Limit at Infinity of Sine Integral Function

Theorem

Let $\Si: \R \to \R$ denote the sine integral function.

Then $\Si$ has a (finite) limit at infinity:

$\ds \lim_{x \mathop \to +\infty} \map \Si x = \frac \pi 2$

Corollary

$\ds \lim_{x \mathop \to -\infty} \map \Si x = -\frac \pi 2$

Proof

The limit:

$\ds \lim_{x \mathop \to +\infty} \map \Si x = \lim_{x \mathop \to +\infty} \int_{t \mathop \to 0}^{t \mathop = x} \frac {\sin t} t \rd t$

is the Dirichlet Integral.

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 35$: Miscellaneous Special Functions: Sine Integral $\ds \map {Si} x = \int_0^x \frac {\sin u} u \rd u$: $35.13$
• 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 36$: Miscellaneous and Riemann Zeta Functions: Sine Integral $\ds \map \Si x = \int_0^x \frac {\sin u} u \rd u$: $36.13.$