6

公式書付

219
0
$$\newcommand{BA}[0]{\begin{align*}} \newcommand{BE}[0]{\begin{equation}} \newcommand{bl}[0]{\boldsymbol} \newcommand{BM}[0]{\begin{matrix}} \newcommand{D}[0]{\displaystyle} \newcommand{EA}[0]{\end{align*}} \newcommand{EE}[0]{\end{equation}} \newcommand{EM}[0]{\end{matrix}} \newcommand{h}[0]{\boldsymbol{h}} \newcommand{k}[0]{\boldsymbol{k}} \newcommand{L}[0]{\left} \newcommand{l}[0]{\boldsymbol{l}} \newcommand{m}[0]{\boldsymbol{m}} \newcommand{n}[0]{\boldsymbol{n}} \newcommand{R}[0]{\right} \newcommand{vep}[0]{\varepsilon} $$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\tiny\rm series$

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

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

記法

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

投稿日:317
更新日:24日前

この記事を高評価した人

高評価したユーザはいません

この記事に送られたバッジ

バッジはありません。

投稿者

コメント

他の人のコメント

コメントはありません。
読み込み中...
読み込み中