公式書付

$\tiny\D\int f(x)\,dx$

$\hspace{5pt}\D \int_0^1 \frac{K(x)^2K'(x)^3}{K(x)^2+K'(x)^2}\,dx=\frac{1}{40}\pi^4A^2 $

$\hspace{5pt}\D \int_0^1 \frac{K'(x)^9}{K(x)^2+K'(x)^2}(1-x^2-4x^4)\,dx=\frac{8}{15}\pi^8A^4 $

$\hspace{5pt}\D \int_0^1 \frac{K(x)^2K'(x)^7}{K(x)^2+K'(x)^2}(1-x^2-4x^4)\,dx=\frac{7}{240}\pi^8A^4 $

$\hspace{5pt}\D \int_0^1 K'(x)^7(1-x^2-4x^4)\,dx=\frac{9}{16}\pi^8A^4 $

$\hspace{5pt}\D \int_0^1 \frac{K(x)^6K'(x)^7}{K(x)^2+K'(x)^2}(1-2x^2+93x^4-92x^6+16x^8)\,dx=\frac{29}{16640}\pi^{12}A^6 $

$\hspace{5pt}\D \int_0^1 \frac{K(x)^2K'(x)^{11}}{K(x)^2+K'(x)^2}(1-2x^2+93x^4-92x^6+16x^8)\,dx=\frac{267}{520}\pi^{12}A^6 $

$\hspace{5pt}\D \int_0^1 \frac{K(x)^8K'(x)^9}{K(x)^2+K'(x)^2}(1-3x^2+1291x^4-2577x^6+2136x^8-848x^{10}-64x^{12})\,dx=\frac{31}{43520}\pi^{16}A^8 $

$\hspace{5pt}\D \int_0^1 \frac{K(x)^4K'(x)^{13}}{K(x)^2+K'(x)^2}(1-3x^2+1291x^4-2577x^6+2136x^8-848x^{10}-64x^{12})\,dx=\frac{1013}{1360}\pi^{16}A^8 $

$\hspace{5pt}\D \int_0^1 \frac{K'(x)^4}{K(x)^2+K'(x)^2}x\,dx=\zeta(3) $

$\hspace{5pt}\D \int_0^1 \frac{K'(x)^6}{K(x)^2+K'(x)^2}x(1-2x^2)\,dx=\frac{39}{8}\zeta(5) $

$\hspace{5pt}\D \int_0^1 \frac{K'(x)^8}{K(x)^2+K'(x)^2}x(2-17x^2+17x^4)\,dx=\frac{675}{8}\zeta(7) $

$\hspace{5pt}\D \int_0^1 \frac{K'(x)^{10}}{K(x)^2+K'(x)^2}x(1-2x^2)(1-31x^2+31x^4)\,dx=\frac{79065}{128}\zeta(9) $

$\tiny\D\int x^nf(x)\,dx$

$\hspace{5pt}\D \int_0^1 \frac{(4x(1-x))^n}{\sqrt{1+\sqrt{x}}}\kappa\left(\frac{2\sqrt{x}}{1+\sqrt{x}}\right)\,dx=\frac{8\sqrt{2}}{3\pi}\frac{n!^2}{\left(\frac{9}{8},\frac{11}{8}\right)_n^{}}$

$\hspace{5pt}\D \int_0^1 \frac{(2t-1)^{2n+1}\kappa(t)^2}{\sqrt{t(1-t)}}\,dt=\pi\sum_{k=0}^n \beta_{k}^{}\beta_{n-k}^{}\sum_{j=0}^k \beta_j^2\beta_{k-j}$

$\hspace{5pt}\D \int_0^1 \kappa(x)^2\kappa(1-x)\left(4x(1-x)\right)^n\,dx=\frac{\pi}{3}{\cal A}^2\left(\frac{n!}{{\left(\frac{3}{4}\right)_n^{}}}\right)^4[x^n]\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^3$

$\hspace{5pt}\BA\D \int_0^1 \frac{x^{2n}}{\sqrt{1+x}}\kappa\left(\frac{1-x}{1+x}\right)dx&=\frac{\beta_n^{}}{(4n+1)\beta_{2n}^{}}\\ \int_0^1 \frac{x^{2n-1}}{\sqrt{1+x}}\kappa\left(\frac{1-x}{1+x}\right)dx&=\frac{1}{\pi}\frac{1}{(2n)^2\beta_n^{}\beta_{2n}^{}}\EA$

