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現代数学議論

多重超幾何級数のFourier-Legendre展開

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$\underline{\rm 記法}$

$\qquad\D \beta_n^{}={2}^{-2n}\binom{2n}{n},\quad K(x)=\frac{\pi}{2}\sum_{n=0}^\infty \beta_n^2x^{2n},\quad A=\sum_{n=0}^\infty \beta_n^3,\quad B=\sum_{n=1}^\infty \L(2n\beta_n^{}\R)^{-3}\,\quad C=\sum_{n=0}^\infty \beta_n^4,\quad D=\sum_{n=1}^\infty \L(2n\beta_n^{}\R)^{-4}$
$\qquad\D E=\sum_{n=0}^\infty \beta_n^4\sum_{m=1}^n(2m\beta_m^{})^{-3},\quad F=\sum_{n=1}^\infty {(2n\beta_n^{})}^{-4}\sum_{m=0}^n\beta_m^3,\quad G=\sum_{n=0}^\infty \beta_n^4\sum_{m=1}^n(2m\beta_m^{})^{-4},\quad H=\sum_{n=1}^\infty {(2n\beta_n^{})}^{-4}\sum_{m=0}^n\beta_m^4$
$\qquad\D \gamma_n^{(1)}=\sum_{m=0}^n \beta_m^2\beta_{n-m}^{},\quad \gamma_n^{(2)}=\sum_{m=1}^n \frac{\beta_{n-m}^{}}{\D(2m\beta_m^{})^2},\quad J=\sum_{n=0}^\infty \beta_n^2\gamma_n^{(2)}$

$\rm Legendre\,多項式$

$\Large\boxed{\rm 定義}$$\hspace{5pt}$Legendre多項式$P_n^{}(x)$は次式で定義されます．

$\BA\D \frac{1}{\sqrt{\D 1-2tx+t^2}}=\sum_{n=0}^\infty {t}^nP_n^{}(x) \EA$

　
$\Large\boxed{\rm 漸化式・微分方程式}$$\hspace{5pt}P_n^{}(x)$は次の漸化式・微分方程式を満たします．

$\BA\D&(2n+1)xP_n^{}(x)=(n+1)P_{n+1}^{}(x)+nP_{n-1}^{}(x)\\&(2n+1)(1-x^2)\frac{d}{dx}P_n^{}(x)=n(n+1)\L(P_{n-1}^{}(x)-P_{n+1}^{}(x)\R)\\&\frac{d}{dx}(1-x^2)\frac{d}{dx}P_n^{}(x)+n(n+1)P_n^{}(x)=0\EA$

$\hspace{5pt}$また，これらを用いて次の等式を証明することができます．

$\BA\D&\L(\frac{d}{dx}x(1-x^2)\frac{d}{dx}-x\R)f(x)\cdot P_n^{}(x)=\frac{d}{dx}x(1-x^2)\L(P_n^{}(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}P_n^{}(x)\R)-\frac{\D n^3P_{n-1}^{}(x)+(n+1)^3P_{n+1}^{}(x)}{2n+1}f(x)\\ \Longleftrightarrow\quad&\begin{cases}&\D P_{2n}^{}(x)f(x)={(-1)}^n\beta_n^3\L(f(x)-\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{d}{dx}x(1-x^2)\L(P_{2m-1}^{}(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}P_{2m-1}^{}(x)\R)-\L(\frac{d}{dx}x(1-x^2)\frac{d}{dx}-x\R)f(x)\cdot P_{2m-1}^{}(x)\R)\R) \\ &\D P_{2n-1}^{}(x)f(x)=\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}{(-1)}^m(4m+1)\beta_k^3\L(\frac{d}{dx}x(1-x^2)\L(P_{2m}^{}(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}P_{2m}^{}(x)\R)-\L(\frac{d}{dx}x(1-x^2)\frac{d}{dx}-x\R)f(x)\cdot P_{2m}^{}(x)\R) \end{cases}\quad\tag{{\bf\color{red}{1}}}\EA$

　
$\Large\boxed{\textrm{Fourier\,-\,Legendre\,展開}}$$\hspace{5pt}$$f(x)$のFourier-Legendre展開は次式で定義されます．

$\BA\D f(x)=\frac{1}{2}\sum_{n=0}^\infty (2n+1)P_n^{}(x)\int_{-1}^1 f(t)P_n^{}(t)\,dt \EA$

$\hspace{5pt}$また，本稿では，$f(x)P_n^{}(x)$を$(0,1)$で積分したものをFL係数と呼ぶことにします．$n$の偶奇で$P_n^{}(x)$の偶奇が異なり，$f(x)$の偶奇も分けて考えるのが妥当と思います．

$\rm multiple\,hypergeometric\,series\,$

$\hspace{5pt}$次の微分方程式について考えてみます．

$\BA\D \L(\frac{d}{dx}x(1-x^2)\frac{d}{dx}-x\R)f(x)=g(x) \EA$

$\hspace{5pt}$$f(x),g(x)$を偶奇で分けて考えると，次のように表現できます．

$\BA\D\begin{cases}&\D g(x)=\sum_{n=1}^\infty g_n^{}{x}^{2n-1},&\D f(x)=\sum_{n=0}^\infty \beta_n^2 {x}^{2n}\sum_{m=1}^n \frac{\D g_m^{}}{\D(2m\beta_m^{})^2}\qquad({\rm \color{red}{A}}) \\&\D g(x)=\sum_{n=0}^\infty g_n^{}{x}^{2n},&\D f(x)=\sum_{n=1}^\infty \frac{\D x^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1} \beta_m^2g_m^{}\hspace{25pt}({\rm \color{red}{B}}) \end{cases}\EA$

$({\bf\color{red}{1}})$式より，$g(x)$のFL係数が計算できるならば，$f(x)$のFL係数が計算でき，ひいては多重級数のFL係数も計算できることになると思います．
以降，上記のようなHypergeometric termの多重和のみを扱うこととします．$({\bf\color{red}{1}})$式の両辺を積分すると，

