In Watson's text(2nd Edition), Section 11.41 p.364,
(1) I would like to know how the following relation is derived \begin{equation} \begin{split} &\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}(-1)^p\frac{z^{p+2q}\cos^p\phi}{2^q p!q!}\frac{J_{-\nu-p-q}[Z]}{Z^{\nu+q}} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}\sum ^{q}_{k=0}\frac{(-1)^{p+1}(\nu+p+2k)\Gamma(-\nu-p-q-k)z^{p+2q}\cos^p\phi} {2^{2q} \Gamma(p+1)\Gamma(k+1)\Gamma(q-k+1)\Gamma(1-\nu-p-k)}\frac{J_{-\nu-p-2k}[Z]}{Z^{\nu}} \notag \end{split} \end{equation}
(2)The next line in the text can be derived by simply relacing $q$ with $n+k$ in the above formula. However, I do not understand how the derivation to the line one after that is accomplished, i.e. \begin{equation} \begin{split} &\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\sum ^{\infty}_{n=0}\frac{(-1)^{p+1}(\nu+p+2k)\Gamma(-\nu-p-n-2k)z^{p+2k+2n}\cos^p\phi} {2^{2k+2n} \Gamma(p+1)\Gamma(k+1)\Gamma(n+1)\Gamma(1-\nu-p-k)}\frac{J_{-\nu-p-2k}[Z]}{Z^{\nu}} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}\frac{(-1)^{p+k}2^{\nu+p}(\nu+p+2k)\Gamma(\nu+p+k)\cos^p\phi} { \Gamma(p+1)\Gamma(k+1)}\frac{J_{\nu+p+2k}[z]}{z^{\nu}}\frac{J_{-\nu-p-2k}[Z]}{Z^{\nu}} \notag \end{split} \end{equation}
In the following, I will briefly explain the formulae introduced in Section 11.4 of the text, in order to elaborate the difference from those mentioned in the Questions. Replacing $\zeta$ and $h$ by $Z^2+z^2$ and $-2 Zz \cos\phi$ respectively in the follwing Lommel's expansion formula yields,
\begin{equation} \begin{split} \Omega=\frac{J_{\nu}[\sqrt{\zeta+h}]}{(\zeta+h)^{\nu/2}}=\sum ^{\infty}_{p=0}\frac{(-h/2)^p}{p!}\frac{J_{\nu+p}[\sqrt{\zeta}]}{\zeta^{(\nu+p)/2}} =\sum ^{\infty}_{p=0}\frac{(Zz\cos\phi)^p}{p!}\frac{J_{\nu+p}[\sqrt{Z^2+z^2}]}{(Z^2+z^2)^{(\nu+p)/2}} \\ (1) \end{split} \end{equation} Applying the above method again to the last term of Eq.1, one obtains, \begin{equation} \begin{split} \frac{J_{\nu+p}[\sqrt{Z^2+z^2}]}{(Z^2+z^2)^{(\nu+p)/2}} =\sum ^{\infty}_{q=0}\frac{(-z^2/2)^q}{q!}\frac{J_{\nu+p+q}[\sqrt{Z^2}]}{(Z^2)^{(\nu+p+q)/2}} =\sum ^{\infty}_{q=0}(-1)^q\frac{a^{2q}}{2^qq!}\frac{J_{\nu+p+q}[Z]}{Z^{\nu+p+q}}\\(2) \end{split} \end{equation} Inserting Eq.2 into Eq.1 gives, \begin{equation} \begin{split} \Omega=\sum ^{\infty}_{p=0}\frac{(Zz\cos\phi)^p}{p!}\sum ^{\infty}_{q=0}(-1)^q\frac{z^{2q}}{2^qq!}\frac{J_{\nu+p+q}[Z]}{Z^{\nu+p+q}} =\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}(-1)^q\frac{z^{p+2q}\cos^p\phi}{2^q p!q!