$\hspace{5pt}\D \int_0^1 \frac{x^{2n+1}}{1+x}\kappa\L(\frac{1-x}{1+x}\R)\kappa\L(\frac{2x}{1+x}\R)\,dx=\frac{1}{(2n+1)\beta_n}\sum_{k=0}^n\beta_k^2\beta_{n-k}^{} $

$\hspace{5pt}\BA\D \int_0^1 \left(\kappa\L(\frac{1-\sqrt{1-x}}{2}\R)\kappa\L(\frac{1+\sqrt{1-x}}{2}\R)^3-\kappa\L(\frac{1-\sqrt{1-x}}{2}\R)^3\kappa\L(\frac{1+\sqrt{1-x}}{2}\R)\right)x^{n-1}\,dx=\pi\sum_{k=0}^n \beta_k^3\beta_{n-k}^3 \EA$

$\hspace{5pt}\D q_n:=\sum_{k=0}^n\beta_k^3\beta_{n-k}^3;\quad \int_0^1 (2x-1)\L(4x(1-x)\R)^nK(\sqrt{x})^4\,dx=\frac{3}{8}q_n\!\L(31\zeta(5)-\sum_{k=1}^n\frac{1}{k^5q_{k-1}q_k}\R) $

$\hspace{5pt}\D d_n:=\sum_{k=0}^n\beta_k^2\beta_{n-k}^2;\quad \int_0^1 x^{2n+1}K(x)^2\,dx=\frac{1}{4}d_n\!\L(7\zeta(3)+\sum_{k=1}^n\frac{1}{k^3d_{k-1}d_k}\sum_{j=0}^{k-1}d_j\R) $

$\hspace{5pt}\D \int_0^1 K'(x)\!\L(\frac{28\zeta(3)}{\pi^3}K(x)-K'(x)\R)x^{2n+1}\,dx=\frac{1}{4}d_n\sum_{k=1}^n\frac{1}{k^3d_{k-1}d_k} $

$\tiny\D\int p_n^{}(x)f(x)\,dx$

$\hspace{5pt}\BA\D &\int_0^1 \frac{P_{2n}^{}(x)}{\sqrt{1+x}}\kappa\left(\frac{1-x}{1+x}\right)dx=\frac{(-1)^n\beta_n^{}}{4n+1}\\ &\int_0^1 \frac{P_{2n-1}^{}(x)}{\sqrt{1+x}}\kappa\left(\frac{1-x}{1+x}\right)dx=\frac{2}{\pi}\frac{(-1)^{n-1}}{n(4n-1)\beta_n^{}}\sum_{k=0}^{n-1}\beta_k^2\EA$

$\hspace{5pt}\D\int_0^1 K(x)^2\left(U_{4n+1}^{}(x)-2\,x^{2n+1}\right)\,dx=0$

$\hspace{5pt}\D\int_0^1 K(x)K'(x)^2\left(U_{4n}^{}(x)-3\,x^{2n+1}\right)\,dx=0$

$\hspace{5pt}\D\int_0^1 \left(2K(x)^2-K'(x)^2\right)x^{2n}\,dx=\int_0^1 K(x)^2\,U_{4n-1}^{}(x)\,dx$

$\hspace{5pt}\D \int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,x^{2n-1}\,dx + 4\int_0^1 \cfrac{K(x)K'(x)}{K(x)^2+K'(x)^2}\frac{T_{4n}^{}(x)}{\sqrt{1-x^2}}\,dx=0$

$\hspace{5pt}\D \left(\beta_{0}^{}\beta_{n}^{}+\beta_{1}^{}\beta_{n-1}^{}x+\cdots+\beta_{n}^{}\beta_{0}^{}x^{n}\right)^2=p_{0}^{}+p_{1}^{}x+\cdots+p_{2n}^{}x^{2n}$とき$\D\int_{-1}^1 P_m^{}(x)^2\,U_{n}^{}(x)\,dx=\frac{p_0^{}}{m-n+\frac{1}{2}}+\frac{p_1^{}}{m-n+\frac{3}{2}}+\cdots+\frac{p_{2n}^{}}{m+n+\frac{1}{2}}$

$\hspace{5pt}\D \int_0^1 K(\sqrt{x})P_m^{}(2x-1)P_n^{}(2x-1)\,dx=2\left(\frac{\Gamma\left(\cfrac{2m+2n+3}{4}\right)\Gamma\left(\cfrac{2m-2n+1}{4}\right)}{(2m+2n+1)\Gamma\left(\cfrac{2m+2n+1}{4}\right)\Gamma\left(\cfrac{2m-2n+3}{4}\right)}\right)^2$

