Primitive of Exponential of a x over x

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Theorem

For $x > 0$:

\(\ds \int \frac {e^{a x} \rd x} x\) \(=\) \(\ds \ln x + \sum_{k \mathop \ge 1} \frac {\paren {a x}^k} {k \times k!} + C\)
\(\ds \) \(=\) \(\ds \ln x + \dfrac {a x} {1 \times 1!} + \dfrac {\paren {a x}^2} {2 \times 2!} + \dfrac {\paren {a x}^3} {3 \times 3!} + \cdots + C\)


Proof

\(\ds \int \frac {e^{a x} \rd x} x\) \(=\) \(\ds \int \frac 1 x \paren {\sum_{k \mathop = 0}^\infty \frac {\paren {a x}^k} {k!} } \rd x\) Power Series Expansion for Exponential Function
\(\ds \) \(=\) \(\ds \int \frac 1 x \paren {1 + \sum_{k \mathop = 1}^\infty \frac {\paren {a x}^k} {k!} } \rd x\) extracting the case $k = 0$ from the expansion
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \int \frac {\d x} x + \sum_{k \mathop = 1}^\infty \frac {a^k} {k!} \int x^{k - 1} \rd x\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \ln x + \sum_{k \mathop = 1}^\infty \frac {a^k} {k!} \int x^{k - 1} \rd x\) Primitive of Reciprocal: $x > 0$ so negative argument does not apply
\(\ds \) \(=\) \(\ds \ln x + \sum_{k \mathop = 1}^\infty \frac {a^k} {k!} \frac {x^k} k + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \ln x + \sum_{k \mathop \ge 1} \frac {\paren {a x}^k} {k \times k!} + C\) simplification

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

$\blacksquare$


Also see


Sources

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