x$ $=$ $\displaystyle \int \frac 1 x \paren {\sum_{k \mathop = 0}^\infty \frac {\paren {a x}^k} {k!} } \rd x$ Power Series Expansion for Exponential Function $\displaystyle$ $=$ $\displaystyle \int \frac 1 x \paren {1 + \sum_{k \mathop = 1}^\infty \frac {\paren {a x}^k} {k!} } \rd x$ extracting the case $k = 0$ from the expansion $\text {(1)}: \quad$ $\displaystyle$ $=$ $\displaystyle \int \frac {\d x} x + \sum_{k \mathop = 1}^\infty \frac {a^k} {k!} \int x^{k - 1} \rd x$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \ln \size x + \sum_{k \mathop = 1}^\infty \frac {a^k} {k!} \int x^{k - 1} \rd x$ Primitive of Reciprocal $\displaystyle$ $=$ $\displaystyle \ln \size x + \sum_{k \mathop = 1}^\infty \frac {a^k} {k!} \frac {x^k} k + C$ Primitive of Power $\displaystyle$ $=$ $\displaystyle \ln \size x + \sum_{k \mathop \ge 1} \frac {\paren {a x}^k} {k \times k!} + C$ simplification

The validity of $(1)$ follows from absolute convergence of the power series expansion.

$\blacksquare$