(*)

ベッセル関数の級数表示より、
\begin{align*} J_{\nu}(z) & =\sum_{m=0}^{\infty}\frac{(-1)^{m}}{m!\Gamma(m+\nu+1)}\left(\frac{z}{2}\right)^{2m+\nu}\\ & =\sum_{m=0}^{\infty}\frac{(-1)^{m}}{2^{\nu}\Gamma(m+\nu+1)}\frac{P\left(m-\frac{1}{2},m\right)}{(2m)!}z^{2m+\nu}\\ & =\sum_{m=0}^{\infty}\frac{(-1)^{m}}{2^{\nu}\Gamma(m+\nu+1)}\frac{\Gamma\left(m+\frac{1}{2}\right)}{\Gamma(2m+1)\Gamma\left(\frac{1}{2}\right)}z^{2m+\nu}\\ & =\sum_{m=0}^{\infty}\frac{(-1)^{m}}{2^{\nu}(2m)!\sqrt{\pi}\Gamma\left(\nu+\frac{1}{2}\right)}\frac{\Gamma\left(\nu+\frac{1}{2}\right)\Gamma\left(m+\frac{1}{2}\right)}{\Gamma(m+\nu+1)}z^{2m+\nu}\\ & =\sum_{m=0}^{\infty}\frac{(-1)^{m}}{2^{\nu}(2m)!\sqrt{\pi}\Gamma\left(\nu+\frac{1}{2}\right)}B\left(\nu+\frac{1}{2},m+\frac{1}{2}\right)z^{2m+\nu}\\ & =\sum_{m=0}^{\infty}\frac{(-1)^{m}}{2^{\nu}(2m)!\sqrt{\pi}\Gamma\left(\nu+\frac{1}{2}\right)}z^{2m+\nu}\int_{0}^{1}t^{\nu-\frac{1}{2}}(1-t)^{m-\frac{1}{2}}dt\\ & =\frac{1}{\sqrt{\pi}\Gamma\left(\nu+\frac{1}{2}\right)}\left(\frac{z}{2}\right)^{\nu}\int_{0}^{1}t^{\nu-\frac{1}{2}}(1-t)^{-\frac{1}{2}}\sum_{m=0}^{\infty}\frac{(-1)^{m}}{(2m)!}\left\{ z(1-t)^{\frac{1}{2}}\right\} {}^{2m}dt\\ & =\frac{1}{\sqrt{\pi}\Gamma\left(\nu+\frac{1}{2}\right)}\left(\frac{z}{2}\right)^{\nu}\int_{0}^{1}t^{\nu-\frac{1}{2}}(1-t)^{-\frac{1}{2}}\cos\left\{ z(1-t)^{\frac{1}{2}}\right\} dt \end{align*}

(1)

\(t=\sin^{2}\theta\)とすると、
\[ J_{\nu}(z)=\frac{2}{\sqrt{\pi}\Gamma\left(\nu+\frac{1}{2}\right)}\left(\frac{z}{2}\right)^{\nu}\int_{0}^{\frac{\pi}{2}}\sin^{2\nu}\theta\cos(z\cos\theta)d\theta \]

(2)

\(1-t=u^{2}\)とすると、
\[ J_{\nu}(z)=\frac{1}{\sqrt{\pi}\Gamma\left(\nu+\frac{1}{2}\right)}\left(\frac{z}{2}\right)^{\nu}\int_{-1}^{1}(1-u^{2})^{\nu-\frac{1}{2}}e^{izu}du \]