The Laplace transform of $t^{-3/2}e^{-1/t}$ is denoted $F(s)$. Show that $\frac{dF}{ds} = -\frac{F}{s^{1/2}}$.

Question

The Laplace transform of $t^{-3/2}e^{-1/t}$ is denoted $F(s)$. Show that $$\frac{dF}{ds} = -\frac{F}{s^{1/2}}$$

not sure how to get this.

I tried differentiating $F(s) = \int_{0}^{\infty}t^{-3/2}e^{-1/t}e^{-st}dt$ $$ \frac{dF}{ds} = \int_{0}^{\infty}-t^{-1/2}e^{-1/t}e^{-st}dt$$ Maybe better using table of laplace transform but not sure.

Answer 1

Let $ust=1$ or $st=\dfrac{1}{u}$ then $$\frac{dF}{ds} = \int_{0}^{\infty}-t^{-1/2}e^{-1/t}e^{-st}dt=\dfrac{-1}{\sqrt{s}}\int_{0}^{\infty}u^{-3/2}e^{-1/u}e^{-su}\ du=\dfrac{-1}{\sqrt{s}}F$$