Question

(a) Using 15.2 , show that ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{J}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx}\mathbf{=}\mathbf{1}$. (b) Use L23of the Laplace Transform Table (Page 469) to show that ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{J}}_0\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dt}\mathbf{=}\mathbf{1}$. (Also see Problem23.29.) $\frac{\mathbf{d}}{\mathbf{dx}}\left[x^{-p}{J}_p\left(x\right)\right]\mathbf{=}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{p}}{\mathbf{J}}_{\mathbf{p}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

Short answer

(a) The required equation is ${\int }_0^{\infty }{\mathrm{J}}_1\left(\mathrm{x}\right)\mathrm{dx}=1$.

(b) Laplace transform equation is ${\int }_0^{\infty }{\mathrm{J}}_0\left(\mathrm{t}\right)\mathrm{dt}=1$.

Step-by-step solution

Concept of Differential equations and Laplace transform:

Differential equations are equations that connect one or more derivatives of a function. This implies that their answer is a function.

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).

(a) Determine equations and prove:

The given equation is as follow.

$\frac{\mathrm{d}}{\mathrm{dx}}\left[{\left(\mathrm{x}\right)}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)\right]=-{\left(\mathrm{x}\right)}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}+1}\left(\mathrm{x}\right)$ ….. (1)

Let, p = 0.

Then,

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left[{\left(\mathrm{x}\right)}^{\mathrm{o}}{\mathrm{J}}_{\mathrm{o}}\left(\mathrm{x}\right)\right]=-{\left(\mathrm{x}\right)}^{\mathrm{o}}{\mathrm{J}}_{\mathrm{o}+1}\left(\mathrm{x}\right) \\ \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{J}}_{\mathrm{o}}\left(\mathrm{x}\right)\right]=-{\mathrm{J}}_1\left(\mathrm{x}\right) \end{gathered}$$

Integrate from 0 to $\infty$as follow.

$$\begin{gathered} {\int }_0^{\mathrm{x}}\left[\frac{\mathrm{d}\left({\mathrm{J}}_0\right(\mathrm{x}\left)\right)}{\mathrm{dx}}\right]\mathrm{dx}=-{\int }_0^{\mathrm{x}}{\mathrm{J}}_1\left(\mathrm{x}\right)\mathrm{dx} \\ -{\left[{\mathrm{J}}_0\left(\mathrm{x}\right)\right]}^{2}0=-{\int }_0^{\mathrm{x}}{\mathrm{J}}_1\left(\mathrm{x}\right)\mathrm{dx} \\ {\int }_0^{\mathrm{x}}{\mathrm{J}}_1\left(\mathrm{x}\right)\mathrm{dx}=1 \end{gathered}$$

Therefore,

$\left(\mathrm{x}\to \infty ,{\mathrm{J}}_0\left(\mathrm{x}\right)\to 0,{\mathrm{J}}_0\left(0\right)=1\right)$

(b) Determining equations with the help of Laplace transform:

The given equation is,

$\frac{\mathrm{d}}{\mathrm{dx}}\left[{\left(\mathrm{x}\right)}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)\right]=-{\left(\mathrm{x}\right)}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}+1}\left(\mathrm{x}\right)$

Use Laplace transform to calculate the equation as follows:

$$\begin{gathered} \mathrm{L}\left({\mathrm{J}}_0\right(\mathrm{at}\left)\right)=\sqrt{{\mathrm{p}}^2+{\mathrm{a}}^2} \\ ={\int }_0{\left(\mathrm{e}\right)}^{-{\mathrm{\mu J}}_0\left(\mathrm{at}\right)\mathrm{dt}=\sqrt{{\mathrm{p}}^2+{\mathrm{a}}^2}} \end{gathered}$$

Therefore,

${\int }_0^{\overline{0}}{\mathrm{J}}_0\left(\mathrm{x}\right)\mathrm{dt}=1$