Primitive of Hyperbolic Secant of a x over x

From ProofWiki
Jump to navigation Jump to search

Theorem

\(\ds \int \frac {\sech a x \rd x} x\) \(=\) \(\ds \ln \size x + \sum_{k \mathop \ge 1} \frac {E_{2 k} \paren {a x}^{2 k} } {\paren {2 k} \paren {2 k}!} + C\)
\(\ds \) \(=\) \(\ds \ln \size x - \frac {\paren {a x}^2} 4 + \frac {\paren {a x}^4} {96} - \frac {\paren {a x}^6} {4320} + \cdots + C\)

where $E_k$ denotes the $k$th Euler number.


Proof

\(\ds \sech x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {E_{2 k} x^{2 k} } {\paren {2 k}!}\) Power Series Expansion for Hyperbolic Secant Function
\(\ds \leadsto \ \ \) \(\ds \frac {\sech a x} x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {E_{2 k} \paren {a x}^{2 k} } {x \paren {2 k}!}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {E_{2 k} a^{2 k} x^{2 k - 1} } {\paren {2 k}!}\)
\(\ds \) \(=\) \(\ds \frac 1 x + \sum_{n \mathop = 1}^\infty \frac {E_{2 k} a^{2 k} x^{2 k - 1} } {\paren {2 k}!}\) extracting the first term, which needs to be handled separately
\(\ds \leadsto \ \ \) \(\ds \int \frac {\sech a x \rd x} x\) \(=\) \(\ds \int \paren {\frac 1 x + \sum_{n \mathop = 1}^\infty \frac {E_{2 k} a^{2 k} x^{2 k - 1} } {\paren {2 k}!} } \rd x\)
\(\ds \) \(=\) \(\ds \int \frac 1 x \rd x + \sum_{k \mathop = 1}^\infty \int \frac {E_{2 k} a^{2 k} x^{2 k - 1} } {\paren {2 k}!} \rd x\)
\(\ds \) \(=\) \(\ds \ln \size x + \sum_{k \mathop = 1}^\infty \int \frac {E_{2 k} a^{2 k} x^{2 k - 1} } {\paren {2 k}!} \rd x\) Primitive of Reciprocal
\(\ds \) \(=\) \(\ds \ln \size x + \sum_{k \mathop = 1}^\infty \frac {E_{2 k} a^{2 k} x^{2 k} } {\paren {2 k} \paren {2 k}!}\) Primitive of Power
\(\ds \) \(=\) \(\ds \ln \size x + \sum_{k \mathop = 1}^\infty \frac {E_{2 k} \paren {a x}^{2 k} } {\paren {2 k} \paren {2 k}!}\)

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Hyperbolic_Secant_of_a_x_over_x&oldid=741009"