公式書付

${\displaystyle \int x^{n}f(x)dx}$

${\displaystyle {\int }_0^1\frac{(4x(1-x){)}^n}{\sqrt{1+\sqrt{x}}}\kappa \left(\frac{2\sqrt{x}}{1+\sqrt{x}}\right)dx=\frac{8\sqrt{2}}{3\pi }\frac{n{!}^2}{{(\frac{9}{8},\frac{11}{8})}_n^{}}}$

${\displaystyle {\int }_0^1\frac{(2t-1{)}^{2n+1}\kappa (t{)}^2}{\sqrt{t(1-t)}}dt=\pi \sum _{k=0}^n{\beta }_k^{}{\beta }_{n-k}^{}\sum _{j=0}^k{\beta }_j^2{\beta }_{k-j}}$

${\displaystyle {\int }_0^1\kappa (x{)}^2\kappa (1-x){(4x(1-x))}^{n}dx=\frac{\pi }{3}{\mathcal{A}}^2{\left(\frac{n!}{{\left(\frac{3}{4}\right)}_n^{}}\right)}^4[x^n]\kappa {\left(\frac{1-\sqrt{1-x}}{2}\right)}^3}$

$\begin{array}{rl}{\displaystyle {\int }_0^1\frac{x^{2n}}{\sqrt{1+x}}\kappa \left(\frac{1-x}{1+x}\right)dx}& =\frac{{\beta }_n^{}}{(4n+1){\beta }_{2n}^{}}\\ {\int }_0^1\frac{x^{2n-1}}{\sqrt{1+x}}\kappa \left(\frac{1-x}{1+x}\right)dx& =\frac{1}{\pi }\frac{1}{(2n{)}^2{\beta }_n^{}{\beta }_{2n}^{}}\end{array}$

$\begin{array}{r}{\displaystyle {\int }_0^1(\kappa \left(\frac{1-\sqrt{1-x}}{2}\right)\kappa {\left(\frac{1+\sqrt{1-x}}{2}\right)}^3-\kappa {\left(\frac{1-\sqrt{1-x}}{2}\right)}^3\kappa \left(\frac{1+\sqrt{1-x}}{2}\right))x^{n-1}dx=\pi \sum _{k=0}^n{\beta }_k^3{\beta }_{n-k}^3}\end{array}$

${\displaystyle \int p_n^{}(x)f(x)dx}$

$\begin{array}{rl}{\displaystyle }& {\int }_0^1\frac{P_{2n}^{}(x)}{\sqrt{1+x}}\kappa \left(\frac{1-x}{1+x}\right)dx=\frac{(-1{)}^n{\beta }_n^{}}{4n+1}\\ & {\int }_0^1\frac{P_{2n-1}^{}(x)}{\sqrt{1+x}}\kappa \left(\frac{1-x}{1+x}\right)dx=\frac{2}{\pi }\frac{(-1{)}^{n-1}}{n(4n-1){\beta }_n^{}}\sum _{k=0}^{n-1}{\beta }_k^2\end{array}$

${\displaystyle {\int }_0^{1}K(x{)}^2(U_{4n+1}^{}(x)-2x^{2n+1})dx=0}$

${\displaystyle {\int }_0^{1}K(x)K^{\prime }(x{)}^2(U_{4n}^{}(x)-3x^{2n+1})dx=0}$

${\displaystyle {\int }_0^1(2K(x{)}^2-K^{\prime }(x{)}^2)x^{2n}dx={\int }_0^{1}K(x{)}^2{U}_{4n-1}^{}(x)dx}$

${\displaystyle {\int }_0^1\frac{K^{\prime }(x{)}^2}{K(x{)}^2+K^{\prime }(x{)}^2}{x}^{2n-1}dx+4{\int }_0^1\frac{K(x)K^{\prime }(x)}{K(x{)}^2+K^{\prime }(x{)}^2}\frac{T_{4n}^{}(x)}{\sqrt{1-x^2}}dx=0}$

