Power Series Expansion for Real Area Hyperbolic Sine/Lemma 2

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Lemma for Power Series Expansion for Real Area Hyperbolic Sine

$\map \ln {1 + \sqrt {1 + x^2} } = \ln 2 + \dfrac 1 2 \cdot \dfrac {x^2} 2 - \dfrac {1 \times 3} {2 \times 4} \cdot \dfrac {x^4} 4 + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \cdot \dfrac {x^6} 6 - \cdots$

This holds for $x \in \R: \size x < 1$.


Proof

\(\ds \map {\dfrac \d {\d x} } {\map \ln {1 + \sqrt {1 + x^2} } }\) \(=\) \(\ds \dfrac 1 {1 + \sqrt {1 + x^2} } \cdot \dfrac 1 {2 \sqrt {1 + x^2} } \cdot 2 x\)
\(\ds \) \(=\) \(\ds \dfrac x {1 + x^2 + \sqrt {1 + x^2} }\) simplifying and multiplying out
\(\ds \) \(=\) \(\ds \dfrac {x \paren {1 + x^2 - \sqrt {1 + x^2} } } {\paren {1 + x^2 + \sqrt {1 + x^2} } \paren {1 + x^2 - \sqrt {1 + x^2} } }\) multiplying top and bottom by $1 + x^2 - \sqrt {1 + x^2}$
\(\ds \) \(=\) \(\ds \dfrac {x \paren {1 + x^2 - \sqrt {1 + x^2} } } {\paren {1 + x^2}^2 - \paren {1 + x^2} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \dfrac {x \paren {1 + x^2} - x \sqrt {1 + x^2} } {1 + 2 x^2 + x^4 - 1 - x^2}\) multiplying out
\(\ds \) \(=\) \(\ds \dfrac {x \paren {1 + x^2} - x \sqrt {1 + x^2} } {x^2 \paren {1 + x^2} }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 x \paren {1 - \dfrac 1 {\sqrt {1 + x^2} } }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 x \paren {1 - \paren {1 - \frac 1 2 x^2 + \frac {1 \times 3} {2 \times 4} x^4 - \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^6 + \cdots} }\) Lemma $1$: for $0 < \size x < 1$
\(\ds \) \(=\) \(\ds \frac 1 2 x - \frac {1 \times 3} {2 \times 4} x^3 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^5 - \cdots\) simplifying
\(\ds \leadsto \ \ \) \(\ds \map \ln {1 + \sqrt {1 + x^2} }\) \(=\) \(\ds C + \frac 1 2 \cdot \dfrac {x^2} 2 - \frac {1 \times 3} {2 \times 4} \cdot \dfrac {x^4} 4 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} \cdot \dfrac {x^6} 6 - \cdots\) Integrating with respect to $x$: Power Series is Termwise Integrable within Radius of Convergence

It remains to determine the value of the arbitrary constant $C$.

Let $x \to 0$.

Then we have:

\(\ds \lim_{x \mathop \to 0} \map \ln {1 + \sqrt {1 + x^2} }\) \(=\) \(\ds \map {\lim_{x \mathop \to 0} } {C + \frac 1 2 \cdot \dfrac {x^2} 2 - \frac {1 \times 3} {2 \times 4} \cdot \dfrac {x^4} 4 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} \cdot \dfrac {x^6} 6 - \cdots}\)
\(\ds \leadsto \ \ \) \(\ds \ln 2\) \(=\) \(\ds C\)

$\blacksquare$