$\hspace{5pt}\D \int_0^1 K'(x)P_{2m}^{}(x)P_{2n}^{}(x)\,dx=\frac{\pi^2}{4}\frac{(-1)^{m+n}\beta_{n-m}^2\beta_m^{}}{\beta_n^{}}\left(\sum_{k=0}^m \frac{\binom{n}{k}\beta_{m-k}^2\beta_{k+n}^{}}{\binom{m}{k}\binom{m+n}{k+n}}\right)^2\qquad(m\le n)$

$\hspace{5pt}\D\int_0^1 P_m^{}(1-2x)P_n^{}(1-2x)\frac{dx}{\sqrt{x}}=\int_{-1}^1 P_m^{}(x)^2U_{2n}^{}(x)\,dx$

$\hspace{5pt}\D\int_{-1}^1 \kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)^3U_{2n+1}^{}(x)\,dx=\pi[x^{n}]\kappa(x)^4$

$\hspace{5pt}\D \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}U_{4n}^{}(x)\,dx=\frac{\delta_{0,n}^{}}{4}$

$\hspace{5pt}\BA\D &\int_0^1 \frac{Q_{2n}^{}(x)T_{2n}^{}(x)}{\sqrt{1-x^2}}\,dx=\beta_{2n}^{}\left(2G-\frac{1}{2}\sum_{k=0}^{n-1}\cfrac{1}{(2k+1)^2\beta_k^{}\beta_{2k+1}^{}}+\frac{1}{2}\sum_{k=1}^n\cfrac{1}{(2k)^2\beta_k^{}\beta_{2k}^{}}\right)=\frac{\beta_{2n}^{}}{2\beta_n^{}}\int_0^1 \L(4x(1-x)\R)^{n-\frac{1}{2}}\tanh^{-1}\sqrt{x}\,dx\\ &\int_0^1 \frac{Q_{2n+1}^{}(x)T_{2n+1}^{}(x)}{\sqrt{1-x^2}}\,dx=\beta_{2n+1}^{}\left(2G-\frac{1}{2}\sum_{k=0}^{n}\cfrac{1}{(2k+1)^2\beta_k^{}\beta_{2k+1}^{}}+\frac{1}{2}\sum_{k=1}^n\cfrac{1}{(2k)^2\beta_k^{}\beta_{2k}^{}}\right) \EA$

$\hspace{5pt}\D2\int_0^1 \frac{P_n^{}(x)Q_n^{}(x)}{\sqrt{1-x^2}}\,dx=\int_0^1 \frac{K({\small\sqrt{1-x}})}{\sqrt{1-x}}P_n^{}(2x-1)\,dx$

$\hspace{5pt}\D \int_0^1 \kappa(x)\kappa(1-x)P_n(2x-1)^2\,dx=\frac{\pi}{2}\sum_{m=0}^n\beta_m^2\beta_{n-m}^{}\sum_{k=0}^m\beta_k^2\beta_{m-k}^{}\beta_{n-k}^{} $

$\tiny\D\int f(tx)g(x)\,dx$

$\hspace{5pt}\D \int_0^1 \frac{\kappa\left(\frac{2tx}{1+tx}\right)}{\sqrt{x(1-x)}\sqrt{1-t^2x^2}}\,dx=\frac{\pi}{\sqrt{1-t}}\sum_{n=0}^\infty \beta_n^3\left(\frac{2t}{t-1}\right)^n$

$\hspace{5pt}\D \int_0^1 \frac{(2x-1)\kappa(x)^3\kappa(1-x)}{1-t^2(2x-1)^2}\,dx=\frac{\pi}{4}\left(\frac{1}{\sqrt{1+t}}\kappa\left(\frac{2t}{1+t}\right)\right)^4$

$\hspace{5pt}\D \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\frac{1}{1-t^2x^2}\,dx=\frac{1}{t^2}\left(1-\frac{\pi}{2}\frac{1}{K(t)}\right)$

$\hspace{5pt}\D \int_0^1 \cfrac{K(x)}{K(x)^2+K'(x)^2}\frac{1}{x\sqrt{1-x^2}\,(1-t^2x^2)}\,dx=\frac{\pi}{2}\frac{1}{\sqrt{1-t^2}\,K(t)}$

$\hspace{5pt}\D \begin{cases}&\D\Re\int_0^1 \frac{x^{n-2m}}{(1-t^2x^2)\sqrt{1-x^2}}\L(K(x)+\sqrt{-1}K'(x)\R)^n\,dx=\frac{\pi}{2}\frac{t^{2m}}{\sqrt{1-t^2}}K(t)^n\qquad (m,n\in{\mathbb Z_{\ge 0}^{}},2m\le n)\\ &\D\Im\int_0^1 \frac{x^{n-1-2m}}{1-t^2x^2}\L(K(x)+\sqrt{-1}K'(x)\R)^n\,dx=\frac{\pi}{2}t^{2m}K(t)^n\qquad (m,n\in{\mathbb Z_{\ge 0}^{}},2m\le n-1)\end{cases}$

$\hspace{5pt}\D \int_0^1 \L(3K'(x)^3+10K'(x)^2K(x)+6K'(x)K(x)^2\R)\!\!\L(\frac{1}{t^2\L(1-t^2x^2\R)}+\frac{1}{(1-t^2)\L(1-(1-t^2)x^2\R)}\R)\,dx=\frac{5}{4}\frac{\pi^4A^2}{t^2(1-t^2)}+2K'(t)^3+5K'(t)^2K(t)+5K'(t)K(t)^2+2K(t)^3 $

$\hspace{5pt}\D \frac{\pi^3}{192}\frac{\vartheta_2(q)^2}{q^\frac{1}{2}}\!\int_0^1\frac{(1-2x)\L(6\kappa(x)\kappa(1-x)^2-5\kappa(x)^3\R)}{\L(\vartheta_3(q)^4+\vartheta_4(q)^4\R)^2-\vartheta_2(q)^8(1-2x)^2}\,dx=\sum_{n=0}^\infty\L( \frac{1}{(4n+1)^4}\frac{q^{2n}}{1-q^{8n+2}}-\frac{1}{(4n+3)^4}\frac{q^{6n+4}}{1-q^{8n+6}} \R) $

$\hspace{5pt}\D \int_0^\frac{1}{2} {\rm sd}^2\!\L(\frac{\pi}{2}z\L.\sqrt{\kappa(1-t)^2-\kappa(t)^2}\,\R|\,1-t\R)\,dt=z^2\sum_{n=1}^\infty\L(\frac{4}{m\L(m^2-z^2\R)}-\frac{2}{m\L(4m^2-z^2\R)}\R)-\frac{4}{\pi}\sum_{n=1}^\infty \frac{2^{n+1}-1}{\binom{2n}{n}}\zeta(n+1)L_{-4}(n+1)z^{2n}$

$\tiny\rm series$

$\hspace{5pt}\D \frac{1}{\sqrt{1-x}}\sum_{n=0}^\infty\beta_n^4\left(\frac{x}{x-1}\right)^n=\sum_{n=0}^\infty \beta_n^{}x^n \sum_{k=0}^n \beta_{k}^{}\beta_{n-k}^{}\sum_{j=0}^k \beta_j^2\beta_{k-j}$

$\hspace{5pt}\D \sum_{n=0}^\infty t^nP_n^{}(x)P_n^{}(y)=\frac{1}{\sqrt{1+2t\L({\tiny\sqrt{1-x^2}\sqrt{1-y^2}}-xy\R)+t^2}}\kappa\L(\frac{4t{\tiny\sqrt{1-x^2}\sqrt{1-y^2}}}{1+2t({\tiny\sqrt{1-x^2}\sqrt{1-y^2}}-xy)+t^2}\R)$

$\hspace{5pt}\D \sum_{n=0}^\infty (-1)^n(4n+1)\beta_n^5P_{2n}(x)=S_1+\sum_{n=0}^\infty \beta_n x^{2n}\L(S_1B_n-\frac{1}{2}\L(S_1-S_2+\frac{8}{\pi^3}\R)A_n\R)-\frac{4}{\pi^2}\sum_{n=0}^\infty \frac{|x|^{2n+1}}{(2n+1)\beta_n}\sum_{k=0}^n \beta_k^2\beta_{n-k}^{}$

記法

$\hspace{5pt}\BA\D &\beta_r^{}=2^{-2r}\binom{2r}{r}\\ &\kappa(x)=\sum_{n=0}^\infty \beta_n^2x^n\\ &K(x)=\frac{\pi}{2}\kappa(x^2)\\ &K'(x)=K({\small \sqrt{1-x^2}})\\ &P_n^{}(x):n\text{次第一種Legendre多項式}\\ &Q_n^{}(x):n\text{次第二種Legendre多項式}\\ &T_n^{}(x):n\text{次第一種Chebyshev多項式}\\ &U_n^{}(x):n\text{次第二種Chebyshev多項式}\\ &{\cal A}=\cfrac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}\\ &G=\sum_{n=0}^\infty \cfrac{(-1)^n}{(2n+1)^2}\\ &L_{-4}(s)=\sum_{n=0}^\infty \frac{(-1)^n}{\qty(2n+1)^s} \EA$