}\frac{J_{\nu+p+q}[Z]}{Z^{\nu}Z^q}\\(3) \end{split} \end{equation} On the other hand, the formula shown in Section 5.21 of the Watson's text stipulates, \begin{equation} \begin{split} \frac{J_{\mu+m}[Z]}{Z^m}=\sum ^{m}_{n=0}\frac{\Gamma(m+1)}{2^m \Gamma(n+1)\Gamma(m-n+1)}\frac{(\mu+2n)\Gamma(\mu+n)}{\Gamma(\mu+m+n+1)}J_{\mu+2 n}[Z]\\(4) \end{split} \end{equation} Replacing $\mu$ with $\nu+p$, $m$ with $q$ and $n$ with $k$ in Eq.4 yields, \begin{equation} \begin{split} \frac{J_{\nu+p+q}[Z]}{Z^q}=\sum ^{q}_{k=0}\frac{\Gamma(q+1)}{2^q \Gamma(n+1)\Gamma(q-k+1)}\frac{(\nu+p+2k)\Gamma(\nu+p+k)}{\Gamma(\nu+p+q+k+1)}J_{\nu+p+2 k}[Z]\\(5) \end{split} \end{equation} Inserting Eq.5 to Eq.3 gives, \begin{equation} \begin{split} \Omega &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}(-1)^q\frac{z^{p+2q}\cos^p\phi}{2^q p!q!}\frac{J_{\nu+p+q}[Z]}{Z^{\nu}Z^q} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}(-1)^q\frac{z^{p+2q}\cos^p\phi}{2^q p!q!}\sum ^{q}_{k=0} \frac{\Gamma(q+1)}{2^q \Gamma(k+1)\Gamma(q-k+1)}\frac{(\nu+p+2k)\Gamma(\nu+p+k)}{\Gamma(\nu+p+q+k+1)}\frac{J_{\nu+p+2 k}[Z]}{Z^\nu} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}\sum ^{q}_{k=0}(-1)^q\frac{z^{p+2q}\cos^p\phi}{2^{2q} \Gamma(p+1)} \frac{(\nu+p+2k)\Gamma(\nu+p+k)}{ \Gamma(k+1)\Gamma(q-k+1)\Gamma(\nu+p+q+k+1)}\frac{J_{\nu+p+2 k}[Z]}{Z^\nu} \\(6) \end{split} \end{equation} Replacing $q$ by $k+n$, \begin{equation} \begin{split} &\Omega =\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\sum ^{\infty}_{n=0}(-1)^{k+n}\frac{z^{p+2k+2n}\cos^p\phi}{2^{2k+2n} \Gamma(p+1)} \frac{(\nu+p+2k)\Gamma(\nu+p+k)}{ \Gamma(k+1)\Gamma(n+1)\Gamma(\nu+p+2k+n+1)}\frac{J_{\nu+p+2 k}[Z]}{Z^\nu} =\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\sum ^{\infty}_{n=0}\frac{(-1)^k(\nu+p+2k)\Gamma(\nu+p+k)\cos^p\phi}{ \Gamma(k+1)\Gamma(p+1)} \frac{(-1)^n z^{p+2k+2n}}{2^{2k+2n}\Gamma(n+1)\Gamma(\nu+p+2k+n+1)}\frac{J_{\nu+p+2 k}[Z]}{Z^\nu} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\frac{(-1)^k 2^{\nu+p}(\nu+p+2k)\Gamma(\nu+p+k)\cos^p\phi}{ \Gamma(k+1)\Gamma(p+1)} \sum ^{\infty}_{n=0}\frac{(-1)^n z^{\nu+p+2k+2n}}{2^{\nu+p+2k+2n}\Gamma(n+1)\Gamma(\nu+p+2k+n+1)}\frac{1}{z^\nu}\frac{J_{\nu+p+2 k}[Z]}{Z^\nu} \\(7) \end{split} \end{equation} Now since from the definition \begin{equation} \begin{split} J_{\nu}[z]=\sum ^{\infty}_{n=0}(-1)^n\frac{z^{\nu+2n}}{2^{\nu+2n}}\frac{1}{\Gamma(n+1)\Gamma(\nu+n+1)}\\(8) \end{split} \end{equation} the second sum in Eq.7 becomes \begin{equation} \begin{split} \sum ^{\infty}_{n=0}\frac{(-1)^n z^{\nu+p+2k+2n}}{2^{\nu+p+2k+2n}\Gamma(n+1)\Gamma(\nu+p+2k+n+1)}=J_{\nu+p+2k}[z]\\(9) \end{split} \end{equation} Therefore, \begin{equation} \begin{split} &\Omega =\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\frac{(-1)^k 2^{\nu+p}(\nu+p+2k)\Gamma(\nu+p+k)\cos^p\phi}{ \Gamma(k+1)\Gamma(p+1)} \frac{J_{\nu+p+2k}[z]}{z^\nu}\frac{J_{\nu+p+2 k}[Z]}{Z^\nu} \\(10) \end{split} \end{equation} The rest of Section 11.4 can be deduced by just following the instructions written in the text. In the case of Section 11.41 however, where the modified form of Gegenbauer's addition theorem is discussed, it seems that the result cannot be straightforwardly obtained as in the above case. Section 5.22 of the Watson's text tells us that, \begin{equation} \begin{split} (z+h)^{\nu/2}J_{\nu}[\sqrt{z+h}]=\sum ^{\infty}_{p=0}\frac{(h/2)^p}{p!}z^{(\nu-p)/2}J_{\nu-p}[\sqrt{z}]\\(11) \end{split} \end{equation} If we transform $\nu\rightarrow -\nu$, and $z\rightarrow \zeta$, then equation above becomes, \begin{equation} \begin{split} \frac{J_{-\nu}[\sqrt{\zeta+h}]}{(\zeta+h)^{\nu/2}}=\sum ^{\infty}_{p=0}\frac{(h/2)^p}{p!}\zeta^{-(\nu+p)/2}J_{-\nu-p}[\sqrt{\zeta}]\\(12) \end{split} \end{equation} Now by replacing $\zeta$ and $h$ by $Z^2+z^2$ and $-2 Zz \cos\phi$ respectively, we obtain \begin{equation} \begin{split} \frac{J_{-\nu}[\sqrt{\zeta+h}]}{(\zeta+h)^{\nu/2}}=\sum ^{\infty}_{p=0}\frac{(-Zz\cos\phi)^p}{p!}\frac{J_{-\nu-p}[\sqrt{Z^2+z^2}]}{(Z^2+z^2)^{(\nu+p)/2}}\\(13) \end{split} \end{equation} By using the relation of Eq.12, \begin{equation} \begin{split} \frac{J_{-\nu-p}[\sqrt{Z^2+z^2}]}{(Z^2+z^2)^{(\nu+p)/2}}=\sum ^{\infty}_{q=0}\frac{z^{2q}}{2^qq!}\frac{J_{-\nu-p-q}[Z]}{Z^{\nu+p+q}}\\(14) \end{split} \end{equation} Then Eq.13 is transformed as \begin{equation} \begin{split} \frac{J_{-\nu}[\sqrt{\zeta+h}]}{(\zeta+h)^{\nu/2}}=\sum ^{\infty}_{p=0}\frac{(-Zz\cos\phi)^p}{p!}\frac{J_{-\nu-p}[\sqrt{Z^2+z^2}]}{(Z^2+z^2)^{(\nu+p)/2}} =\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}(-1)^p\frac{z^{p+2q}\cos^p\phi}{2^q p!q!}\frac{J_{-\nu-p-q}[Z]}{Z^{\nu+q}}\\(15) \end{split} \end{equation} However, further derivations cannot be achieved by simply replacing the result obtained for $\nu$, $p$, $q$ in Eq.5 through Eq.10 above with $-\nu$, $-p$, $-q$ respectively, which leads to the Questions(1) and (2) I have raised in the beginning of this questionnaire.
I think I may have solved the issue myself. Followings are the result.