$\BA\D\begin{cases}&\D \int_0^1 P_{2n}^{}(x)f(x)\,dx={(-1)}^n\beta_n^3\L(\int_0^1 f(x)\,dx-\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\lim_{x\to1}\L[x(1-x^2)\L(P_{2m-1}^{}(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}P_{2m-1}^{}(x)\R)\R]-\lim_{x\to0}\L[x(1-x^2)\L(P_{2m-1}^{}(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}P_{2m-1}^{}(x)\R)\R]-\int_0^1 P_{2m-1}^{}(x)g(x)\,dx\R)\R) \\ &\D \int_0^1 P_{2n-1}^{}(x)f(x)\,dx=\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}{(-1)}^m(4m+1)\beta_m^3\L(\lim_{x\to1}\L[x(1-x^2)\L(P_{2m}^{}(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}P_{2m}^{}(x)\R)\R]-\lim_{x\to0}\L[x(1-x^2)\L(P_{2m}^{}(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}P_{2m}^{}(x)\R)\R]-\int_0^1 P_{2m}^{}(x)g(x)\,dx\R) \end{cases}\EA$

ここで，$\D \lim_{x\to1}x(1-x^2)f(x)\frac{d}{dx}P_{k}^{}(x)=\lim_{x\to0}x(1-x^2)P_{k}^{}(x)\frac{d}{dx}f(x)=\lim_{x\to0}x(1-x^2)f(x)\frac{d}{dx}P_{k}^{}(x)=0$であり，

$\BA\D &({\rm \color{red}{A}})\,{\rm の場合：}\lim_{x\to1}x(1-x^2)P_{k}^{}(x)\frac{d}{dx}f(x)=\lim_{x\to1}(1-x^2)f'(x)=\frac{2}{\pi}\sum_{n=1}^\infty \frac{\D g_n^{}}{\D(2n\beta_n^{})^2}\\ &({\rm \color{red}{B}})\,{\rm の場合：}\lim_{x\to1}x(1-x^2)P_{k}^{}(x)\frac{d}{dx}f(x)=\lim_{x\to1}(1-x^2)f'(x)=\frac{\pi}{2}\sum_{n=0}^\infty \beta_n^2g_n^{}\EA$

となります．

$\BA\D \begin{cases}&\D \int_0^1 P_{2n}^{}(x)f(x)\,dx={(-1)}^n\beta_n^3\L(\int_0^1 f(x)\,dx-\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\lim_{x\to1}(1-x^2)f'(x)-\int_0^1 P_{2m-1}^{}(x)g(x)\,dx\R)\R) \\ &\D \int_0^1 P_{2n-1}^{}(x)f(x)\,dx=\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}{(-1)}^m(4m+1)\beta_m^3\L(\lim_{x\to1}(1-x^2)f'(x)-\int_0^1 P_{2m}^{}(x)g(x)\,dx\R) \end{cases}\quad\tag{{\color{red}{2}}} \EA$

$\textrm{Fourier-Legendre\,展開\,}$

$\hspace{5pt}$具体的な$g_n^{}$について考えていきます．

$\underline{{\bf 1.}\,g_n^{}=\beta_n^{}{\,\rm の場合}}$

$\BA\D g(x)=\sum_{n=0}^\infty \beta_n^{}{x}^{2n}=\frac{1}{\sqrt{1-x^2}},\qquad f(x)=\sum_{n=1}^\infty \frac{\D x^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1} \beta_m^3\EA$

とすれば，

$\BA\D\begin{cases}&\D \int_0^1 P_{2n}^{}(x)f(x)\,dx={(-1)}^n\beta_n^3\L(\int_0^1 f(x)\,dx-\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{\pi}{2}A-\frac{1}{\D(2m\beta_m)^2}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3\R)\R) \\ &\D \int_0^1 P_{2n-1}^{}(x)f(x)\,dx=\frac{\pi}{2}\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}{(-1)}^m(4m+1)\beta_m^3\L(A-\beta_m^2\R) \end{cases}\EA$
$\BA\D\\f(x)=\frac{\pi}{2}\sum_{n=1}^\infty \frac{\D(-1)^{n-1}(4n-1)P_{2n-1}^{}(x)}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}{(-1)}^m(4m+1)\beta_m^3\L(A-\beta_m^2\R)\EA$

と書けることになります．いま，右辺の$m$に関する和は$n\to\infty$で$0$に収束することより$\D-\sum_{n\le m}$と書き換えることができ，$\D\sum_{n=1}^\infty \frac{\D x^{2n-1}}{\D(2n\beta_n^{})^2}$のFL展開を思い出すと，$\D\hat{f}(x)=\sum_{n=0}^\infty \beta_n^3\sum_{m=1}^n \frac{\D x^{2m-1}}{\D(2m\beta_m^{})^2}$と書くことで

$\BA\D\hat{f}(x)=-\frac{\pi}{2}\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^5\sum_{m=1}^n\frac{\D(-1)^{m-1}(4m-1)P_{2m-1}^{}(x)}{\D(2m\beta_m^{})^3}\EA$

となります．

$\underline{{\bf 2.}\,\D g_n^{}=\frac{1}{2n\beta_n^{}}{\,\rm の場合}}$

$\BA\D g(x)=\sum_{n=1}^\infty \frac{\D x^{2n-1}}{2n\beta_n^{}}=\frac{\sin^{-1}x}{\sqrt{1-x^2}},\qquad f(x)=\sum_{n=0}^\infty \beta_n^2{x}^{2n}\sum_{m=1}^{n} \frac{1}{\D(2m\beta_m^{})^3},\qquad \hat{f}(x)=\sum_{n=1}^\infty \frac{1}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}\beta_m^2{x}^{2m}\EA$

