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:

\(\ds \lim _{x \mathop \to +\infty} \size {\map \Si x}\)

\(=\)

\(\ds \lim _{x \mathop \to -\infty} \size {\map \Si x}\)

\(\ds \)

\(=\)

\(\ds \dfrac \pi 2\)

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:

\(\ds \size {\map \Si x}\)

\(=\)

\(\ds \size {\int_0 ^x \dfrac {\sin t} t \rd t }\)

\(\ds \)

\(\le\)

\(\ds \int _0 ^{\size x} \size { \dfrac {\sin t} t} \rd t\)

\(\ds \)

\(\le\)

\(\ds \int _0 ^{\size x} 1 \rd t\)

\(\ds \)

\(=\)

\(\ds \size x\)

\(\ds \)

\(\le\)

\(\ds M\)

Thus:

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

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