Title:Commentary on Subsection 11.41 of Watson's text;"The Theory of Bessel Functions"
In Subsection 11.41, the formula derivation of \begin{equation} \Omega=\frac{J_{-\nu}[w]}{w^{\nu}} \tag{1}\label{eq1} \end{equation} where $$w=\sqrt{Z^2+z^2-2Zz\cos\phi} $$ is discussed. Subsection 5.22 of the text tells us that, \begin{equation} (z+h)^{\nu/2}J_{\nu}[\sqrt{z+h}]=\sum ^{\infty}_{p=0}\frac{(h/2)^p}{p!}z^{(\nu-p)/2}J_{\nu-p}[\sqrt{z}] \tag{2}\label{eq2} \end{equation} If we transform $\nu\rightarrow -\nu$, $z\rightarrow \zeta$, then equation above becomes, \begin{equation} \frac{J_{-\nu}[\sqrt{\zeta+h}]}{(\zeta+h)^{\nu/2}}=\sum ^{\infty}_{p=0}\frac{(h/2)^p}{p!}\zeta^{-(\nu+p)/2}J_{-\nu-p}[\sqrt{\zeta}] \tag{3}\label{eq3} \end{equation} Now by replacing $\zeta$ and $h$ by $Z^2+z^2$ and $-2 Zz \cos\phi$ respectively, we obtain \begin{equation} \Omega=\frac{J_{-\nu}[\sqrt{\zeta+h}]}{(\zeta+h)^{\nu/2}}=\sum ^{\infty}_{p=0}\frac{(-Zz\cos\phi)^p}{p!}\frac{J_{-\nu-p}[\sqrt{Z^2+z^2}]}{(Z^2+z^2)^{(\nu+p)/2}} \tag{4} \end{equation} By adopting the relation of \eqref{eq3} \begin{equation} \frac{J_{-\nu-p}[\sqrt{Z^2+z^2}]}{(Z^2+z^2)^{(\nu+p)/2}}=\sum ^{\infty}_{q=0}\frac{z^{2q}}{2^qq!}\frac{J_{-\nu-p-q}[Z]}{Z^{\nu+p+q}} \tag{4}\label{eq4} \end{equation} Then \eqref{eq3} is eventually transformed as \begin{equation} \begin{aligned} &\Omega=\frac{J_{-\nu}[\sqrt{\zeta+h}]}{(\zeta+h)^{\nu/2}} \\ &=\sum ^{\infty}_{p=0}\frac{(-Zz\cos\phi)^p}{p!}\frac{J_{-\nu-p}[\sqrt{Z^2+z^2}]}{(Z^2+z^2)^{(\nu+p)/2}} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}(-1)^p\frac{z^{p+2q}\cos^p\phi}{2^q p!q!}\frac{J_{-\nu-p-q}[Z]}{Z^{\nu+q}} \end{aligned} \tag{5}\label{eq5} \end{equation} According to the definition of Bessel function, the ascending series of $J_{-\nu}[z]$ is expressed as, \begin{equation} J_{-\nu}(z)=\sum^{\infty}_{m=0}\frac{(-1)^m}{\Gamma(m+1)\Gamma(-\nu+m+1)}\left(\frac{1}{2}z\right)^{-\nu+2m} \tag{6}\label{eq6} \end{equation} Multiplying $(z/2)^{\mu-\nu}$ to the above equation gives \begin{equation} \left(\frac{1}{2}z\right)^{\mu-\nu}J_{-\nu}(z) =\sum^{\infty}_{m=0}\frac{(-1)^m}{\Gamma(m+1)\Gamma(-\nu+m+1)}\left(\frac{1}{2}z\right)^{\mu-2\nu+2m} \tag{7}\label{eq7} \end{equation} Using the relation introduced in Section 5.2 of the text, \begin{equation} \left(\frac{1}{2}z\right)^{\mu-2\nu+2m}= \sum^{\infty}_{p=0}\frac{(\mu-2\nu+2m+2p)\Gamma(\mu-2\nu+2m+p)}{\Gamma(p+1)}J_{\mu-2\nu+2m+2p}(z) \tag{8}\label{eq8} \end{equation} Inserting \eqref{eq8} to \eqref{eq7}, the relation corresponding to Bessel function