Laplace Transform of Exponential Integral Function/Proof 1

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Theorem

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

where:

$\laptrans f$ denotes the Laplace transform of the function $f$
$\Ei$ denotes the exponential integral function.


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 $\dfrac 1 {a x + b}$


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:

\(\ds s \laptrans {\map f t}\) \(=\) \(\ds \map \ln {s + 1}\)
\(\ds \leadsto \ \ \) \(\ds \laptrans {\map f t}\) \(=\) \(\ds \dfrac {\map \ln {s + 1} } s\)

$\blacksquare$


Sources