inverse Laplace transform of derivatives

It may be shown that the Laplace transform  $F(s)={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)𝑑t$  is always differentiable and that its derivative can be formed by differentiating under the integral sign (http://planetmath.org/DifferentiationUnderIntegralSign), i.e. one has

$$F^{\prime }(s)={\int }_0^{\mathrm{\infty }}\frac{\partial (e^{-st}f(t))}{\partial s}𝑑t={\int }_0^{\mathrm{\infty }}{e}^{-st}(-t)f(t)𝑑t.$$

This gives the rule

${ℒ}^{-1}\{F^{\prime }(s)\}=-tf(t).$(1)

Applying (1) to $F^{\prime }(s)$ instead of $F(s)$ gives

$${ℒ}^{-1}\{F^{\prime \prime }(s)\}=t^{2}f(t).$$

Continuing this way we can obtain the general rule

${ℒ}^{-1}\{F^{(n)}(s)\}={(-1)}^n{t}^{n}f(t),$(2)

or equivalently

$ℒ\{t^{n}f(t)\}={(-1)}^n\cdot {\displaystyle \frac{d^nℒ\{f(t)\}}{d{s}^n}},$(3)

for any  $n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots }$ (and of course for  $n=0$).

Example.  Let’s find the Laplace transform of the first kind and 0th Bessel function

$$J_0(t):=\sum _{m=0}^{\mathrm{\infty }}\frac{{(-1)}^m}{{(m!)}^2}{\left(\frac{t}{2}\right)}^{2m},$$

which is the solution $y(t)$ of the Bessel’s equation

$t{y}^{\prime \prime }(t)+y^{\prime }(t)+ty(t)=0$(4)

satisfying the initial condition  $y(0)=1$.  The equation implies that  $y^{\prime }(0)=0$.

By (3), the Laplace transform of the differential equation (4) is

$$-\frac{dℒ\{y^{\prime \prime }(t)\}}{ds}+ℒ\{y^{\prime }(t)\}-\frac{dℒ\{y(t)\}}{ds}=0.$$

Using here twice the rule 5 in the parent (http://planetmath.org/LaplaceTransform) entry gives us

$$-\frac{d(s^{2}Y(s)-s)}{ds}+sY(s)-1-\frac{dY(s)}{ds}=0,$$

which is simplified to

$$(s^2+1)\frac{dY}{ds}+sY=0,$$

i.e. to

$$\frac{dY}{Y}=-\frac{sds}{s^2+1}.$$

Integrating this gives

$$\ln Y=-\frac{1}{2}\ln (s^2+1)+\ln C=\ln \frac{C}{\sqrt{s^2+1}},$$

i.e.

$$Y(s)=\frac{C}{\sqrt{s^2+1}}.$$

The initial condition enables to justify that the integration constant $C$ must be 1.  Thus we have the result

$$ℒ\{J_0(t)\}=\frac{1}{\sqrt{s^2+1}}.$$