Question

In Problems 29 and 30 use (22) or (23) to obtain the given result.

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}{\mathbf{rJ}}_{\mathbf{0}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{dr}\mathbf{=}\mathbf{x}\mathbf{\times }{\mathbf{J}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Short answer

The obtained integral is ${\int }_0^{\mathrm{x}}{\mathrm{rJ}}_0\left(\mathrm{r}\right)\mathrm{dr}={\mathrm{xJ}}_1\left(\mathrm{x}\right)$.

Step-by-step solution

Define differential recurrence relation.

Recurrence formulas that relate Bessel functions of different orders are important in theory and in applications.

$\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}{\mathbf{x}}^{\mathbf{v}}{\mathbf{J}}_{\mathbf{v}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\mathbf{=}{\mathbf{x}}^{\mathbf{v}}{\mathbf{J}}_{\mathbf{v}\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$… (1)

Obtain the integration

Substitute the value$\mathrm{v}=1$in the equation (1).

${\mathrm{xJ}}_0\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{xJ}}_1\left(\mathrm{x}\right)\right]$

Integrate both sides of the expression.

$\begin{array}{c}{\int }_0^{\mathrm{x}}{\mathrm{rJ}}_0\left(\mathrm{r}\right)\mathrm{dr}={\int }_0^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{xJ}}_1\left(\mathrm{x}\right)\right]\\ {\int }_0^{\mathrm{x}}{\mathrm{rJ}}_0\left(\mathrm{r}\right)\mathrm{dr}={\mathrm{xJ}}_1\left(\mathrm{x}\right)\end{array}$