Sine Integral Function is Bounded

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

Then $\Si$ is bounded.

Proof
By Limit at Infinity of Sine Integral Function and its corollary:

Thus there is a $M > 0$ such that for all $x \in \R$:
 * $\size x > M \implies \size {\map \Si x} \le \dfrac \pi 2 + 1$

On the other hand, if $\size x \le M$, then:

Thus:
 * $\forall x \in \R : \size {\map \Si x} \le \max \set {\dfrac \pi 2 + 1, M}$