とすれば，

$\BA\D\begin{cases}&\D \int_0^1 P_{2n}^{}(x)f(x)\,dx={(-1)}^n\beta_n^3\sum_{n< m} \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{2}{\pi}B-\frac{1}{\D(2m\beta_m)^2}\R) \\ &\D \int_0^1 P_{2n-1}^{}(x)f(x)\,dx=-\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{2}{\pi}B-\beta_m^2\L(\frac{\pi^2}{8}+\sum_{l=1}^m \frac{\D(-1)^{l-1}(4l-1)}{\D(2l)^2(2l-1)\beta_l^{}}\R)\R) \end{cases}\EA$
$\BA\D\\&f(x)=\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)\sum_{n< m} \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{2}{\pi}B-\frac{1}{\D(2m\beta_m)^2}\R)\\&\hat{f}(x)=\sum_{n=1}^\infty \frac{\D(-1)^{n-1}(4n-1)}{\D(2n\beta_n^{})^5}\sum_{m=0}^{n-1}{(-1)}^m(4m+1)\beta_m^3P_{2m}^{}(x) \EA$

$\underline{{\bf 3.}\,\D g_n^{}=\beta_n^2{\,\rm の場合}}$

$\BA\D g(x)=\sum_{n=0}^\infty \beta_n^2{x}^{2n}=\frac{2}{\pi}K(x),\qquad f(x)=\sum_{n=1}^\infty \frac{\D x^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1} \beta_m^4,\qquad\hat{f}(x)=\sum_{n=0}^\infty \beta_n^4\sum_{m=1}^n \frac{\D x^{2m-1}}{\D(2m\beta_m^{})^2}\EA$

とすれば，

$\BA\D\begin{cases}&\D \int_0^1 P_{2n}^{}(x)f(x)\,dx={(-1)}^n\beta_n^3\sum_{n< m} \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{\pi}{2}C-\frac{2}{\pi}\frac{\D(-1)^{m-1}}{\D(2m\beta_m^{})^3}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3\R)\\ &\D\int_0^1 P_{2n-1}^{}(x)f(x)\,dx=-\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{\pi}{2}C-\frac{2}{\pi}{(-1)}^m\beta_m^3\sum_{m< l}\frac{\D(-1)^{l-1}(4l-1)}{\D(2l\beta_l^{})^3}\R)\end{cases}\EA$
$\BA\D\\&f(x)=-\sum_{n=1}^\infty \frac{\D(-1)^{n-1}(4n-1)P_{2n-1}^{}(x)}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{\pi}{2}C-\frac{2}{\pi}{(-1)}^m\beta_m^3\sum_{m< l}\frac{\D(-1)^{l-1}(4l-1)}{\D(2l\beta_l^{})^3}\R)\\&\hat{f}(x)=-\frac{2}{\pi}\sum_{n=1}^\infty \frac{\D(-1)^{n-1}(4n-1)}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^6\sum_{l=1}^m\frac{\D(-1)^{l-1}(4l-1)P_{2l-1}^{}(x)}{\D(2l\beta_l^{})^3}\EA$

$\underline{{\bf 4.}\,\D g_n^{}=\frac{1}{\D(2n\beta_n^{})^2}{\,\rm の場合}}$

$\BA\D g(x)=\sum_{n=1}^\infty \frac{\D x^{2n-1}}{\D(2n\beta_n^{})^2},\qquad f(x)=\sum_{n=0}^\infty \beta_n^2{x}^{2n}\sum_{m=1}^{n} \frac{1}{\D(2m\beta_m^{})^4},\qquad\hat{f}(x)=\sum_{n=1}^\infty \frac{1}{\D(2n\beta_n^{})^4}\sum_{m=0}^{n-1}\beta_m^2{x}^{2m}\EA$

とすれば，

$\BA\D\begin{cases}&\D\int_0^1 P_{2n}^{}(x)f(x)\,dx={(-1)}^n\beta_n^3\sum_{n< m} \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{2}{\pi}D+\frac{\pi}{2}\frac{\D(-1)^{m-1}}{\D(2m\beta_m^{})^3}\sum_{m\le l} {(-1)}^l(4l+1)\beta_l^3\R) \\ &\D \int_0^1 P_{2n-1}^{}(x)f(x)\,dx=-\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{2}{\pi}D-{(-1)}^m\beta_m^3\sum_{m< l} \frac{\D(-1)^{l-1}(4l-1)}{\D(2l\beta_l^{})^3}\L(\frac{\pi}{2}-\frac{\D(-1)^{l-1}\beta_l^{}}{2l-1}\R)\R) \end{cases}\EA$
$\BA\D\\&f(x)=\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)\sum_{n< m} \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{2}{\pi}D+\frac{\pi}{2}\frac{\D(-1)^{m-1}}{\D(2m\beta_m^{})^3}\sum_{m\le l} {(-1)}^l(4l+1)\beta_l^3\R)\\&\hat{f}(x)=-\frac{\pi}{2}\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3\sum_{m=1}^n\frac{4m-1}{\D(2m\beta_m^{})^6}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3P_{2l}^{}(x)\EA$

$\underline{{\bf 5.}\,\D g_n^{}=\beta_n^2\sum_{m=1}^n\frac{1}{\D(2m\beta_m^{})^3}{\,\rm の場合}}$

$\BA\D g(x)=\sum_{n=0}^\infty \beta_n^2x^{2n}\sum_{m=1}^n\frac{1}{\D(2m\beta_m^{})^3},\qquad f(x)=\sum_{n=1}^\infty \frac{\D x^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_n^4\sum_{l=1}^m\frac{1}{\D(2l\beta_l^{})^3},\quad \hat{f}(x)=\sum_{n=1}^\infty \frac{1}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}\beta_m^4\sum_{l=1}^m\frac{x^{2l-1}}{\D(2l\beta_l^{})^2}\EA$