having the order of $-\nu$ therefore becomes \begin{equation} \begin{aligned} &\left(\frac{1}{2}z\right)^{\mu-\nu}J_{-\nu}(z) \\ &=\sum^{\infty}_{m=0}\frac{(-1)^m}{\Gamma(m+1)\Gamma(-\nu+m+1)} \times \\ &\sum^{\infty}_{p=0}\frac{(\mu-2\nu+2m+2p)\Gamma(\mu-2\nu+2m+p)}{\Gamma(p+1)}J_{\mu-2\nu+2m+2p}(z) \end{aligned} \tag{9}\label{eq9} \end{equation} Putting $m+p=n$, \begin{equation} \begin{aligned} \left(\frac{1}{2}z\right)^{\mu-\nu}J_{-\nu}(z) &=\sum^{\infty}_{m=0}\frac{(-1)^m}{\Gamma(m+1)\Gamma(-\nu+m+1)} \sum^{\infty}_{n=0}\frac{(\mu-2\nu+2n)\Gamma(\mu-2\nu+m+n)}{\Gamma(p+1)}J_{\mu-2\nu+2n}(z) \\ &=\sum^{\infty}_{n=0} \left[\sum^{\infty}_{m=0}\frac{(-1)^m\Gamma(\mu-2\nu+m+n)}{\Gamma(m+1)\Gamma(-\nu+m+1)\Gamma(n-m+1)} \right](\mu-2\nu+2n)J_{\mu-2\nu+2n}(z) \end{aligned} \tag{10}\label{eq10} \end{equation} Adopting the following summation relation of Gamma function, \begin{equation} \sum^{n}_{m=0}(-1)^m\frac{\Gamma(n+1)}{\Gamma(n-m+1)\Gamma(m+1)}\frac{\Gamma(m+b)}{\Gamma(m+a)} =\frac{\Gamma(n+a-b)\Gamma(b)}{\Gamma(a-b)\Gamma(n+a)} \notag \end{equation} summation of the formula inside the parenthesis of \eqref{eq10}, second line, is derived as \begin{equation} \begin{aligned} &\sum^{\infty}_{m=0}\frac{(-1)^m\Gamma(\mu-2\nu+m+n)}{\Gamma(m+1)\Gamma(-\nu+m+1)\Gamma(n-m+1)} \\ &=\frac{1}{\Gamma(n+1)}\sum^{\infty}_{m=0}\frac{(-1)^m\Gamma(n+1)}{\Gamma(m+1)\Gamma(n-m+1)} \frac{\Gamma(m+\mu-2\nu+n)}{\Gamma(m-\nu+1)} \\ &=\frac{1}{\Gamma(n+1)}\frac{\Gamma(\nu+1-\mu)\Gamma(\mu-2\nu+n)}{\Gamma(\nu+1-\mu-n)\Gamma(n-\nu+1)} \end{aligned} \tag{11}\label{eq11} \end{equation} Inserting \eqref{eq11} into \eqref{eq10} yields \begin{equation} \begin{aligned} &\left(\frac{1}{2}z\right)^{\mu-\nu}J_{-\nu}(z) =\sum^{\infty}_{n=0} \left[ \frac{1}{\Gamma(n+1)}\frac{\Gamma(\nu+1-\mu)\Gamma(\mu-2\nu+n)}{\Gamma(\nu+1-\mu-n)\Gamma(n-\nu+1)} \right]\times \\ &(\mu-2\nu+2n)J_{\mu-2\nu+2n}(z) \end{aligned} \tag{12}\label{eq12} \end{equation} By putting $\nu=\mu+m$, the equation above becomes \begin{equation} \begin{aligned} &\left(\frac{1}{2}z\right)^{-m}J_{-\mu-m}(z) =\sum^{m}_{n=0} \left[ \frac{\Gamma(m+1)}{\Gamma(n+1)\Gamma(m-n+1)}\frac{\Gamma(n-\mu-2m)}{\Gamma(n-\mu-m+1)} \right]\times \\ &(-\mu-2m+2n)J_{-\mu-2m+2n}(z) \end{aligned} \tag{13}\label{eq13} \end{equation} under the condition $n\leq m$. Repalcing $\mu$ with $\nu+p$, $m$ with $q$, and $z$ with $Z$, the equation above is transformed to \begin{equation} \begin{aligned} &\frac{J_{-\nu-p-q}(Z)}{Z^q}=\sum^{q}_{n=0} \left[ \frac{\Gamma(q+1)}{\Gamma(n+1)\Gamma(q-n+1)}\frac{\Gamma(n-\nu-p-2q)}{\Gamma(n-\nu-p-q+1)} \right]\times \\ &(-\nu-p-2q+2n)J_{-\nu-p-2q+2n}(Z) \end{aligned} \tag{14}\label{eq14} \end{equation} Now when $n$ is replaced by $k$, accompanied by applying the relation above to \eqref{eq5}, \begin{equation} \begin{aligned} &\Omega=\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}(-1)^p\frac{z^{p+2q}\cos^p\phi}{2^q p!q!}\frac{J_{-\nu-p-q}[Z]}{Z^{\nu+q}} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}(-1)^p\frac{z^{p+2q}\cos^p\phi}{2^q p!q!}\sum^{q}_{k=0} \left[ \frac{\Gamma(q+1)}{\Gamma(k+1)\Gamma(q-k+1)}\frac{\Gamma(k-\nu-p-2q)}{\Gamma(k-\nu-p-q+1)} \right] \times \\ &(-\nu-p-2q+2k)\frac{J_{-\nu-p-2q+2k}(Z)}{Z^\nu} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{q=0}\sum^{q}_{k=0}\frac{(-1)^{p+1}z^{p+2q}\cos^p\phi}{2^{2q}\Gamma(p+1)\Gamma(k+1)\Gamma(q-k+1)} \frac{(\nu+p+2q-2k)\Gamma(k-\nu-p-2q)}{\Gamma(k-\nu-p-q+1)}\times \\ &\frac{J_{-\nu-p-2q+2k}(Z)}{Z^\nu} \end{aligned} \tag{15}\label{eq15} \end{equation} Replacing $q$ by $n+k$ gives, \begin{equation} \begin{aligned} &\Omega=\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\sum^{\infty}_{n=0}\frac{(-1)^{p+1}z^{p+2n+2k}\cos^p\phi}{2^{2n+2k}\Gamma(p+1)\Gamma(k+1)\Gamma(n+1)} \frac{(\nu+p+2n)\Gamma(-k-\nu-p-2n)}{\Gamma(-\nu-p-n+1)}\times \\ &\frac{J_{-\nu-p-2n}(Z)}{Z^\nu} \end{aligned} \tag{16}\label{eq16} \end{equation} We can exchange $n$ and $k$, yielding \begin{equation} \begin{aligned} \Omega&=\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\sum^{\infty}_{n=0}\frac{(-1)^{p+1}z^{p+2n+2k}\cos^p\phi}{2^{2n+2k}\Gamma(p+1)\Gamma(k+1)\Gamma(n+1)} \frac{(\nu+p+2k)\Gamma(-n-\nu-p-2k)}{\Gamma(-\nu-p-k+1)}\frac{J_{-\nu-p-2k}(Z)}{Z^\nu} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\frac{(-1)^{p+1}\cos^p\phi}{\Gamma(p+1)\Gamma(k+1)}\frac{\nu+p+2k}{\Gamma(-\nu-p-k+1)} \left[ \sum^{\infty}_{n=0}\frac{\Gamma(-n-\nu-p-2k)}{\Gamma(n+1)}\frac{z^{p+2n+2k}}{2^{2n+2k}} \right]\times \\ &\frac{J_{-\nu-p-2k}(Z)}{Z^\nu} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\frac{(-1)^{p+1}\cos^p\phi}{\Gamma(p+1)\Gamma(k+1)}\frac{\nu+p+2k}{\Gamma(-\nu-p-k+1)} \left[ \frac{2^{\nu+p}}{z^\nu}\sum^{\infty}_{n=0}\frac{\Gamma(-n-\nu-p-2k)}{\Gamma(n+1)}\frac{z^{\nu+p+2n+2k}}{2^{\nu+p+2n+2k}} \right]\\ &\times\frac{J_{-\nu-p-2k}(Z)}{Z^\nu}\\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\frac{(-1)^{p+1}\cos^p\phi}{\Gamma(p+1)\Gamma(k+1)} \frac{(\nu+p+2k)\Gamma(\nu+p+k)}{\Gamma(\nu+p+k)\Gamma(1-\nu-p-k)}\times \\ &\left[ \frac{2^{\nu+p}}{z^\nu}\sum^{\infty}_{n=0}\frac{\Gamma(-n-\nu-p-2k)\Gamma(1+n+\nu+p+2k)}{\Gamma(n+1)\Gamma(1+n+\nu+p+2k)}\frac{z^{\nu+p+2n+2k}}{2^{\nu+p+2n+2k}} \right] \frac{J_{-\nu-p-2k}(Z)}{Z^\nu}\\ \end{aligned} \tag{17}\label{eq17} \end{equation} Since according to the relation, \begin{equation} \Gamma(1-a)\Gamma(a)=\frac{\pi}{\sin\pi a} \notag \end{equation} following relations are deduced. One is \begin{equation} \Gamma(1-\nu-p-k)\Gamma(\nu+p+k)=\frac{\pi}{\sin(\nu+p+k)\pi}=(-1)^k\frac{\pi}{\sin(\nu+p)\pi} \tag{18}\label{eq18} \end{equation} and also, \begin{equation} \begin{aligned} &\Gamma(-\nu-p-2k-n)\Gamma(1+\nu+p+2k+n)=\frac{\pi}{-\sin(\nu+p+2k+n)\pi} =\frac{\pi}{-\sin(\nu+p+n)\pi} \\ &=(-1)^{n+1}\frac{\pi}{\sin(\nu+p)\pi} \end{aligned} \tag{19}\label{eq19} \end{equation} Inserting the relations of \eqref{eq18} and \eqref{eq19} to \eqref{eq17} yields \begin{equation} \begin{aligned} &\Omega=\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\frac{(-1)^{p+1}\cos^p\phi}{\Gamma(p+1)\Gamma(k+1)} \frac{(\nu+p+2k)\Gamma(\nu+p+k)}{\Gamma(\nu+p+k)\Gamma(1-\nu-p-k)} \times \\ &\left[ \frac{2^{\nu+p}}{z^\nu}\sum^{\infty}_{n=0} \frac{(-1)^n}{\Gamma(n+1)\Gamma(1+n+\nu+p+2k)}\frac{z^{\nu+p+2n+2k}}{2^{\nu+p+2n+2k}} \frac{\Gamma(-n-\nu-p-2k)\Gamma(1+n+\nu+p+2k)}{(-1)^n} \right] \frac{J_{-\nu-p-2k}(Z)}{Z^\nu} \\ &=\sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\frac{(-1)^{p+1}\cos^p\phi}{\Gamma(p+1)\Gamma(k+1)} (\nu+p+2k)\Gamma(\nu+p+k)(-1)^k\frac{\sin(\nu+p)\pi}{\pi} \times \\ &\left[ \frac{2^{\nu+p}}{z^\nu}J_{\nu+p+2k}(z)\sum^{\infty}_{n=0}\frac{(-1)^{n+1}\pi}{(-1)^n\sin(\nu+p)\pi} \right] \frac{J_{-\nu-p-2k}(Z)}{Z^\nu} \\ &=2^\nu \sum ^{\infty}_{p=0}\sum ^{\infty}_{k=0}\frac{(-1)^{p+k}2^{p}\cos^p\phi}{\Gamma(p+1)\Gamma(k+1)} (\nu+p+2k)\Gamma(\nu+p+k) \frac{J_{\nu+p+2k}(z)}{z^\nu} \frac{J_{-\nu-p-2k}(Z)}{Z^\nu}\\ \end{aligned} \tag{20}\label{eq20} \end{equation} Replacing $p+2k$ by $m$, the above equation is transformed as \begin{equation} \begin{aligned} &\Omega=2^\nu \sum ^{\infty}_{m=0}\sum ^{k\leq m/2}_{k=0}\frac{(-1)^{m-k}2^{m-2k}\cos^{m-2k}\phi}{\Gamma(m-2k+1)\Gamma(k+1)} (\nu+m)\Gamma(\nu+m-k) \frac{J_{\nu+m}(z)}{z^\nu}\frac{J_{-\nu-m}(Z)}{Z^\nu}\\ &=2^\nu \Gamma(\nu)\sum ^{\infty}_{m=0}(-1)^{m} (\nu+m)\Gamma(\nu+m-k) \frac{J_{\nu+m}(z)}{z^\nu}\frac{J_{-\nu-m}(Z)}{Z^\nu}C_m^{\nu}(\cos\phi)\\ \end{aligned} \tag{21}\label{eq21} \end{equation} where \begin{equation} C_m^{\nu}(\cos\phi)=\sum ^{k\leq m/2}_{k=0}\frac{(-1)^k 2^{m-2k}\Gamma(\nu+m-k)\cos^{m-2k}\phi}{\Gamma(\nu)\Gamma(k+1)\Gamma(m-2k+1)} \tag{22}\label{eq22} \end{equation}