Primitive of Inverse Hyperbolic Secant of x over a over x

From ProofWiki
Jump to navigation Jump to search

Theorem

\(\ds \int \dfrac 1 x \arsech \dfrac x a \rd x\) \(=\) \(\ds -\frac 1 2 \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} - \sum_{n \mathop \ge 1} \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n}^2} \paren {\frac x a}^{2 n} + C\)
\(\ds \) \(=\) \(\ds -\dfrac 1 2 \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} - \dfrac 1 {2 \times 2^2} \paren {\dfrac x a}^2 - \dfrac {1 \times 3} {2 \times 4 \times 4^2} \paren {\dfrac x a}^4 - \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6^2} \paren {\dfrac x a}^6 + C\)

where $\arsech$ denotes the real area hyperbolic secant.


Corollary

\(\ds \int \dfrac 1 x \paren {-\sech^{-1} \dfrac x a} \rd x\) \(=\) \(\ds \frac 1 2 \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} + \sum_{n \mathop \ge 1} \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n}^2} \paren {\frac x a}^{2 n} + C\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} + \dfrac 1 {2 \times 2^2} \paren {\dfrac x a}^2 - \dfrac {1 \times 3} {2 \times 4 \times 4^2} \paren {\dfrac x a}^4 - \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6^2} \paren {\dfrac x a}^6 + C\)

where $-\sech^{-1}$ denotes the negative branch of the real inverse hyperbolic secant multifunction.


Proof

\(\ds \arsech \dfrac x a\) \(=\) \(\ds \ln \frac {2 a} x - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \paren {\frac x a}^{2 n} }\) Power Series Expansion for Real Area Hyperbolic Secant
\(\ds \leadsto \ \ \) \(\ds \frac 1 x \arsech \dfrac x a\) \(=\) \(\ds \dfrac 1 x \ln \frac {2 a} x - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \frac 1 {a^{2 n} } x^{2 n - 1} }\)
\(\ds \leadsto \ \ \) \(\ds \int \frac 1 x \arsech \dfrac x a \rd x\) \(=\) \(\ds \int \paren {\dfrac 1 x \ln \frac {2 a} x - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \frac 1 {a^{2 n} } x^{2 n - 1} } } \rd x\)
\(\ds \) \(=\) \(\ds \int -\frac 1 x \ln \frac x {2 a} \rd x - \paren {\sum_{n \mathop = 1}^\infty \int \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \frac 1 {a^{2 n} } x^{2 n - 1} \rd x}\) Fubini's Theorem
\(\ds \) \(=\) \(\ds -\frac 1 2 \map {\ln^2} {\dfrac x {2 a} } - \paren {\sum_{n \mathop = 1}^\infty \int \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} } \frac 1 {a^{2 n} } x^{2 n - 1} \rd x} + C\) Primitive of $\dfrac {\ln x} x$: Corollary
\(\ds \) \(=\) \(\ds -\frac 1 2 \map {\ln^2} {\dfrac x {2 a} } - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 n \paren {2 n} } \frac 1 {a^{2 n} } \dfrac {x^{2 n} } {2 n} } + C\) Primitive of Power
\(\ds \) \(=\) \(\ds -\frac 1 2 \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} + \map {\ln^2} 2 - \sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n}^2} \paren {\frac x a}^{2 n} + C\) rearranging
\(\ds \) \(=\) \(\ds -\frac 1 2 \map \ln {\dfrac a x} \map \ln {\dfrac {4 a} x} - \sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n}^2} \paren {\frac x a}^{2 n} + C\) subsuming $\map {\ln^2} 2$ into the constant of integration

$\blacksquare$


Also see


Sources