Primitive of Reciprocal of Logarithm of x

From ProofWiki
Jump to navigation Jump to search

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 1

\(\ds u\) \(=\) \(\ds \ln x\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds e^u\)
\(\ds \leadsto \ \ \) \(\ds \d x\) \(=\) \(\ds e^u \d u\)
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\ln x}\) \(=\) \(\ds \int \frac {e^u \rd u} u\)
\(\ds \) \(=\) \(\ds \ln u + \sum_{k \mathop \ge 1} \frac {u^k} {k \times k!} + C\) 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 $u = \ln x$

$\blacksquare$


Proof 2

From Primitive of $\dfrac {x^m} {\ln x}$:

$\ds \int \frac {x^m \rd x} {\ln x} = \map \ln {\ln x} + \paren {m + 1} \ln x + \sum_{k \mathop \ge 2}^n \frac {\paren {m + 1}^k \paren {\ln x}^k} {k \times k!} + C$

The result follows by setting $m = 0$.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Reciprocal_of_Logarithm_of_x&oldid=565781"