とすれば，

$\BA\D\begin{cases}&\D\int_0^1 P_{2n}^{}(x)f(x)\,dx={(-1)}^n\beta_n^3\sum_{n< m}\frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{\pi}{2}E+\frac{\D(-1)^{m-1}}{\D(2m\beta_m^{})^3}\sum_{m\le l}{(-1)}^l(4l+1)\beta_l^3\L(\frac{2}{\pi}B-\beta_l^2\L(\frac{\pi^2}{8}+\sum_{k=1}^l \frac{\D(-1)^{k-1}(4k-1)}{\D(2k)^2(2k-1)\beta_k^{}}\R)\R)\R) \\&\D\int_0^1 P_{2n-1}^{}(x)f(x)\,dx=-\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{\pi}{2}E-{(-1)}^m\beta_m^3\sum_{m< l} \frac{\D(-1)^{l-1}(4l-1)}{\D(2l\beta_l^{})^3}\L(\frac{2}{\pi}B-\frac{1}{\D(2l\beta_l)^2}\R)\R)\end{cases}\EA$
$\BA\D\\&f(x)=-\sum_{n=1}^\infty\frac{\D(-1)^{n-1}(4n-1)P_{2n-1}^{}(x)}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{\pi}{2}E-{(-1)}^m\beta_m^3\sum_{m< l} \frac{\D(-1)^{l-1}(4l-1)}{\D(2l\beta_l^{})^3}\L(\frac{2}{\pi}B-\frac{1}{\D(2l\beta_l)^2}\R)\R)\\&\hat{f}(x)=-\sum_{n=1}^\infty\frac{\D(-1)^{n-1}(4n-1)}{\D(2n\beta_n^{})^5}\sum_{m=0}^{n-1}(4m+1)\beta_m^6\sum_{l=1}^m\frac{\D(-1)^{l-1}(4l-1)P_{2l-1}^{}(x)}{\D(2l\beta_l^{})^3}\EA$

$\underline{{\bf 6.}\,\D g_n^{}=\frac{1}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_m^3{\,\rm の場合}}$

$\BA\D g(x)=\sum_{n=1}^\infty \frac{x^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_m^3,\qquad f(x)=\sum_{n=0}^\infty \beta_n^2x^{2n}\sum_{m=1}^n\frac{1}{\D(2m\beta_m^{})^4}\sum_{l=0}^{m-1}\beta_l^3,\qquad\hat{f}(x)=\sum_{n=0}^\infty \beta_n^3\sum_{m=1}^n\frac{1}{\D(2m\beta_m^{})^4}\sum_{l=0}^{m-1}\beta_l^2x^{2l}\EA$

とすれば，

$\BA\D\begin{cases}&\D\int_0^1 P_{2n}^{}(x)f(x)\,dx={(-1)}^n\beta_n^3\sum_{n< m}\frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{2}{\pi}F+\frac{\pi}{2}\frac{\D(-1)^{m-1}}{\D(2m\beta_m^{})^3}\sum_{m\le l}{(-1)}^l(4l+1)\beta_l^3\L(A-\beta_l^2\R)\R) \\&\D\int_0^1 P_{2n-1}^{}(x)f(x)\,dx=-\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{2}{\pi}F-{(-1)}^m\beta_m^3\sum_{m< l}\frac{\D(-1)^{l-1}(4l-1)}{\D(2l\beta_l^{})^3}\L(\frac{\pi}{2}A-\frac{1}{\D(2l\beta_l)^2}\sum_{k=0}^{l-1}{(-1)}^k(4k+1)\beta_k^3\R)\R)\end{cases}\EA$
$\BA\D\\&f(x)=\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)\sum_{n< m}\frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{2}{\pi}F+\frac{\pi}{2}\frac{\D(-1)^{m-1}}{\D(2m\beta_m^{})^3}\sum_{m\le l}{(-1)}^l(4l+1)\beta_l^3\L(A-\beta_l^2\R)\R)\\&\hat{f}(x)=-\frac{\pi}{2}\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^5\sum_{m=1}^n\frac{4m-1}{\D(2m\beta_m^{})^6}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3P_{2l}^{}(x)\EA$

$\underline{{\bf 7.}\,\D g_n^{}=\beta_n^2\sum_{m=1}^n\frac{1}{\D(2m\beta_m^{})^4}{\,\rm の場合}}$

$\BA\D g(x)=\sum_{n=0}^\infty \beta_n^2x^{2n}\sum_{m=1}^n\frac{1}{\D(2m\beta_m^{})^4},\qquad f(x)=\sum_{n=1}^\infty \frac{\D x^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_n^4\sum_{l=1}^m\frac{1}{\D(2l\beta_l^{})^4},\quad \hat{f}(x)=\sum_{n=1}^\infty\frac{1}{\D(2n\beta_n^{})^4}\sum_{m=0}^{n-1}\beta_m^4\sum_{l=1}^m\frac{x^{2l-1}}{\D(2l\beta_l^{})^2}\EA$

であり，

$\BA\D\begin{cases}&\D\int_0^1 P_{2n}^{}(x)f(x)\,dx={(-1)}^n\beta_n^3\sum_{n< m}\frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{\pi}{2}G-\frac{\D(-1)^{m-1}}{\D(2m\beta_m^{})^3}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3\L(\frac{2}{\pi}D-{(-1)}^l\beta_l^3\sum_{l< k} \frac{\D(-1)^{k-1}(4k-1)}{\D(2k\beta_k^{})^3}\L(\frac{\pi}{2}-\frac{\D(-1)^{k-1}\beta_k^{}}{2k-1}\R)\R)\R)\\&\D\int_0^1 P_{2n-1}^{}(x)f(x)\,dx=-\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{\pi}{2}G-{(-1)}^n\beta_n^3\sum_{n< m} \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{2}{\pi}D+\frac{\pi}{2}\frac{\D(-1)^{m-1}}{\D(2m\beta_m^{})^3}\sum_{m\le l} {(-1)}^l(4l+1)\beta_l^3\R)\R)\end{cases}\EA$
$\BA\D\\&f(x)=-\sum_{n=1}^\infty \frac{\D(-1)^{n-1}(4n-1)P_{2n-1}^{}(x)}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{\pi}{2}G-{(-1)}^n\beta_n^3\sum_{n< m} \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{2}{\pi}D+\frac{\pi}{2}\frac{\D(-1)^{m-1}}{\D(2m\beta_m^{})^3}\sum_{m\le l} {(-1)}^l(4l+1)\beta_l^3\R)\R)\\&\hat{f}(x)=\sum_{n=0}^\infty{(-1)}^n(4n+1)\beta_n^3\sum_{m=1}^n\frac{4m-1}{\D(2m\beta_m^{})^6}\sum_{l=0}^{m-1}(4l+1)\beta_l^6\sum_{k=1}^l\frac{\D(-1)^{k-1}(4k-1)P_{2k-1}(x)}{\D(2k\beta_k^{})^3}\EA$

