便利さんの積分・級数botの積分,級数一覧

積分,級数一覧

一覧としてGitHubの これ があるのですがわたしはいちいち開いてみるのがめんどくさいので、ここでまとめておこうと思います。

積分

$${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx=\frac{\pi }{2}$$

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{1}{e^x+e^y+e^z-1}dxdydz=\frac{13}{4}\zeta (3)$$

$${\int }_0^1\frac{\arctan \sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}dx=\frac{5{\pi }^2}{96}$$

$${\int }_0^1\frac{1}{x^2+1}dx=\frac{\pi }{4}$$

$${\int }_0^{\mathrm{\infty }}\frac{{\log }^{4}x}{1+x^2}dx=\frac{5{\pi }^5}{64}$$

$${\int }_0^{\frac{\pi }{4}}\cos \log \tan xdx=\frac{\pi }{4\cosh \frac{\pi }{2}}$$

$${\int }_0^1\frac{{\tanh }^{-1}\frac{x}{2+\sqrt{5}}}{x}dx=\frac{{\pi }^2}{24}-\frac{3}{4}{\log }^2\varphi$$

$${\int }_0^{\mathrm{\infty }}\frac{x\log (1-\frac{1}{\cosh x})}{\cosh x}dx=-2\beta (2)\log 2$$

$${\int }_0^{\frac{\pi }{2}}\log (2\sin \frac{x}{2})\log (2\cos \frac{x}{2})dx=-\frac{{\pi }^3}{48}$$

$${\int }_0^2\frac{1}{\sqrt{1+x^3}}dx=\frac{\mathrm{\Gamma }\left(\frac{1}{3}\right)}{2^{\frac{4}{3}}\sqrt{3}\pi }$$

$${\int }_0^{\mathrm{\infty }}\frac{x\sinh x}{(1+{\cosh }^{2}x{)}^2}dx=\frac{\pi }{4}(\log (1+\sqrt{2})-\frac{1}{\sqrt{2}})$$

$${\int }_0^{\mathrm{\infty }}\frac{\sinh x}{x\cosh \frac{x}{a}}dx=\log \cot \frac{\pi (1-a)}{4}$$

$${\int }_0^1\frac{x{\tanh }^{-1}x}{3+x^2}dx=\frac{{\pi }^2}{72}$$

$${\int }_0^{\mathrm{\infty }}\log \left(\frac{1+x^2+x^4+x^6}{1+x^6}\right)dx=\pi (\sqrt{2}-1)$$

$${\int }_0^{\frac{\pi }{2}}\arctan (2{\tan }^{2}x)dx=\pi \arctan \frac{1}{2}$$

$${\int }_0^{\frac{\pi }{2}}\arctan \frac{\sin x}{2}dx=\frac{{\pi }^2}{12}-\frac{3{\log }^2\varphi }{2}$$

$${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}{\int }_0^x\frac{\cos y}{y}dydx=0$$

$${\int }_0^1\frac{{\tanh }^{-1}\sqrt{1-x^2}}{1-x}dx=\frac{3{\pi }^2}{8}$$

$${\int }_0^{\frac{\pi }{2}}\log (\frac{9}{16}+{\cos }^{2}x)dx=0$$

$${\int }_0^{\pi }\arctan \frac{\log \sin x}{x}dx=-\pi \arctan \frac{2\log 2}{\pi }$$

$${\int }_1^{\sqrt{2}}\frac{\arctan \sqrt{2-x^2}}{1+x^2}dx=\frac{{\pi }^2}{96}$$

$${\int }_0^{\mathrm{\infty }}\frac{e^{-kx}\sin x}{1+x^2}dx={\int }_0^{\mathrm{\infty }}\frac{\sin x}{1+(x+k{)}^2}dx$$

$${\int }_0^1\log \mathrm{\Gamma }(x)dx=\frac{1}{2}\log 2\pi$$

$${\int }_0^{\frac{1}{2}}\frac{1}{x}\log (1+x)\log (\frac{1}{x}-1)dx=\frac{13}{24}\zeta (3)$$

$${\int }_0^{\pi }\log (\frac{5}{4}+\cos x)dx=0$$

$${\int }_0^{\frac{\pi }{2}}\frac{1+\cos 2x}{3+\cos 4x}dx=\frac{\pi }{4\sqrt{2}}$$

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\coth }^{-1}\sqrt{x^2+{\pi }^2}}{\sqrt{x^2+{\pi }^2}}dx=\pi \arcsin \frac{1}{\pi }$$

$${\int }_0^1\frac{x}{{\tanh }^{-1}x}dx=\frac{7\zeta (3)}{{\pi }^2}$$

$${\int }_0^{\mathrm{\infty }}x\log \tanh xdx=-\frac{7}{16}\zeta (3)$$

$${\int }_0^{\frac{\pi }{2}}\frac{1}{1+{\cos }^{4}x}dx=\frac{\sqrt{1+\sqrt{2}}}{4}\pi$$

$$\mathrm{Re}{\int }_0^{\mathrm{\infty }}{H}_{\frac{i}{x}}dx=\frac{{\pi }^3}{12}$$

$${\int }_0^{\frac{\pi }{2}}\frac{\log \cos x}{x^2+{\log }^2\cos x}dx=\frac{\pi }{2}(1-\frac{1}{\log 2})$$

