Key to "Determine the Laplace transform of the Laguerre polynomials,..."

nagasenaz

Answered question

2021-09-03

Determine the Laplace transform of the Laguerre polynomials, defined by
$L_{n} (t) = \frac{e^{t}}{n!} \frac{d^{n}}{d t^{n}} (t^{n} e^{- t}), n = 0, 1, 2, . . .$

Answer & Explanation

Dora

Skilled2021-09-04Added 98 answers

Consider the provided question
$L_{n} (t) = \frac{e^{t}}{n!} \frac{d^{n}}{d t^{n}} (t^{n} e^{- t}), n = 0, 1, 2, . . .$
This is Laguerre polynomials,
The Laplace transformation of any function f(t) is given by,
$L [f (t)] = F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$
The Laplace transform of Laguerre's polynomial is,
$L = [L_{n} (t)] = \int_{0}^{\infty} e^{s t} (\frac{e^{t}}{n!}) \cdot \frac{d^{n}}{{d t}^{n}} (t^{n} \cdot e^{- t}) d t$
$= \frac{1}{n!} \int_{0}^{\infty} e^{- (s - 1) t} \cdot \frac{d^{n}}{{d t}^{n}} (t^{n} \cdot e^{- t}) d t$
Now solve the above integral,
$\frac{1}{n} \int_{0}^{\infty} e^{- (s - 1) t} \cdot \frac{d^{n}}{{d t}^{n}} (t^{n} \cdot e^{- t}} d t = \frac{1}{n!} e^{- (s - 1) t} \int_{0}^{\infty} \cdot \frac{d^{n}}{{d t}^{n}} (t^{n} \cdot e^{- t}) d t - \frac{1}{n!} \int_{0}^{\infty} [\frac{d}{d t} e^{- (s - 1) t} \cdot \int \frac{d^{n}}{{d t}^{n}} (t^{n} \cdot e^{- t})] d t$
$= [\frac{e^{- (s - 1) t}}{n!} \cdot \frac{d^{n - 1}}{}$

Do you have a similar question?Recalculate according to your conditions!

New Questions in Precalculus

Didn't find what you were looking for?