${\displaystyle {({\beta }_0^{}{\beta }_n^{}+{\beta }_1^{}{\beta }_{n-1}^{}x+\cdots +{\beta }_n^{}{\beta }_0^{}{x}^n)}^2=p_0^{}+p_1^{}x+\cdots +p_{2n}^{}{x}^{2n}}$とき${\displaystyle {\int }_{-1}^1{P}_m^{}(x{)}^2{U}_n^{}(x)dx=\frac{p_0^{}}{m-n+\frac{1}{2}}+\frac{p_1^{}}{m-n+\frac{3}{2}}+\cdots +\frac{p_{2n}^{}}{m+n+\frac{1}{2}}}$

${\displaystyle {\int }_0^{1}K(\sqrt{x})P_m^{}(2x-1)P_n^{}(2x-1)dx=2{\left(\frac{\mathrm{\Gamma }\left(\frac{2m+2n+3}{4}\right)\mathrm{\Gamma }\left(\frac{2m-2n+1}{4}\right)}{(2m+2n+1)\mathrm{\Gamma }\left(\frac{2m+2n+1}{4}\right)\mathrm{\Gamma }\left(\frac{2m-2n+3}{4}\right)}\right)}^2}$

${\displaystyle {\int }_0^1{P}_m^{}(1-2x)P_n^{}(1-2x)\frac{dx}{\sqrt{x}}={\int }_{-1}^1{P}_m^{}(x{)}^2{U}_{2n}^{}(x)dx}$

${\displaystyle {\int }_{-1}^1\kappa \left(\frac{1-x}{2}\right)\kappa {\left(\frac{1+x}{2}\right)}^3{U}_{2n+1}^{}(x)dx=\pi [x^n]\kappa (x{)}^4}$

${\displaystyle {\int }_0^1\frac{K^{\prime }(x)}{K(x{)}^2+K^{\prime }(x{)}^2}{U}_{4n}^{}(x)dx=\frac{{\delta }_{0,n}^{}}{4}}$

$\begin{array}{rl}{\displaystyle }& {\int }_0^1\frac{Q_{2n}^{}(x)T_{2n}^{}(x)}{\sqrt{1-x^2}}dx={\beta }_{2n}^{}(2G-\frac{1}{2}\sum _{k=0}^{n-1}\frac{1}{(2k+1{)}^2{\beta }_k^{}{\beta }_{2k+1}^{}}+\frac{1}{2}\sum _{k=1}^n\frac{1}{(2k{)}^2{\beta }_k^{}{\beta }_{2k}^{}})=\frac{{\beta }_{2n}^{}}{2{\beta }_n^{}}{\int }_0^1{(4x(1-x))}^{n-\frac{1}{2}}{\tanh }^{-1}\sqrt{x}dx\\ & {\int }_0^1\frac{Q_{2n+1}^{}(x)T_{2n+1}^{}(x)}{\sqrt{1-x^2}}dx={\beta }_{2n+1}^{}(2G-\frac{1}{2}\sum _{k=0}^n\frac{1}{(2k+1{)}^2{\beta }_k^{}{\beta }_{2k+1}^{}}+\frac{1}{2}\sum _{k=1}^n\frac{1}{(2k{)}^2{\beta }_k^{}{\beta }_{2k}^{}})\end{array}$

${\displaystyle 2{\int }_0^1\frac{P_n^{}(x)Q_n^{}(x)}{\sqrt{1-x^2}}dx={\int }_0^1\frac{K(\sqrt{1-x})}{\sqrt{1-x}}{P}_n^{}(2x-1)dx}$

${\displaystyle \int f(tx)g(x)dx}$

${\displaystyle {\int }_0^1\frac{\kappa \left(\frac{2tx}{1+tx}\right)}{\sqrt{x(1-x)}\sqrt{1-t^2{x}^2}}dx=\frac{\pi }{\sqrt{1-t}}\sum _{n=0}^{\mathrm{\infty }}{\beta }_n^3{\left(\frac{2t}{t-1}\right)}^n}$

