Limit at Infinity of Sine Integral Function

Theorem
Let $\operatorname{Si}: \R \to \R$ denote the sine integral function.

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


 * $\displaystyle \lim_{x \mathop \to +\infty} \operatorname{Si} \left({ x }\right) = \frac \pi 2$.

Corollary

 * $\displaystyle \lim_{x \mathop \to -\infty} \operatorname{Si} \left({ x }\right) = -\frac \pi 2$.

Proof
The limit:


 * $\displaystyle \lim_{x \mathop \to +\infty} \operatorname{Si} \left({ x }\right) = \lim_{x \mathop \to +\infty} \int_{t \mathop \to 0}^{t \mathop = x} \frac {\sin t} t \, \mathrm d t$

is the Dirichlet Integral.