# Evaluation of Integral using Laplace Transform/Examples/Example 1

## Examples of Use of Evaluation of Integral using Laplace Transform

$\ds \int_0^\infty \dfrac {e^{-t} - e^{-3 t} } t \rd t = \ln 3$

## Proof

Let $\map f t = e^{-t} - e^{-3 t}$.

Then from Laplace Transform of Exponential:

$\laptrans {\map f t} = \dfrac 1 {s + 1} - \dfrac 1 {s + 3}$

Hence:

 $\ds \laptrans {\dfrac {e^{-t} - e^{-3 t} } t}$ $=$ $\ds \int_0^\infty \paren {\dfrac 1 {u + 1} - \dfrac 1 {u + 3} } \rd u$ Integral of Laplace Transform $\ds \leadsto \ \$ $\ds \int_0^\infty e^{-s t} \paren {\dfrac {e^{-t} - e^{-3 t} } t} \rd t$ $=$ $\ds \map \ln {\dfrac {s + 3} {s + 1} }$ Definition of Laplace Transform, Primitive of Reciprocal $\ds \leadsto \ \$ $\ds \int_0^\infty \dfrac {e^{-t} - e^{-3 t} } t \rd t$ $=$ $\ds \ln 3$ taking the limit as $s \to 0^+$

$\blacksquare$