級数

$$\sum _{n=0}^{\mathrm{\infty }}\frac{{(\genfrac{}{}{0ex}{}{2n}{n})}^3}{2^{6n}}=\frac{\pi }{\mathrm{\Gamma }{\left(\frac{3}{4}\right)}^4}$$

$$\sum _{m,n\in \mathbb{Z}}{q}^{n^2+m^2}=1+4\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{q}^{2n+1}}{1-q^{2n+1}}$$

$${\left(\sum _{n\in \mathbb{Z}}{q}^{n^2}\right)}^4=1+8\sum _{\begin{array}{}0<n\\ n\not\equiv 0\mathrm{mod}4\end{array}}\frac{n{q}^n}{1-q^n}$$

$$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2n+1}=\frac{\pi }{4}$$

$$\sum _{0\le n_1<n_2\le n_3}\frac{1}{(n_1+\frac{1}{2})n_2(n_3+\frac{1}{2})}\frac{(\genfrac{}{}{0ex}{}{2n_3}{n_3})}{2^{2n_3}}=\frac{{\pi }^3}{3}$$

$$\sum _{0<n,m}\frac{1}{2^{n}n{m}^2}{\left(\frac{n!m!}{(n+m)!}\right)}^2=\frac{{\pi }^2\log 2}{6}-\frac{13}{16}\zeta (3)$$

$$\sum _{0<n,m}\frac{1}{2^{m}n{m}^2}{\left(\frac{n!m!}{(n+m)!}\right)}^2=\frac{{\log }^{3}2}{12}-\frac{{\pi }^2\log 2}{8}+\frac{27}{32}\zeta (3)$$

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$

$$\prod _{n\in \mathbb{Z}}(1+\frac{1}{\cosh 2\pi n})=2^{\frac{3}{8}}\sqrt{1+\sqrt{2}}$$

$$\sum _{n\in \mathbb{Z}}\frac{1}{{\cosh }^2\pi n}=\frac{1}{\pi }+\frac{\mathrm{\Gamma }{\left(\frac{1}{4}\right)}^4}{8{\pi }^3}$$

$$\sum _{n\in \mathbb{Z}}\frac{1}{{\cosh }^2(\sqrt{3}\pi n)}=\frac{1}{\sqrt{3}\pi }+\frac{\mathrm{\Gamma }{\left(\frac{1}{3}\right)}^6}{2^{\frac{11}{3}}{\pi }^4}(1+\sqrt{3})$$

$$\sum _{n\in \mathbb{Z}}\frac{1}{(n+x{)}^2}=\frac{{\pi }^2}{{\sin }^2\pi x}$$

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(e^{2\pi n}-1)}=\frac{\log 2}{2}+\log \mathrm{\Gamma }\left(\frac{3}{4}\right)-\frac{\log \pi }{4}-\frac{\pi }{12}$$

$$\sum _{n=0}^{\mathrm{\infty }}\frac{((2n)!{)}^3}{(n!{)}^6}\frac{6n+1}{2^{8n}}=\frac{4}{\pi }$$

$$\sum _{n=0}^{\mathrm{\infty }}\frac{((2n)!{)}^3}{(n!{)}^6}\frac{42n+1}{2^{12n}}=\frac{16}{\pi }$$

$$\sum _{0<m<n}\frac{1}{m}\frac{(1-x{)}_{m}n!}{m!(1-x{)}_n)}\frac{1}{n^2}\frac{n!a!}{(a+n)!}=\frac{(1-x{)}_a}{a!}\sum _{a<n}\frac{1}{n^3(1-x{)}_n}$$

$$\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\log (\frac{1}{n}+1)=\log \frac{2}{\pi }$$

$$\sum _{n=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{2n}{n})\frac{F_n}{2^{3n}}=\sqrt{\frac{2}{5}}$$

$$\sum _{n=0}^{\mathrm{\infty }}\frac{2^{2n}}{(2n+1{)}^2(\genfrac{}{}{0ex}{}{2n}{n})}=2\beta (2)$$

$$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^3}=\frac{{\pi }^3}{32}$$

$$\sum _{n=1}^{\mathrm{\infty }}\frac{H_n^2}{{\phi }^{2n-1}}=\frac{{\pi }^2}{15}$$

$$\sum _{k=0}^n\frac{k}{n^k(n-k)!}=\frac{1}{(n-1)!}$$

$$\sum _{n=0}^{\mathrm{\infty }}{\mathrm{Li}}_3(-e^{-(2n-1)\pi })=-\frac{{\pi }^3}{720}$$

$$\sum _{n=0}^{\mathrm{\infty }}{(\genfrac{}{}{0ex}{}{2n}{n})}^2\frac{1}{(2n+1)2^{4n}}=\frac{4\beta (2)}{\pi }$$

$$\sum _{n=0}^{\mathrm{\infty }}\frac{2^{2n}}{(2n+1{)}^2(\genfrac{}{}{0ex}{}{2n}{n})}=2\beta (2)$$

$$\sum _{n\in \mathbb{Z}}{e}^{-\pi n^2}=\frac{\sqrt[4]{\pi }}{\mathrm{\Gamma }\left(\frac{4}{3}\right)}$$

$$\sum _{n=1}^{\mathrm{\infty }}\frac{H_n}{n^2}=2\zeta (3)$$

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}=\frac{{\pi }^2}{18}$$

$$\sum _{0<n<m}\frac{1}{2^{n}n{m}^2}\frac{(\genfrac{}{}{0ex}{}{2n}{n})}{(\genfrac{}{}{0ex}{}{2m}{m})}=\frac{{\pi }^2\log 2}{18}-\frac{13}{48}\zeta (3)$$