Laplace Transform of Exponential Integral Function
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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 1
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
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $10$