Question

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

${\mathbf{J}}_{\mathbf{0}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{J}}_{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{J}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Short answer

The obtained integral is ${\mathrm{J}}_0^{\text{'}}\left(\mathrm{x}\right)=-{\mathrm{J}}_1\left(\mathrm{x}\right)={\mathrm{J}}_{-1}\left(\mathrm{x}\right)$.

Step-by-step solution

Define differential recurrence relation

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

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

$\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^{\mathrm{v}}{\mathrm{J}}_{\mathrm{v}}\left(\mathrm{x}\right)\right]={\mathrm{x}}^{\mathrm{v}}{\mathrm{J}}_{\mathrm{v}-1}\left(\mathrm{x}\right)$… (2)

Obtain the integration

Substitute the value$\mathrm{v}=0$in the equation (1).

$\begin{array}{c}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^0{\mathrm{J}}_0\left(\mathrm{x}\right)\right]=-{\mathrm{x}}^0{\mathrm{J}}_{0+1}\left(\mathrm{x}\right)\\ \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{J}}_0\left(\mathrm{x}\right)\right]={\mathrm{J}}_1\left(\mathrm{x}\right)\end{array}$

${\mathrm{J}}_0^{\text{'}}\left(\mathrm{x}\right)=-{\mathrm{J}}_1\left(\mathrm{x}\right)$… (3)

Substitute the value$\mathrm{v}=0$in the equation (2).

$\begin{array}{c}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^0{\mathrm{J}}_0\left(\mathrm{x}\right)\right]={\mathrm{x}}^0{\mathrm{J}}_{0-1}\left(\mathrm{x}\right)\\ \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{J}}_0\left(\mathrm{x}\right)\right]={\mathrm{J}}_{-1}\left(\mathrm{x}\right)\end{array}$

${\mathrm{J}}_0^{\text{'}}\left(\mathrm{x}\right)={\mathrm{J}}_{-1}\left(\mathrm{x}\right)$… (4)

From (3) and (4) yields the result,

${\mathrm{J}}_0^{\text{'}}\left(\mathrm{x}\right)=-{\mathrm{J}}_1\left(\mathrm{x}\right)={\mathrm{J}}_{-1}\left(\mathrm{x}\right)$