Usuário(a):Bernardo Francio/Testes

Transformada de Laplace para séries de potências

Temos J0(t) a função de Bessel de ordem zero dada por:

$J_{0}(at)=1-\left({\frac {at}{2}}\right)^{2}+{\frac {1}{(2!)^{2}}}\left({\frac {at}{2}}\right)^{4}-{\frac {1}{(3!)^{2}}}\left({\frac {at}{2}}\right)^{6}+...$

Calculando a transformada, temos:

${\mathcal {L}}\left\{J_{0}(at)\right\}={\mathcal {L}}\left\{1\right\}-\left({\frac {a}{2}}\right)^{2}{\mathcal {L}}\left\{t^{2}\right\}+{\frac {1}{(2!)^{2}}}\left({\frac {a}{2}}\right)^{4}{\mathcal {L}}\left\{t^{4}\right\}-{\frac {1}{(3!)^{2}}}\left({\frac {a}{2}}\right)^{6}{\mathcal {L}}\left\{t^{6}\right\}+...$

${\mathcal {L}}\left\{J_{0}(at)\right\}={\frac {1}{s}}-\left({\frac {a}{2}}\right)^{2}{\frac {2!}{s^{3}}}+{\frac {1}{(2!)^{2}}}\left({\frac {a}{2}}\right)^{4}{\frac {4!}{s^{5}}}-{\frac {1}{(3!)^{2}}}\left({\frac {a}{2}}\right)^{6}{\frac {6!}{s^{7}}}+...$

${\mathcal {L}}\left\{J_{0}(at)\right\}={\frac {1}{s}}\left[1-{\frac {1}{2}}\left({\frac {a}{s}}\right)^{2}+{\frac {1}{2}}\cdot {\frac {3}{2}}\cdot {\frac {1}{2!}}\left({\frac {a}{s}}\right)^{4}-{\frac {1}{2}}\cdot {\frac {3}{2}}\cdot {\frac {5}{2}}\cdot {\frac {1}{3!}}\left({\frac {a}{s}}\right)^{6}+...\right]$

Logo,

${\mathcal {L}}\left\{J_{0}(at)\right\}={\frac {1}{s}}\left(1+\left({\frac {a}{s}}\right)^{2}\right)^{-{\frac {1}{2}}}$

Podemos, também, demonstrar a Transformada de Laplace de uma função t(k) que leva à função gama $\Gamma$. Para k>0, temos que:

${\mathcal {L}}\left\{t^{k-1}\right\}={\frac {\Gamma (k)}{s^{k}}}$

Onde:

$\Gamma (k)=\int _{0}^{\infty }e^{-x}x^{k-1}dx$

Aplicando a Transformada de Laplace, temos:

${\mathcal {L}}\left\{t^{k-1}\right\}=\int _{0}^{\infty }t^{k-1}e^{-st}dx$

Fazendo uma mudança de variável $x=st$ para obter:

${\mathcal {L}}\left\{t^{k-1}\right\}=\int _{0}^{\infty }{\frac {x^{k-1}}{s^{k-1}}}e^{-x}{\frac {dx}{s}}$

${\mathcal {L}}\left\{t^{k-1}\right\}={\frac {1}{s^{k}}}\int _{0}^{\infty }x^{k-1}e^{-x}dx$

${\mathcal {L}}\left\{t^{k-1}\right\}={\frac {\Gamma (k)}{s^{k}}}$