Power Series Expansion for Logarithm of Sine of x

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Theorem

\(\ds \ln \size {\sin x}\) \(=\) \(\ds \ln \size x - \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n 2^{2 n - 1} B_{2 n} \, x^{2 n} } {n \paren {2 n}!}\)
\(\ds \) \(=\) \(\ds \ln \size x - \frac {x^2} 6 - \frac {x^4} {180} - \frac {x^6} {2835} + \cdots\)


for all $x \in \R$ such that $0 < \size x < \pi$.


Proof

From Power Series Expansion for Cotangent Function:

\(\text {(1)}: \quad\) \(\ds \cot x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}\)
\(\ds \) \(=\) \(\ds \frac 1 x - \frac x 3 - \frac {x^3} {45} - \frac {2 x^5} {945} + \cdots\)

for $0 < \size x < \pi$.


From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term:

\(\ds \int_0^x \cot x \rd x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \int_0^x \frac {\paren {-1}^{n - 1} 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!} \rd x\)
\(\ds \) \(=\) \(\ds \int_0^x \frac 1 x \rd x + \sum_{n \mathop = 1}^\infty \int_0^x \frac {\paren {-1}^{n - 1} 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!} \rd x\) extracting the zeroth term
\(\ds \leadsto \ \ \) \(\ds \ln \size {\sin x}\) \(=\) \(\ds \ln \size x + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} B_{2 n} \, x^{2 n} } {\paren {2 n} \paren {2 n}!}\) Primitive of $\cot x$, Integral of Power, Primitive of Reciprocal
\(\ds \leadsto \ \ \) \(\ds \ln \size {\sin x}\) \(=\) \(\ds \ln \size x - \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n 2^{2 n - 1} B_{2 n} \, x^{2 n} } {n \paren {2 n}!}\) simplification

$\blacksquare$


Sources

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