Question

How to get the laplace transform of this equation?

f(t)=t^-3/2

Answer 1

`${L}\{t^(-3/2)}\}$` does not exists

Explanation:

First lets introduce the definition of the Laplace transform (unitary version):

`$F(s) = \int\limits_0^\infty {f(t) e^( - st) dt}$`;

Let's demonstrate why
`${L}\{t^(-3/2)}\}$` does not exists:

`$F(s) = \int\limits_0^\infty {t^(-3/2) e^( - st) dt}$`

Writing st=u

⇒
`$F(s) = \int\limits_0^\infty {(u/s)^(-3/2) e^( - u) {du/s}} = \int\limits_0^\infty {(s/u)^(3/2) e^( - u) {du/s}} = s^(1/2) \int\limits_0^\infty {(1/u)^(3/2) e^( - u) {du}} =$`

⇒

Where:

`$\int\limits_0^\infty {(1/u)^(3/2) e^( - u) {du}$`

can be written as:

`$\int\limits_0^\infty {u^((-1/2)-1) e^( - u) {du} = \Gamma (-1/2)$`

Where:

`$ \Gamma (z) = \int\limits_0^\infty {u^(z - 1) e^( - u) du} $`

Is the Gamma Function

But:

`$ \Gamma \left( z \right) $` ; is only defined for z>0

Getting Demonstrated that
`${L}\{t^(-3/2)}\}$` does not exists