Laplace Transform of Sine Integral Function/Proof 1

Proof
Let $\map f t := \map \Si t = \ds \int_0^t \dfrac {\sin u} u \rd u$.

Then:
 * $\map f 0 = 0$

and:

By the Initial Value Theorem of Laplace Transform:


 * $\ds \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$

which leads to:
 * $c = \dfrac \pi 2$

Thus: