Question

Use L23of the Laplace Transform Table (page 469 ) to evaluate ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{tJ}}_{\mathbf{0}}\mathbf{\left(}\mathbf{2}\mathbf{t}\mathbf{\right)}\mathbf{e}\widehat{\mathbf{}}\mathbf{-}\mathbf{tdt}$.

Short answer

The equation by use of Laplace theorem is ${\int }_0^{\infty }\left[\left({\mathrm{tJ}}_0\right)\left(2\mathrm{t}\right)\right]{\mathrm{e}}^{-1}\mathrm{dt}=\sqrt{5}$.

Step-by-step solution

Concept of Laplace Transform

The Laplace transform is an integral transform in mathematics that turns a function of a real variable (typically time) to a function of a complex variable. It is named after its inventor Pierre-Simon Laplace. (Complex periodicity)

Determine equations with proof:

The given equation is as follow.

${\int }_0^{\infty }\left[{\mathrm{tJ}}_0\left(2\mathrm{t}\right)\right]{\mathrm{e}}^{-1}\mathrm{dt}$ ….. (1)

Let $\mathrm{g}\left(\mathrm{t}\right)=\left({\mathrm{tJ}}_0\right)-\left(2\mathrm{t}\right)$

Therefore,

${\int }_0^{\infty }\left[{\mathrm{tJ}}_0\left(2\mathrm{t}\right)\right]{\mathrm{e}}^{-1}\mathrm{dt}=\mathrm{L}\left(\mathrm{g}\right(\mathrm{t}\left)\right)$

Thus,

$$\begin{gathered} \mathrm{L}\left(\mathrm{g}\right(\mathrm{t}\left)\right)=\mathrm{L}\left[\left({\mathrm{tJ}}_0\right)\left(2\mathrm{t}\right)\right] \\ =-\frac{\mathrm{d}}{\mathrm{dp}}\mathrm{L}\left[\frac{1}{\sqrt{{\mathrm{P}}^2+4}}\right] \\ =\mathrm{p}\sqrt{{\mathrm{P}}^2+4} \end{gathered}$$

At p = 1

$$\begin{gathered} \mathrm{L}\left(\mathrm{g}\right(\mathrm{t}\left)\right)=1\times \sqrt{{\left(1\right)}^2+4} \\ =\sqrt{5} \end{gathered}$$

Hence, the solution is role="math" localid="1659263208798" ${\int }_0^{\infty }\left[\left({\mathrm{tJ}}_0\right)\left(2\mathrm{t}\right)\right]{\mathrm{e}}^{-1}\mathrm{dt}=\sqrt{5}$.