Primitive of Reciprocal of Logarithm of x/Proof 1

Theorem

For $x > 1$:

\(\ds \int \frac {\d x} {\ln x}\)

\(=\)

\(\ds \map \ln {\ln x} + \sum_{k \mathop \ge 1} \frac {\paren {\ln x}^k} {k \times k!} + C\)

\(\ds \)

\(=\)

\(\ds \map \ln {\ln x} + \dfrac {\ln x} {1 \times 1!} + \dfrac {\paren {\ln x}^2} {2 \times 2!} + \dfrac {\paren {\ln x}^3} {3 \times 3!} + \cdots + C\)

Proof

\(\ds u\)

\(=\)

\(\ds \ln x\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds e^u\)

\(\ds \leadsto \ \ \)

\(\ds \d x\)

\(=\)

\(\ds e^u \d u\)

\(\ds \leadsto \ \ \)

\(\ds \int \frac {\d x} {\ln x}\)

\(=\)

\(\ds \int \frac {e^u \rd u} u\)

\(\ds \)

\(=\)

\(\ds \ln u + \sum_{k \mathop \ge 1} \frac {u^k} {k \times k!} + C\)

Primitive of $\dfrac {e^{a x} } x$: $u > 0$ for $x > 1$

\(\ds \)

\(=\)

\(\ds \map \ln {\ln x} + \sum_{k \mathop \ge 1} \frac {\paren {\ln x}^k} {k \times k!} + C\)

substituting $u = \ln x$

$\blacksquare$