$\underline{{\bf 8.}\,\D g_n^{}=\frac{1}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_m^4{\,\rm の場合}}$

$\BA\D g(x)=\sum_{n=1}^\infty \frac{x^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_m^4,\qquad f(x)=\sum_{n=0}^\infty \beta_n^2x^{2n}\sum_{m=1}^n\frac{1}{\D(2m\beta_m^{})^4}\sum_{l=0}^{m-1}\beta_l^4,\qquad\hat{f}(x)=\frac{\pi}{2}\sum_{n=0}^\infty \beta_n^4\sum_{m=1}^n\frac{1}{\D(2m\beta_m^{})^4}\sum_{l=0}^{m-1}\beta_l^2x^{2l}\EA$

とすれば，

$\BA\D\begin{cases}&\D\int_0^1 P_{2n}^{}(x)f(x)\,dx={(-1)}^n\beta_n^3\sum_{n< m}\frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{2}{\pi}H+\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{\pi}{2}C-\frac{2}{\pi}{(-1)}^m\beta_m^3\sum_{m< l}\frac{\D(-1)^{l-1}(4l-1)}{\D(2l\beta_l^{})^3}\R)\R)\\&\D\int_0^1 P_{2n-1}^{}(x)f(x)\,dx=-\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{2}{\pi}H-{(-1)}^n\beta_n^3\sum_{n< m} \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{\pi}{2}C-\frac{2}{\pi}\frac{\D(-1)^{m-1}}{\D(2m\beta_m^{})^3}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3\R)\R)\end{cases}\EA$
$\BA\D\\&f(x)=\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)\sum_{n< m}\frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\L(\frac{2}{\pi}H+\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{n\le m}{(-1)}^m(4m+1)\beta_m^3\L(\frac{\pi}{2}C-\frac{2}{\pi}{(-1)}^m\beta_m^3\sum_{m< l}\frac{\D(-1)^{l-1}(4l-1)}{\D(2l\beta_l^{})^3}\R)\R)\\&\hat{f}(x)=-\frac{2}{\pi}\sum_{n=1}^\infty\frac{\D(-1)^{n-1}(4n-1)}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^6\sum_{l=1}^m\frac{4l-1}{\D(2l\beta_l^{})^6}\sum_{k=0}^{l-1}{(-1)}^k(4k+1)\beta_k^3P_{2k}^{}(x)\EA$

となります．
$\hspace{5pt}$次に，$x\to\sqrt{1-x^2}$とした場合のFL展開を計算します．
まず，$\D h(x)=f(\sqrt{1-x^2}),\,j(x)=g(\sqrt{1-x^2})$とし，微分方程式は

$\BA\D\L(\frac{d}{dx}x(1-x^2)\frac{d}{dx}-x\R)h(x)=\frac{x}{\sqrt{1-x^2}}j(x)\EA$

であり，以下の場合分けもします．

$\BA\D\begin{cases}&\D j(x)=\sum_{n=1}^\infty g_n^{}\L(\sqrt{1-x^2}\R)^{2n-1},&\D h(x)=\sum_{n=0}^\infty \beta_n^2{(1-x^2)}^n\sum_{m=1}^n \frac{\D g_m^{}}{\D(2m\beta_m^{})^2}&({\color{red}{\rm C}})\\&\D j(x)=\sum_{n=0}^\infty g_n^{}{(1-x^2)}^n,&\D h(x)=\sum_{n=1}^\infty \frac{\L(\sqrt{1-x^2}\R)^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_m^2g_m^{}&({\color{red}{\rm D}})\end{cases}\EA$

また，$\D \lim_{x\to1}x(1-x^2)h(x)\frac{d}{dx}P_k^{}(x)=\lim_{x\to1}x(1-x^2)P_k^{}(x)\frac{d}{dx}h(x)=\lim_{x\to0}x(1-x^2)h(x)\frac{d}{dx}P_k^{}(x)=0$，

$\BA\D\begin{cases}&\D ({\color{red}{\rm C}})\,{\rm の場合：}\lim_{x\to0}x(1-x^2)P_k^{}(x)\frac{d}{dx}h(x)=P_k^{}(0)\lim_{x\to0}xh'(x)=-\frac{2}{\pi}P_k^{}(0)\sum_{n=1}^\infty \frac{\D g_n^{}}{\D(2n\beta_n^{})^2}\\&\D ({\color{red}{\rm D}})\,{\rm の場合：}\lim_{x\to0}x(1-x^2)P_k^{}(x)\frac{d}{dx}h(x)=P_k^{}(0)\lim_{x\to0}xh'(x)=-\frac{\pi}{2}P_k^{}(0)\sum_{n=0}^\infty \beta_n^2g_n^{}\end{cases}\EA$

であるので，

$\BA\D\begin{cases}&\D \int_0^1 P_{2n}^{}(x)h(x)\,dx={(-1)}^n\beta_n^3\L(\int_0^1 h(x)\,dx+\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}\int_0^1 \frac{xP_{2m-1}^{}(x)j(x)}{\sqrt{1-x^2}}\,dx\R)\\&\D\int_0^1 P_{2n-1}^{}(x)h(x)\,dx=\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}{(-1)}^m(4m+1)\beta_m^3\L(-{(-1)}^m\beta_m^{}\lim_{x\to0}xh'(x)-\int_0^1 \frac{xP_{2m}^{}(x)j(x)}{\sqrt{1-x^2}}\,dx\R)\end{cases}\EA$

となります．$\D\int_0^1 \frac{xP_{n}^{}(x)j(x)}{\sqrt{1-x^2}}\,dx$をうまく計算するために制限がかかりそうです．

