Use o problema 4 e o fato que \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} para encontrar a transformada de Laplace da função
f(t)=2 t^{\frac{1}{2}}+8 t^{\frac{5}{2}} .

Passo 1

O enunciado pede a transformada de Laplace de
f(t)=2\,t^{\frac12}+8\,t^{\frac52},
usando a fórmula geral (vista no problema 4)
\mathcal{L}\{t^{\alpha}\}(s)=\frac{\Gamma(\alpha+1)}{s^{\alpha+1}},
e o fato de que \Gamma\bigl(\tfrac12\bigr)=\sqrt\pi.

Passo 2

Calculamos primeiro
\mathcal{L}\{2\,t^{\frac12}\}(s) =2\,\frac{\Gamma\bigl(\tfrac12+1\bigr)}{s^{\tfrac12+1}} =2\,\frac{\Gamma\bigl(\tfrac32\bigr)}{s^{\tfrac32}}.
Usando a relação \Gamma(\alpha+1)=\alpha\,\Gamma(\alpha), obtemos
\Gamma\bigl(\tfrac32\bigr) =\tfrac12\,\Gamma\bigl(\tfrac12\bigr) =\tfrac12\,\sqrt\pi,
logo
\mathcal{L}\{2\,t^{\frac12}\}(s) =2\cdot\frac{\tfrac12\,\sqrt\pi}{s^{\tfrac32}} =\frac{\sqrt\pi}{s^{\tfrac32}}.

Passo 3

Agora calculamos
\mathcal{L}\{8\,t^{\frac52}\}(s) =8\,\frac{\Gamma\bigl(\tfrac52+1\bigr)}{s^{\tfrac52+1}} =8\,\frac{\Gamma\bigl(\tfrac72\bigr)}{s^{\tfrac72}}.
De novo usando a relação de recorrência várias vezes,
\Gamma\bigl(\tfrac72\bigr) =\tfrac52\,\Gamma\bigl(\tfrac52\bigr) =\tfrac52\cdot\tfrac32\,\Gamma\bigl(\tfrac32\bigr) =\tfrac52\cdot\tfrac32\cdot\tfrac12\,\Gamma\bigl(\tfrac12\bigr) =\frac{15}{8}\,\sqrt\pi.
Portanto
\mathcal{L}\{8\,t^{\frac52}\}(s) =8\cdot\frac{\tfrac{15}{8}\,\sqrt\pi}{s^{\tfrac72}} =\frac{15\,\sqrt\pi}{s^{\tfrac72}}.

Passo 4

Somando as duas partes, a transformada de Laplace de f(t) é
F(s)=\frac{\sqrt\pi}{s^{\tfrac32}} +\frac{15\,\sqrt\pi}{s^{\tfrac72}}.

Resposta

\mathcal{L}\{f(t)\}(s) = \frac{\sqrt\pi}{s^{\tfrac32}} + \frac{15\,\sqrt\pi}{s^{\tfrac72}}.

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