Answer:
[tex]${L}\{t^{-3/2}}\}$[/tex] does not exists
Step-by-step explanation:
First lets introduce the definition of the Laplace transform (unitary version):
[tex]$F(s) = \int\limits_0^\infty {f(t) e^{ - st} dt}$[/tex];
Let's demonstrate why [tex]${L}\{t^{-3/2}}\}$[/tex] does not exists:
[tex]$F(s) = \int\limits_0^\infty {t^{-3/2} e^{ - st} dt}$[/tex]
Writing st=u
⇒[tex]$F(s) = \int\limits_0^\infty {(u/s)^{-3/2} e^{ - u} {du/s}} = \int\limits_0^\infty {(s/u)^{3/2} e^{ - u} {du/s}} = s^{1/2} \int\limits_0^\infty {(1/u)^{3/2} e^{ - u} {du}} =$[/tex]
⇒
Where:
[tex]$\int\limits_0^\infty {(1/u)^{3/2} e^{ - u} {du}$ [/tex]
can be written as:
[tex]$\int\limits_0^\infty {u^{(-1/2)-1} e^{ - u} {du} = \Gamma (-1/2)$[/tex]
Where:
[tex]$ \Gamma (z) = \int\limits_0^\infty {u^{z - 1} e^{ - u} du} $[/tex]
Is the Gamma Function
But:
[tex]$ \Gamma \left( z \right) $[/tex] ; is only defined for z>0
Getting Demonstrated that [tex]${L}\{t^{-3/2}}\}$[/tex] does not exists