$\underline{{\bf 1.}\,g_n^{}=\beta_n^{}{\,\rm の場合}}$

$\BA\D j(x)=\sum_{n=0}^\infty \beta_n^{}{(1-x^2)}^n=\frac{1}{x},\qquad h(x)=\sum_{n=1}^\infty \frac{\L(\sqrt{1-x^2}\R)^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_m^3\EA$

とすれば，

$\BA\D\begin{cases}&\D \int_0^1 P_{2n}^{}(x)h(x)\,dx={(-1)}^n\beta_n^3\L(\int_0^1 h(x)\,dx+\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^5}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3\R)\\&\D\int_0^1 P_{2n-1}^{}(x)h(x)\,dx=\frac{\pi}{2}\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^4\L(A-{(-1)}^m\beta_m\R)\end{cases}\EA$
$\BA\D\\h(x)=\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)\L(B+\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^5}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3\R)\EA$

$\underline{{\bf 2.}\,\D g_n^{}=\frac{1}{2n\beta_n^{}}{\,\rm の場合}}$

$\BA\D j(x)=\sum_{n=1}^\infty \frac{\L(\sqrt{1-x^2}\R)^{2n-1}}{2n\beta_n^{}}=\frac{1}{x}\L(\frac{\pi}{2}-\sin^{-1}x\R),\qquad h(x)=\sum_{n=0}^\infty \beta_n^2{(1-x^2)}^n\sum_{m=1}^{n} \frac{1}{\D(2m\beta_m^{})^3},\qquad \hat{h}(x)=\sum_{n=1}^\infty \frac{1}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}\beta_m^2{(1-x^2)}^m\EA$

とすれば，

$\BA\D\begin{cases}&\D \int_0^1 P_{2n}^{}(x)h(x)\,dx={(-1)}^n\beta_n^3\L(\int_0^1 h(x)\,dx+\frac{\pi}{2}\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^5}\sum_{m\le l}{(-1)}^l(4l+1)\beta_l^3\R)\\&\D\int_0^1 P_{2n-1}^{}(x)h(x)\,dx=\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^4\L(\frac{2}{\pi}B-{(-1)}^m\beta_m\sum_{m< l}\frac{\D(-1)^{m-1}(4m-1)}{\D(2m)^2(2m-1)\beta_m^{}}\R)\\&\D\int_0^1 P_{2n-1}^{}(x)\hat{h}(x)\,dx=\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}{(-1)}^m(4m+1)\beta_m^5\sum_{m< l}\frac{\D(-1)^{m-1}(4m-1)}{\D(2m)^2(2m-1)\beta_m^{}}\end{cases}\EA$
$\BA\D\\h(x)=\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)\L(\int_0^1 h(x)\,dx+\frac{\pi}{2}\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^5}\sum_{m\le l}{(-1)}^l(4l+1)\beta_l^3\R)\EA$

$\underline{{\bf 3.}\,\D g_n^{}=\gamma_{n-1}^{(1)}{\,\rm の場合}}$

$\BA\D j(x)=\sum_{n=1}^\infty \gamma_{n-1}^{(1)}\L(\sqrt{1-x^2}\R)^{2n-1}=\frac{\sqrt{1-x^2}}{x}\sum_{n=0}^\infty \beta_n^2{(1-x^2)}^n,\qquad h(x)=\sum_{n=0}^\infty \beta_n^2\L(1-x^2\R)^n\sum_{m=1}^n \frac{\gamma_{m-1}^{(1)}}{\D(2m\beta_m^{})^2},\qquad \hat{h}(x)=\sum_{n=1}^\infty\frac{\gamma_{n-1}^{(1)}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_m^2{(1-x^2)}^m\EA$

とすれば，

$\BA\D\begin{cases}&\D \int_0^1 P_{2n}^{}(x)h(x)\,dx={(-1)}^n\beta_n^3\L(\int_0^1 h(x)\,dx+\frac{2}{\pi}\sum_{m=1}^n \frac{4m-1}{\D(2m\beta_m^{})^6}\sum_{l=0}^{m-1}(4l+1)\beta_l^4\R)\\&\D\int_0^1 P_{2n-1}^{}(x)h(x)\,dx=\frac{\pi}{2}\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^4\L(C-\beta_m^2\R)\\&\D\int_0^1 P_{2n-1}^{}(x)\hat{h}(x)\,dx=\frac{\pi}{2}\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^6\end{cases}\EA$
$\BA\D\\h(x)=\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)\L(\int_0^1 h(x)\,dx+\frac{2}{\pi}\sum_{m=1}^n \frac{4m-1}{\D(2m\beta_m^{})^6}\sum_{l=0}^{m-1}(4l+1)\beta_l^4\R)\EA$

$\underline{{\bf 4.}\,\D g_n^{}=\gamma_{n}^{(2)}{\,\rm の場合}}$

$\BA\D j(x)=\sum_{n=0}^\infty \gamma_{n}^{(2)}{(1-x^2)}^n=\frac{\sqrt{1-x^2}}{x}\sum_{n=1}^\infty \frac{\L(\sqrt{1-x^2}\R)^{2n-1}}{\D(2n\beta_n^{})^2},\qquad h(x)=\sum_{n=1}^\infty \frac{\L(\sqrt{1-x^2}\R)^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_m^2\gamma_m^{(2)}\EA$

$\BA\D\begin{cases}&\D \int_0^1 P_{2n}^{}(x)h(x)\,dx={(-1)}^n\beta_n^3\L(\int_0^1 h(x)\,dx+\sum_{m=1}^n \frac{4m-1}{\D(2m\beta_m^{})^6}\L(\frac{\pi}{2}\sum_{m=0}^{n-1}(4m+1)\beta_m^4-\sum_{m=0}^{n-1}\frac{\D(-1)^m\beta_m^{}}{2m+1}\sum_{l=0}^{m}{(-1)}^l(4l+1)\beta_l^3+\sum_{m=1}^{n-1}\frac{\D(-1)^{m-1}\beta_m^{}}{2m}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3\R)\R)\\&\D\int_0^1 P_{2n-1}^{}(x)h(x)\,dx=\frac{\pi}{2}\frac{\D(-1)^{n-1}}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^4\L(J-\beta_m^2\L(\frac{\pi^2}{8}+\sum_{l=1}^m\frac{(-1)^{l-1}(4l-1)}{\D(2l)^2(2l-1)\beta_l^{}}\R)\R)\end{cases}\EA$

