It may be shown that the Laplace transform $F(s) = \int _0^\infty e^{-st}f(t)\,dt$ is always differentiable and that its derivative can be formed by differentiating under the integral sign, i.e. one has $$F'(s) = \int_0^\infty\frac{\partial(e^{-st}f(t))}{\partial s}\,dt = \int_0^\infty e^{-st}(-t)f(t)\,dt.$$ This gives the rule

(1)

(2)

(3)

Example. Let's find the Laplace transform of the first kind and 0th order Bessel function $$ J_{0}(t) := \sum_{m=0}^\infty \frac{(-1)^m}{(m!)^2}\left(\frac{t}{2}\right)^{2m}, $$ which is the solution $y(t)$ of the Bessel's equation

(4)

By (3), the Laplace transform of the differential equation (4) is $$-\frac{d\mathcal{L}\{y''(t)\}}{ds}+\mathcal{L}\{y'(t)\} -\frac{d\mathcal{L}\{y(t)\}}{ds} = 0.$$ Using here twice the rule 5 in the parent entry gives us $$-\frac{d(s^2Y(s)-s)}{ds}+sY(s)-1-\frac{dY(s)}{ds} = 0,$$ which is simplified to $$(s^2+1)\frac{dY}{ds}+sY = 0,$$ i.e. to $$\frac{dY}{Y} = -\frac{s\,ds}{s^2+1}.$$ 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 $$\mathcal{L}\{J_0(t)\} = \frac{1}{\sqrt{s^2+1}}.$$

Bibliography

1

K. V¨AISÄLÄ: Laplace-muunnos. Handout Nr. 163.    Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1968).