Derivative of Sine Integral Function

Theorem

$\dfrac \d {\d x} \paren {\map \Si x} = \dfrac {\sin x} x$

where $\Si$ denotes the sine integral function.

Proof

We have, by the definition of the sine integral function:

$\ds \map \Si x = \int_0^x \frac {\sin t} t \rd t$

By Fundamental Theorem of Calculus (First Part): Corollary, we have:

$\ds \frac \d {\d x} \paren {\map \Si x} = \frac {\sin x} x$

$\blacksquare$

Also see

• Derivative of Cosine Integral Function