Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Cosine of a x/Mistake
Source Work
1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables
- Chapter $14$: Indefinite Integrals
- Integrals involving $\cosh a x$: $14.584$
This mistake can be seen in the edition as published by Schaum: ISBN 0-07-060224-7 (unknown printing).
Mistake
- $\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \begin {cases}
\dfrac 1 {2 a p \sqrt {p^2 + q^2} } \map \ln {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } } \\ \dfrac 1 {a p \sqrt {p^2 + q^2} } \arctan \dfrac {p \tanh a x} {\sqrt {p^2 + q^2} } \end {cases}$
Correction
As demonstrated in Primitive of $\dfrac {\d x} {p^2 + q^2 \cosh^2 a x}$, the second expression is incorrect.
It should just be:
- $\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \dfrac 1 {2 a p \sqrt {p^2 + q^2} } \map \ln {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }$
The second expression is not completely wrong, however.
The result can also be presented as:
- $\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \dfrac 1 {a p \sqrt {p^2 + q^2} } \tanh^{-1} \dfrac {p \tanh a x} {\sqrt {p^2 + q^2} }$
but it is a genuine equivalence which is valid over the whole range.
Hence presenting it as two separate cases, each expected to valid over a subdomain of $\R$ (in this case unspecified), is misleading.
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\cosh a x$: $14.584$