I'm asked to prove the following result:
If $F(s)$ is the Laplace transform of $f(t) = t^{-3/2} \mathrm{e}^{-1/t}$, show that $F'(s)=-s^{-1/2}F(s)$.
I'm having a lot of troubles to prove this result without relying on computing the Laplace transform directly. As a hint, I've been suggested to compute an equation involving $F$, $F'$ and $F''$ by succesive integration by parts but got me nowhere (or to wrong results). At some point, I end up with a 2nd order ODE with non-constant coefficients which seems rather difficult.
Could somebody illustrate me about the best choice to integrate by parts or tell if there's an easier way to prove the result above?
Thanks in advance.
$$\begin{align} F(s) &= \int_0^{\infty} dt \, t^{-3/2} e^{-(s t+1/t)} \\ &= \int_0^{\infty} dt \, t^{-3/2} e^{-s^{1/2}[s^{1/2} t+1/(s^{1/2} t)]} \\ &= s^{1/4} \int_0^{\infty} du \, u^{-3/2} e^{-s^{1/2}(u+1/u)}\\ &= s^{1/4} \int_0^{\infty} dv \, v^{-1/2} e^{-s^{1/2}(v+1/v)} \\ &= s^{1/2} \int_0^{\infty} dt \, t^{-1/2} e^{-(s t+1/t)}\end{align}$$
In the second line, I subbed $t = s^{-1/2} u$. In the third line, I subbed $u = 1/v$. In the fourth line I subbed $v = s^{1/2} t$. Now we take the derivative of the original integral wrt $s$:
$$F'(s) = - \int_0^{\infty} dt \, t^{-1/2} e^{-1/t} e^{-s t} $$
Thus,
$$F'(s) = -s^{-1/2} F(s) $$