Asymptotic Expansion for Sine Integral Function

Theorem

$\ds \map \Si x \sim \frac \pi 2 - \frac {\sin x} x \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {2 n + 1}!} {x^{2 n + 1} } - \frac {\cos x} x \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {2 n}!} {x^{2 n} }$

where:

$\Si$ denotes the sine integral function

$\sim$ denotes asymptotic equivalence as $x \to \infty$.

Proof

This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 35$: Sine Integral $\ds \map \Si x = \int_0^x \frac {\sin u} u \rd u$: $35.12$