0次のBessel関数, Struve関数に関する等式メモ

この記事において, $\stackrel{\text{?}}{=}$が付いているものはまだ証明ができていないことを意味する. まず, 以下のように定義する.
$$\begin{array}{rl}\gamma (x)& :=\sum _{0\le n}\frac{x^n}{n{!}^2}\\ {\gamma }^{\mathrm{♯}}(x)& :=\sum _{0\le n}\frac{x^{n+\frac{1}{2}}}{\mathrm{\Gamma }{(n+\frac{3}{2})}^2}\\ {\gamma }_1(x)& :=-\frac{1}{\pi }{\frac{d}{dw}\sum _{0\le n}\frac{x^{n+w}}{\mathrm{\Gamma }(n+w+1{)}^2}|}_{w=0}\\ \bar{\gamma }(x)& :=\sum _{0\le n}(-1{)}^n\frac{x^n}{n{!}^2}\\ {\bar{\gamma }}^{\mathrm{♯}}(x)& :=\sum _{0\le n}(-1{)}^n\frac{x^{n+\frac{1}{2}}}{\mathrm{\Gamma }{(n+\frac{3}{2})}^2}\\ {\bar{\gamma }}_1(x)& :=-\frac{1}{\pi }{\frac{d}{dw}\sum _{0\le n}(-1{)}^n\frac{x^{n+w}}{\mathrm{\Gamma }(n+w+1{)}^2}|}_{w=0}\end{array}$$
これらの記法と, 通常のBessel関数とStruve関数との間の関係は以下のようになる.

$$\begin{array}{rl}{J}_0(x)& =\bar{\gamma }\left(\frac{x^2}{4}\right)\\ Y_0(x)& =-{\bar{\gamma }}_1\left(\frac{x^2}{4}\right)\\ I_0(x)& =\gamma \left(\frac{x^2}{4}\right)\\ K_0(x)& =\frac{\pi }{2}{\gamma }_1\left(\frac{x^2}{4}\right)\\ {\mathbf{H}}_0(x)& ={\bar{\gamma }}^{\mathrm{♯}}\left(\frac{x^2}{4}\right)\\ {\mathbf{L}}_0(x)& ={\gamma }^{\mathrm{♯}}\left(\frac{x^2}{4}\right)\\ {\mathbf{K}}_0(x)& ={\bar{\gamma }}^{\mathrm{♯}}\left(\frac{x^2}{4}\right)+{\bar{\gamma }}_1\left(\frac{x^2}{4}\right)\\ {\mathbf{M}}_0(x)& ={\gamma }^{\mathrm{♯}}\left(\frac{x^2}{4}\right)-\gamma \left(\frac{x^2}{4}\right)\end{array}$$

以下の積分表示が成り立つ.
$$\begin{array}{rl}{\bar{\gamma }}^{\mathrm{♯}}(x)& =\frac{1}{\pi }{\int }_0^x\frac{\sin \left(2\sqrt{t}\right)}{\sqrt{t(x-t)}}dt\\ \bar{\gamma }(x)& =\frac{1}{\pi }{\int }_0^x\frac{\cos \left(2\sqrt{t}\right)}{\sqrt{t(x-t)}}dt\\ \bar{\gamma }(x)& =\frac{1}{\pi }{\int }_x^{\mathrm{\infty }}\frac{\sin \left(2\sqrt{t}\right)}{\sqrt{t(t-x)}}dt\\ {\bar{\gamma }}_1(x)& =\frac{1}{\pi }{\int }_x^{\mathrm{\infty }}\frac{\cos \left(2\sqrt{t}\right)}{\sqrt{t(t-x)}}dt\\ {\gamma }_1\left(x\right)& =\frac{1}{\pi }{\int }_x^{\mathrm{\infty }}\frac{e^{-2\sqrt{t}}}{\sqrt{t(t-x)}}dt\\ {\gamma }_1\left(x\right)& =\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\cos \left(2\sqrt{t}\right)}{\sqrt{t(t+x)}}dt\\ \gamma (x)-{\gamma }^{\mathrm{♯}}(x)& =\frac{1}{\pi }{\int }_0^x\frac{e^{-2\sqrt{t}}}{\sqrt{t(x-t)}}dt\\ \gamma (x)-{\gamma }^{\mathrm{♯}}(x)& =\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\sin \left(2\sqrt{t}\right)}{\sqrt{t(x+t)}}dt\\ {\bar{\gamma }}^{\mathrm{♯}}(x)+{\bar{\gamma }}_1(x)& =\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{e^{-2\sqrt{t}}}{\sqrt{t(x+t)}}dt\end{array}$$

