# Laplace Transform of Exponential Integral Function/Proof 1

## Theorem

Let $\Ei: \R_{>0} \to \R$ denote the exponential integral function:

$\map \Ei x = \ds \int_{t \mathop = x}^{t \mathop \to +\infty} \frac {e^{-t} } t \rd t$

Then:

$\laptrans {\map \Ei t} = \dfrac {\map \ln {s + 1} } s$

where $\laptrans f$ denotes the Laplace transform of the function $f$

## Proof

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

Then:

 $\ds \map {f'} t$ $=$ $\ds -\dfrac {e^{-t} } t$ $\ds \leadsto \ \$ $\ds t \map {f'} t$ $=$ $\ds -e^{-t}$ $\ds \leadsto \ \$ $\ds \laptrans {t \map {f'} t}$ $=$ $\ds -\laptrans {e^{-t} }$ $\ds$ $=$ $\ds -\dfrac 1 {s + 1}$ Laplace Transform of Exponential $\ds \leadsto \ \$ $\ds -\dfrac \d {\d s} \laptrans {\map {f'} t}$ $=$ $\ds -\dfrac 1 {s + 1}$ Derivative of Laplace Transform $\ds \leadsto \ \$ $\ds \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}$ $=$ $\ds \dfrac 1 {s + 1}$ Laplace Transform of Derivative $\ds \leadsto \ \$ $\ds s \laptrans {\map f t}$ $=$ $\ds \int \dfrac 1 {s + 1} \rd s$ $\map f 0 = 0$, and integrating both sides with respect to $s$ $\ds \leadsto \ \$ $\ds s \laptrans {\map f t}$ $=$ $\ds \map \ln {s + 1} + C$ Primitive of $\dfrac 1 {a x + b}$
$\ds \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$

$c = 0$
 $\ds s \laptrans {\map f t}$ $=$ $\ds \map \ln {s + 1}$ $\ds \leadsto \ \$ $\ds \laptrans {\map f t}$ $=$ $\ds \dfrac {\map \ln {s + 1} } s$
$\blacksquare$