となります．
$\hspace{5pt}$ここまでをまとめると

$f(x)$FL展開
$\D\sum_{n=0}^\infty \beta_n^{}x^{2n}$$\D\frac{\pi}{2}\sum_{n=0}^\infty (4n+1)\beta_n^2P_{2n}^{}(x)$
$\D\sum_{n=1}^\infty \frac{x^{2n-1}}{2n\beta_n^{}}$$\D\sum_{n=1}^\infty \frac{4n-1}{\D(2n\beta_n^{})^2}P_{2n-1}^{}(x)$
$\D\sum_{n=0}^\infty \beta_n^2x^{2n}$$\D\frac{2}{\pi}\sum_{n=1}^\infty\frac{\D(-1)^{n-1}(4n-1)}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}{(-1)}^m(4m+1)\beta_m^3P_{2m}^{}(x)$
$\D\sum_{n=1}^\infty \frac{x^{2n-1}}{\D(2n\beta_n^{})^2}$$\D-\frac{\pi}{2}\sum_{n=0}^\infty{(-1)}^n(4n+1)\beta_n^3\sum_{m=1}^n\frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^3}P_{2m-1}^{}(x)$
$\D\sum_{n=0}^\infty \beta_n^3\sum_{m=1}^n\frac{x^{2m-1}}{\D(2m\beta_m^{})^2}$$\D-\frac{\pi}{2}\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^5\sum_{m=1}^n\frac{\D(-1)^{m-1}(4m-1)P_{2m-1}^{}(x)}{\D(2m\beta_m^{})^3}$
$\D\sum_{n=1}^\infty \frac{1}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}\beta_m^2{x}^{2m}$$\D\sum_{n=1}^\infty \frac{\D(-1)^{n-1}(4n-1)}{\D(2n\beta_n^{})^5}\sum_{m=0}^{n-1}{(-1)}^m(4m+1)\beta_m^3P_{2m}^{}(x)$
$\D\sum_{n=0}^\infty \beta_n^4\sum_{m=1}^n \frac{\D x^{2m-1}}{\D(2m\beta_m^{})^2}$$\D-\frac{2}{\pi}\sum_{n=1}^\infty \frac{\D(-1)^{n-1}(4n-1)}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^6\sum_{l=1}^m\frac{\D(-1)^{l-1}(4l-1)P_{2l-1}^{}(x)}{\D(2l\beta_l^{})^3}$
$\D\sum_{n=1}^\infty \frac{1}{\D(2n\beta_n^{})^4}\sum_{m=0}^{n-1}\beta_m^2{x}^{2m}$$\D-\frac{\pi}{2}\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3\sum_{m=1}^n\frac{4m-1}{\D(2m\beta_m^{})^6}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3P_{2l}^{}(x)$
$\D\sum_{n=1}^\infty \frac{1}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}\beta_m^4\sum_{l=1}^m\frac{x^{2l-1}}{\D(2l\beta_l^{})^2}$$\D-\sum_{n=1}^\infty\frac{\D(-1)^{n-1}(4n-1)}{\D(2n\beta_n^{})^5}\sum_{m=0}^{n-1}(4m+1)\beta_m^6\sum_{l=1}^m\frac{\D(-1)^{l-1}(4l-1)P_{2l-1}^{}(x)}{\D(2l\beta_l^{})^3}$
$\D\sum_{n=0}^\infty \beta_n^3\sum_{m=1}^n\frac{1}{\D(2m\beta_m^{})^4}\sum_{l=0}^{m-1}\beta_l^2x^{2l}$$\D-\frac{\pi}{2}\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^5\sum_{m=1}^n\frac{4m-1}{\D(2m\beta_m^{})^6}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3P_{2l}^{}(x)$
$\D\sum_{n=1}^\infty\frac{1}{\D(2n\beta_n^{})^4}\sum_{m=0}^{n-1}\beta_m^4\sum_{l=1}^m\frac{x^{2l-1}}{\D(2l\beta_l^{})^2}$$\D\frac{\pi}{2}\sum_{n=0}^\infty{(-1)}^n(4n+1)\beta_n^3\sum_{m=1}^n\frac{4m-1}{\D(2m\beta_m^{})^6}\sum_{l=0}^{m-1}(4l+1)\beta_l^6\sum_{k=1}^l\frac{\D(-1)^{k-1}(4k-1)P_{2k-1}(x)}{\D(2k\beta_k^{})^3}$
$\D\sum_{n=0}^\infty \beta_n^4\sum_{m=1}^n\frac{1}{\D(2m\beta_m^{})^4}\sum_{l=0}^{m-1}\beta_l^2x^{2l}$$\D-\frac{2}{\pi}\sum_{n=1}^\infty\frac{\D(-1)^{n-1}(4n-1)}{\D(2n\beta_n^{})^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^6\sum_{l=1}^m\frac{4l-1}{\D(2l\beta_l^{})^6}\sum_{k=0}^{l-1}{(-1)}^k(4k+1)\beta_k^3P_{2k}^{}(x)$
$\D\sum_{n=0}^\infty \beta_n^2{(1-x^2)}^n$$\D \frac{\pi}{2}\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)$
$\D \sum_{n=1}^\infty \frac{\L(\sqrt{1-x^2}\R)^{2n-1}}{\D(2n\beta_n^{})^2}\sum_{m=0}^{n-1}\beta_m^3$$\D\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)\L(B+\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^5}\sum_{l=0}^{m-1}{(-1)}^l(4l+1)\beta_l^3\R)$
$\D\sum_{n=0}^\infty \beta_n^2{(1-x^2)}^n\sum_{m=1}^{n} \frac{1}{\D(2m\beta_m^{})^3}$$\D\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)\L(\int_0^1 f(x)\,dx+\frac{\pi}{2}\sum_{m=1}^n \frac{\D(-1)^{m-1}(4m-1)}{\D(2m\beta_m^{})^5}\sum_{m\le l}{(-1)}^l(4l+1)\beta_l^3\R)$
$\D\sum_{n=0}^\infty \beta_n^2\L(1-x^2\R)^n\sum_{m=1}^n \frac{\gamma_{m-1}^{(1)}}{\D(2m\beta_m^{})^2}$$\D\sum_{n=0}^\infty {(-1)}^n(4n+1)\beta_n^3P_{2n}^{}(x)\L(\int_0^1 f(x)\,dx+\frac{2}{\pi}\sum_{m=1}^n \frac{4m-1}{\D(2m\beta_m^{})^6}\sum_{l=0}^{m-1}(4l+1)\beta_l^4\R)$