以下のような積に関する級数表示がある.
$$\begin{array}{rl}\gamma (x)\bar{\gamma }(x)& =\sum _{0\le n}\frac{(-1{)}^n}{n{!}^2(2n)!}{x}^{2n}\\ {\gamma }_1(x)\bar{\gamma }(x)+\gamma (x){\bar{\gamma }}_1(x)& \stackrel{\text{?}}{=}-\frac{1}{\pi }{\frac{d}{dw}\sum _{0\le n}\frac{(-1{)}^n}{\mathrm{\Gamma }(n+w+1{)}^2\mathrm{\Gamma }(2n+2w+1)}{x}^{2n+2w}|}_{w=0}\\ 2{\gamma }_1(x){\bar{\gamma }}_1(x)-\frac{1}{2}\gamma (x)\bar{\gamma }(x)& \stackrel{\text{?}}{=}\frac{1}{2{\pi }^2}{{\left(\frac{d}{dw}\right)}^2\sum _{0\le n}\frac{(-1{)}^n}{\mathrm{\Gamma }(n+w+1{)}^2\mathrm{\Gamma }(2n+2w+1)}{x}^{2n+2w}|}_{w=0}\\ {\gamma }_1(x)\bar{\gamma }(x)-\gamma (x){\bar{\gamma }}_1(x)& \stackrel{\text{?}}{=}\sum _{0\le n}\frac{(-1{)}^n}{\mathrm{\Gamma }{(n+\frac{3}{2})}^2(2n+1)!}{x}^{2n+1}\end{array}$$

${\beta }_w:=\frac{\mathrm{\Gamma }(\frac{1}{2}+w)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(w+1)}$として, $\gamma (x)$の類似として, 以下のような関数を導入する.

$$\begin{array}{rl}\alpha (x)& :=\sum _{0\le n}\frac{{\beta }_n}{n{!}^2}{x}^n\\ {\alpha }_k(x)& :=\frac{1}{k!}{{(-\frac{1}{\pi }\frac{d}{dw})}^k\sum _{0\le n}\frac{{\beta }_{n+w}{x}^{n+w}}{\mathrm{\Gamma }(n+w+1{)}^2}|}_{w=0}\\ \bar{\alpha }(x)& :=\sum _{0\le n}(-1{)}^n\frac{{\beta }_n}{n{!}^2}{x}^n\\ {\bar{\alpha }}_k(x)& :=\frac{1}{k!}{{(-\frac{1}{\pi }\frac{d}{dw})}^k\sum _{0\le n}(-1{)}^n\frac{{\beta }_{n+w}{x}^{n+w}}{\mathrm{\Gamma }(n+w+1{)}^2}|}_{w=0}\end{array}$$

このとき, 積に関して以下のような関係式がある.

$$\begin{array}{rl}\gamma (x{)}^2& =\alpha (4x)\\ \gamma (x){\gamma }_1(x)& \stackrel{\text{?}}{=}{\alpha }_1(4x)\\ {\gamma }_1(x{)}^2& \stackrel{\text{?}}{=}2{\alpha }_2(4x)\\ \bar{\gamma }(x{)}^2& =\bar{\alpha }(4x)\\ \bar{\gamma }(x){\bar{\gamma }}_1(x)& \stackrel{\text{?}}{=}{\bar{\alpha }}_1(4x)\\ {\bar{\gamma }}_1(x{)}^2& \stackrel{\text{?}}{=}2{\bar{\alpha }}_2(4x)\end{array}$$

積に関する積分表示としては, 以下のようなものがある.
$$\begin{array}{rl}\gamma (x{)}^2& =\frac{1}{\pi }{\int }_0^x\frac{\gamma (4t)}{\sqrt{t(x-t)}}dt\\ \gamma (x){\gamma }_1(x)& =\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\bar{\gamma }(4t)}{\sqrt{t(x+t)}}dt\\ {\gamma }_1(x{)}^2& =\frac{2}{\pi }{\int }_x^{\mathrm{\infty }}\frac{{\gamma }_1(4t)}{\sqrt{t(t-x)}}dt\\ \frac{1}{2}(\bar{\gamma }(x{)}^2+{\bar{\gamma }}_1(x{)}^2)& =\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{{\gamma }_1(4t)}{\sqrt{t(x+t)}}dt\end{array}$$