Verify the formula.

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathit{x}}^{\mathbf{-}\mathbf{p}}{\mathit{J}}_{\mathbf{p}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{2}}^{\mathbf{p}}\mathbf{\Gamma }\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{p}\mathbf{)}}$

A recursion relation is an equation that defines a sequence based on a rule that gives the next term as a function of the previous term(s).

The simplest form of a recursion relation is the case where the next term depends only on the immediately previous term.

As the given formula is ${\int }_0^{\infty }{\mathrm{x}}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}+1}\left(\mathrm{x}\right)\mathrm{dx}=\frac{1}{2^{\mathrm{p}}\mathrm{\Gamma }(1+\mathrm{p})}$.

Use the recursion relation as follows:

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

Integrate between the limits 0 and infinity as follows:

$$\begin{gathered} {\int }_0^{\infty }-x^{-p}{J}_{p+1}\left(x\right)={\int }_0^{\infty }-\frac{d}{dx}\left[x^{-p}{J}_p\left(x\right)\right] \\ {\int }_0^{\infty }-x^{-p}{J}_{p+1}\left(x\right)={\left[-x^{-p}{J}_p\left(x\right)\right]}_0^{\infty } \end{gathered}$$

${\int }_0^{\infty }-x^{-p}{J}_{p+1}\left(x\right)=\underset{x\to \infty }{lim}\left[-x^{-p}{J}_p\left(x\right)\right]-\underset{x\to \infty }{lim}\left[-x^{-p}{J}_p\left(x\right)\right]$.....(1)

For large x, as follows:

role="math" localid="1664366326069" $$\begin{gathered} J_p\left(x\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\cos (x-\frac{2p+1}{4}\mathrm{\pi })+\mathrm{O}\left({\mathrm{x}}^{-\frac{3}{2}}\right) \\ \underset{\mathrm{x}\to \infty }{\lim }{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to \infty }{\lim }\sqrt{\frac{2}{\mathrm{\pi x}}}\cos (\mathrm{x}-\frac{2\mathrm{p}+1}{4}\mathrm{\pi })+\mathrm{O}\left({\mathrm{x}}^{-\frac{3}{2}}\right) \\ \underset{\mathrm{x}\to \infty }{\lim }{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)=0 \end{gathered}$$

Therefore, obtain:

$$\begin{gathered} \underset{\mathrm{x}\to \infty }{\lim }\left[x^{-p}{J}_p\left(x\right)\right]=\underset{\mathrm{x}\to \infty }{\lim }\left({\mathrm{x}}^{-\mathrm{p}}\right)\underset{\mathrm{x}\to \infty }{\lim }{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right) \\ \underset{\mathrm{x}\to \infty }{\lim }\left[{\mathrm{x}}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)\right]=0 \end{gathered}$$......(2)

For small x, as follows:

$$\begin{gathered} {\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{1}{\mathrm{l}(\mathrm{p}+1)}{\left(\frac{\mathrm{x}}{2}\right)}^{\mathrm{p}}+\mathrm{O}\left({\mathrm{x}}^{\mathrm{p}+2}\right) \\ {\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{\mathrm{p}!}{2^{\mathrm{p}}\mathrm{\Gamma }(\mathrm{p}+1)} \end{gathered}$$

Therefore, obtain:

$$\begin{gathered} \underset{x\to 0}{lim}\left[-x^{-p}{J}_p\left(x\right)\right]=\underset{x\to 0}{lim}\left[-\frac{J_p\left(x\right)}{x^p}\right] \\ \underset{x\to 0}{lim}\left[-x^{-p}{J}_p\left(x\right)\right]=\frac{p!}{2^p\Gamma (p+1)p!} \\ \underset{x\to 0}{lim}\left[-x^{-p}{J}_p\left(x\right)\right]=\frac{1}{2^p\Gamma (p+1)}.......\left(3\right) \end{gathered}$$

Use (2) and (3) as follows:

$$\begin{gathered} {\int }_0^{\infty }-x^{-p}{J}_{p+1}\left(x\right)=\underset{x\to \infty }{lim}\left[-x^{-p}{J}_p\left(x\right)\right]-\underset{x\to \infty }{lim}\left[-x^{-p}{J}_p\left(x\right)\right] \\ {\int }_0^{\infty }-x^{-p}{J}_{p+1}\left(x\right)=0-\left[-\frac{1}{2^p\Gamma (p+1)}\right] \\ {\int }_0^{\infty }-x^{-p}{J}_{p+1}\left(x\right)=\frac{1}{2^p\Gamma (p+1)} \\ {\int }_0^{\infty }-x^{-p}{J}_{p+1}\left(x\right)=\frac{1}{2^p\Gamma (1+p)} \end{gathered}$$

Thus, it’s proved that${\int }_0^{\infty }-x^{-p}{J}_{p+1}\left(x\right)=\frac{1}{2^p\Gamma (1+p)}$