91Ó°ÊÓ

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 ${∫}_0^{∞}{\mathrm{x}}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}+1}\left(\mathrm{x}\right)\mathrm{dx}=\frac{1}{2^{\mathrm{p}}\mathrm{Γ}(1+\mathrm{p})}$.

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:

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

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

For large x, as follows:

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

Therefore, obtain:

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

For small x, as follows:

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

Therefore, obtain:

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

Use (2) and (3) as follows:

$$\begin{gathered} {∫}_0^{∞}-x^{-p}{J}_{p+1}\left(x\right)=\underset{x→∞}{lim}\left[-x^{-p}{J}_p\left(x\right)\right]-\underset{x→∞}{lim}\left[-x^{-p}{J}_p\left(x\right)\right] \\ {∫}_0^{∞}-x^{-p}{J}_{p+1}\left(x\right)=0-\left[-\frac{1}{2^p\mathrm{Γ}(p+1)}\right] \\ {∫}_0^{∞}-x^{-p}{J}_{p+1}\left(x\right)=\frac{1}{2^p\mathrm{Γ}(p+1)} \\ {∫}_0^{∞}-x^{-p}{J}_{p+1}\left(x\right)=\frac{1}{2^p\mathrm{Γ}(1+p)} \end{gathered}$$

Thus, it’s proved that${∫}_0^{∞}-x^{-p}{J}_{p+1}\left(x\right)=\frac{1}{2^p\mathrm{Γ}(1+p)}$