となります．

$\textrm{4乗のFourier-Legendre\,展開}$

$\hspace{5pt}$ $({\bf\color{red}{1}})$式を受けて，$4$乗となる式もどうなるか気になりましたので計算してみました．

$\BA\D&(2n+1)xP_n^{}(x)=(n+1)P_{n+1}^{}(x)+nP_{n-1}^{}(x)&[1]\\&(2n+1)(1-x^2)\frac{d}{dx}P_n^{}(x)=n(n+1)\L(P_{n-1}^{}(x)-P_{n+1}^{}(x)\R)&[2]\\&\frac{d}{dx}(1-x^2)\frac{d}{dx}P_n^{}(x)=-n(n+1)P_n^{}(x)&[3]\EA$

$[2],\,[3]$式より

$\BA\D(2n+1)(1-x^2)\frac{d^2}{dx^2}(1-x^2)\frac{d}{dx}P_n^{}(x)=-(n^4+2n^3+n^2)P_{n-1}^{}(x)+(n^4+2n^3+n^2)P_{n+1}^{}(x)\EA$

$[1],\,[3]$式より

$\BA\D(2n+1)x\frac{d}{dx}(1-x^2)\frac{d}{dx}P_n^{}(x)=-(n^3+n^2)P_{n-1}^{}(x)-(n^3+2n^2+n)P_{n+1}^{}(x)\EA$

以上の式を適当に足し引きして整理すれば

$\BA\D\L(\frac{d}{dx}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}-(1-x^2)\frac{d}{dx}+x\R)P_n^{}(x)=\frac{1}{2n+1}\L({(n+1)}^4P_{n+1}^{}(x)-{n}^4P_{n-1}^{}(x)\R)\EA$

となりました．また，これにより

$\BA\D\begin{cases}&\D\int_0^1 f(x)P_{2n-1}^{}(x)\,dx=\frac{1}{\D(2n\beta_n^{})^4}\sum_{m=0}^{n-1}(4m+1)\beta_m^4\L(\lim_{x\to1}D_{2m}^{}(x)-\lim_{x\to0}D_{2m}^{}(x)-\int_0^1 g(x)P_{2m}^{}(x)\,dx\R)\\&\D\int_0^1 f(x)P_{2n}^{}(x)\,dx=\beta_n^4\L(\int_0^1 f(x)\,dx+\sum_{m=1}^n\frac{4m-1}{\D(2m\beta_m^{})^4}\L(\lim_{x\to1}D_{2m-1}^{}(x)-\lim_{x\to0}D_{2m-1}^{}(x)-\int_0^1 g(x)P_{2m-1}^{}(x)\,dx\R)\R)\end{cases}\EA$

$\BA\D D_n^{}(x)\coloneqq(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}P_n^{}(x)\cdot f(x)-(1-x^2)\frac{d}{dx}P_n^{}(x)\cdot(1-x^2)\frac{d}{dx}f(x)+P_n^{}(x)\cdot(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}f(x)-(1-x^2)P_n^{}(x)f(x)\EA$

$\BA\D g(x)\coloneqq\L(\frac{d}{dx}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}-(1-x^2)\frac{d}{dx}+x\R)f(x)\qquad [\spadesuit]\EA$

となりました．また，$[\spadesuit]$式の解として

$\BA\D f(x)=\frac{2}{\pi^2}\int_0^x \L(K\L(\sqrt{\frac{1-x}{2}}\R)K\L(\sqrt{\frac{1+t}{2}}\R)-K\L(\sqrt{\frac{1+x}{2}}\R)K\L(\sqrt{\frac{1-t}{2}}\R)\R)^2g(t)\,dt\EA$

があることがわかりました．これをみると

$\BA\D\lim_{x\to1}D_{n}^{}(x)=\frac{4}{\pi^2}\int_0^1 K\L(\sqrt{\frac{1-t}{2}}\R)^2g(t)\,dt,\qquad \lim_{x\to0}D_{n}^{}(x)=0\EA$

となりそうです．$f(x)$は奇関数なので

$\BA\D f(x)=\sum_{n=1}^\infty \frac{4n-1}{\D(2n\beta_n^{})^4}P_{2n-1}^{}(x)\sum_{m=0}^{n-1}(4m+1)\beta_m^4\L(\frac{4}{\pi^2}\int_0^1 K\L(\sqrt{\frac{1-t}{2}}\R)^2g(t)\,dt-\int_0^1 g(t)P_{2m}^{}(t)\,dt\R)\EA$

となります．

$\D \bar{f}(x)={\rm sign}(x)f(x)$

$\hspace{5pt}$ところで，$\bar{f}(x)={\rm sign}(x)f(x)$と定義するならば，$f(x)$と$\bar{f}(x)$の偶奇は逆になり，また$0< x<1$において両者は一致することにより

$\BA\D f(x)=\sum_{n=0}^\infty (4n+1)P_{2n}^{}(x)\int_0^1 f(t)P_{2n}^{}(t)\,dt=\sum_{n=1}^\infty (4n-1)P_{2n-1}^{}(x)\int_0^1 f(t)P_{2n-1}^{}(t)\,dt=\bar{f}(x)\qquad (0< x<1)\EA$

と書けるのではないでしょうか．

$\hspace{5pt}$

$\BA\D\EA$

投稿日：9月23日
更新日：10月9日

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