# Laplace Transform of Cosine Integral Function/Proof 1

## Theorem

$\laptrans {\map \Ci t} = \dfrac {\map \ln {s^2 + 1} } {2 s}$

where:

$\laptrans f$ denotes the Laplace transform of the function $f$
$\Ci$ denotes the cosine integral function.

## Proof

Let $\map f t := \map \Ci t = \ds \int_t^\infty \dfrac {\cos u} u \rd u$.

Then:

 $\ds \map {f'} t$ $=$ $\ds -\dfrac {\cos t} t$ $\ds \leadsto \ \$ $\ds t \map {f'} t$ $=$ $\ds -\cos t$ $\ds \leadsto \ \$ $\ds \laptrans {t \map {f'} t}$ $=$ $\ds -\laptrans {\cos t}$ $\ds$ $=$ $\ds -\dfrac s {s^2 + 1}$ Laplace Transform of Cosine $\ds \leadsto \ \$ $\ds -\dfrac \d {\d s} \laptrans {\map {f'} t}$ $=$ $\ds -\dfrac s {s^2 + 1}$ Derivative of Laplace Transform $\ds \leadsto \ \$ $\ds \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}$ $=$ $\ds \dfrac s {s^2 + 1}$ Laplace Transform of Derivative $\ds \leadsto \ \$ $\ds s \laptrans {\map f t}$ $=$ $\ds \int \dfrac s {s^2 + 1} \rd s$ $\map f 0 = 0$, and integrating both sides with respect to $s$ $\ds \leadsto \ \$ $\ds s \laptrans {\map f t}$ $=$ $\ds \dfrac 1 2 \map \ln {s^2 + 1} + C$ Primitive of $\dfrac x {x^2 + a^2}$
$\ds \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$

$c = 0$
 $\ds s \laptrans {\map f t}$ $=$ $\ds \dfrac 1 2 \map \ln {s^2 + 1}$ $\ds \leadsto \ \$ $\ds \laptrans {\map f t}$ $=$ $\ds \dfrac {\map \ln {s^2 + 1} } {2 s}$
$\blacksquare$