${\displaystyle {\int }_0^1\frac{(2x-1)\kappa (x{)}^3\kappa (1-x)}{1-t^2(2x-1{)}^2}dx=\frac{\pi }{4}{\left(\frac{1}{\sqrt{1+t}}\kappa \left(\frac{2t}{1+t}\right)\right)}^4}$

${\displaystyle {\int }_0^1\frac{K^{\prime }(x)}{K(x{)}^2+K^{\prime }(x{)}^2}\frac{1}{1-t^2{x}^2}dx=\frac{1}{t^2}(1-\frac{\pi }{2}\frac{1}{K(t)})}$

${\displaystyle {\int }_0^1\frac{K(x)}{K(x{)}^2+K^{\prime }(x{)}^2}\frac{1}{x\sqrt{1-x^2}(1-t^2{x}^2)}dx=\frac{\pi }{2}\frac{1}{\sqrt{1-t^2}K(t)}}$

${\displaystyle \{\begin{array}{ll}& {\displaystyle \mathrm{Re}{\int }_0^1\frac{x^{n-2m}}{(1-t^2{x}^2)\sqrt{1-x^2}}{(K(x)+\sqrt{-1}{K}^{\prime }(x))}^{n}dx=\frac{\pi }{2}\frac{t^{2m}}{\sqrt{1-t^2}}K(t{)}^n(m,n\in {\mathbb{Z}}_{\ge 0}^{},2m\le n)}\\ & {\displaystyle \mathrm{Im}{\int }_0^1\frac{x^{n-1-2m}}{1-t^2{x}^2}{(K(x)+\sqrt{-1}{K}^{\prime }(x))}^{n}dx=\frac{\pi }{2}{t}^{2m}K(t{)}^n(m,n\in {\mathbb{Z}}_{\ge 0}^{},2m\le n-1)}\end{array}}$

$\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{e}\mathrm{s}$

${\displaystyle \frac{1}{\sqrt{1-x}}\sum _{n=0}^{\mathrm{\infty }}{\beta }_n^4{\left(\frac{x}{x-1}\right)}^n=\sum _{n=0}^{\mathrm{\infty }}{\beta }_n^{}{x}^n\sum _{k=0}^n{\beta }_k^{}{\beta }_{n-k}^{}\sum _{j=0}^k{\beta }_j^2{\beta }_{k-j}}$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{t}^n{P}_n^{}(x)P_n^{}(y)=\frac{1}{\sqrt{1+2t(\sqrt{1-x^2}\sqrt{1-y^2}-xy)+t^2}}\kappa \left(\frac{4t\sqrt{1-x^2}\sqrt{1-y^2}}{1+2t(\sqrt{1-x^2}\sqrt{1-y^2}-xy)+t^2}\right)}$

記法

$\begin{array}{rl}{\displaystyle }& {\beta }_r^{}=2^{-2r}(\genfrac{}{}{0ex}{}{2r}{r})\\ & \kappa (x)=\sum _{n=0}^{\mathrm{\infty }}{\beta }_n^2{x}^n\\ & K(x)=\frac{\pi }{2}\kappa (x^2)\\ & K^{\prime }(x)=K(\sqrt{1-x^2})\\ & P_n^{}(x):n\text{次第一種Legendre多項式}\\ & Q_n^{}(x):n\text{次第二種Legendre多項式}\\ & T_n^{}(x):n\text{次第一種Chebyshev多項式}\\ & U_n^{}(x):n\text{次第二種Chebyshev多項式}\\ & \mathcal{A}=\frac{\pi }{\mathrm{\Gamma }{\left(\frac{3}{4}\right)}^4}\\ & G=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}\end{array}$