Laplace Transform of Exponential Integral Function/Proof 1

Proof
Let $\map f t := \map \Ei t = \ds \int_t^\infty \dfrac {e^{-u} } u \rd u$.

Then:

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:
 * $c = 0$

Thus: