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Question Number 218560 by MrGaster last updated on 12/Apr/25

Commented by MrGaster last updated on 12/Apr/25

$$J_0\left(a\sqrt{1-u^2}\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{a}^{2n}}{{(n!)}^2{2}^{2n}}{(1-u^2)}^n$$$$\Rightarrow \underset{m=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{a}^{2n}}{{(n!)}^2{2}^{2n}}{\int }_{\cos \alpha }^1{(1-u^2)}^n(1+u)e^{ilu}du$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^m{a}^{2n}}{{(n!)}^2{2}^{2n}}{\int }_{\cos \alpha }^1{(1-u)}^n{(1+u)}^{u+1}{e}^{idu}du$$$$\mathrm{Handle}\mathrm{internal}\mathrm{integration}:$$$${\int }_{\cos \alpha }^1{(1-u)}^n{(1+u)}^{n+1}{e}^{idu}du$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(ib\right)}^k}{k!}{\int }_{\cos \alpha }^1{(1-u)}^n{(1+u)}^{n+1}{u}^{k}du$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(ib\right)}^k}{k!}{\int }_{\cos \alpha }^1{(1-u)}^n\underset{m=0}{\sum ^{n+1}}{C}_{n+1}^m{u}^{k+m}du$$$$=\underset{\lambda =0}{\sum ^{\mathrm{\infty }}}\underset{m=0}{\sum ^{n+1}}{C}_{n+1}^m\frac{{\left(ib\right)}^k}{k!}{\int }_{\cos \alpha }^1{(1-u)}^n{u}^{k+m}du$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}\underset{m=0}{\sum ^{n+1}}{C}_{n+1}^m\frac{{\left(ib\right)}^k}{k!}[B(k+m+1,n+1)-B(\cos \alpha ;k+m+1,n+1)]$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}\underset{m=0}{\sum ^{n+1}}{C}_{n+1}^m\frac{{(-ib)}^k}{k!}\frac{{(1-\cos \alpha )}^{k+m+n+1}}{k+m+n+1}$$$${}_2{F}_1(-n,k+m+1;k+m+n+2;1-\cos \alpha )$$$$\mathrm{Back}\mathrm{to}\mathrm{the}\mathrm{original}\mathrm{formula}.$$$$.\dots \dots \mathrm{Quadruple}\mathrm{series}.$$$$\overline{)\begin{array}{c}\underset{n=0}{\sum ^{\mathrm{\infty }}}\underset{k=0}{\sum ^{\mathrm{\infty }}}\underset{m=0}{\sum ^{n+1}}\frac{{(-1)}^n{a}^{2n}}{{(n!)}^2{2}^{2n}}\frac{{(-ib)}^k}{k!}\frac{{(1-\cos \alpha )}^{k+m+n+1}}{k+m+n+1}{C}_{n+1}^m\\ {}_2{F}_1(-n,k+m+1;k+m+n+2;1-\cos \alpha )\end{array}}$$$$\dots \dots \dots \mathrm{I}\mathrm{dont}\mathrm{know}\mathrm{how}\mathrm{thee}$$$$\mathrm{convrgence}\mathrm{is}\mathrm{but}\mathrm{I}\mathrm{dontt}$$$$\mathrm{wan}\mathrm{to}\mathrm{simplify}.$$

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