好きな式集1　(級数積分極限)

$$\sum_{n=0}^{\infty}\dfrac{(3n)!(2n)!}{(n!)^{5}3^{3n}2^{2n}}=\dfrac{9 \Gamma\bigl(\frac{1}{3}\bigl)^{6}}{16\sqrt[3]{4}\,\pi^{4}}$$
$$\sum_{n=0}^{\infty}\dfrac{(6n)!}{((3n)!)^2}\dfrac{(-1)^n}{256^n}=\dfrac{\sqrt{(2-\sqrt[3]{-2})(2+\sqrt[3]{2})}+\sqrt{(2-(-1)^{\frac{2}{3}}\sqrt[3]{2})(2+\sqrt[3]{2})}+\sqrt{4-2\sqrt[3]{2}+\sqrt[3]{4}}}{3\sqrt{5}}$$
$$ \sum_{n=0}^{\infty}\dfrac{(6n)!}{(3n)!((2n)!)^2}\dfrac{n!}{729^n}=3\sqrt{\dfrac{3}{23}}\cos{\biggl(\dfrac{2}{3}\sin^{-1}{\biggl(\dfrac{2}{3\sqrt{3}}\biggl)}\biggl)}$$
$$\sum_{n=0}^{\infty}\dfrac{(-1)^n (6n)!}{(3n)!((2n)!)^2}\dfrac{n!}{729^n}=3\sqrt{\dfrac{3}{31}}\cosh{\biggl(\dfrac{2}{3}\sinh^{-1}{\biggl(\dfrac{2}{3\sqrt{3}}\biggl)}\biggl)}$$
$$\sum_{n=0}^{\infty}\dfrac{(4n)!}{((2n)!)^2 4^{4n}}=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{5}}$$
$$\sum_{n=0}^{\infty}\dfrac{(6n)!}{(3n)!((n)!)^3 2^{6n}3^{3n}}=\dfrac{\sqrt{3}\Gamma\bigl(\frac{1}{4}\bigl)^{4}}{8\pi^3}$$
$$\sum_{n=0}^{\infty}\dfrac{(6n)!}{((3n)!)^2}\dfrac{1}{256^n}=\dfrac{2\sqrt{3}}{9}(\sqrt[3]{2}+\sqrt[3]{4})$$
$$\sum_{n=1}^{\infty}\dfrac{1}{n^3(e^{2n\pi}-1)}=\dfrac{7\pi^3}{360}-\dfrac{\zeta(3)}{2}$$
$$\sum_{n=0}^{\infty}\dfrac{1}{n^2+1}=\dfrac{1}{2}(1+\pi \coth\pi)$$
$$\sum_{n=0}^{\infty}\dfrac{1}{n^3+1}=\dfrac{1}{3}(\gamma+)$$
$$\sum_{n=0}^{\infty}\dfrac{(_{2n}C_n)^2}{(2n+1)16^n}=\dfrac{4G}{\pi}$$
$$\sum_{n=1}^{\infty}\dfrac{_{2n}C_n}{n4^n}=2\log{2}$$

$$\sum_{n=0}^{\infty}\dfrac{_{2n}C_n}{(n+1)4^n}=2$$
$$\sum_{n=0}^{\infty}\dfrac{_{2n}C_n}{(n+2)4^n}=\dfrac{4}{3}$$
$$\sum_{n=0}^{\infty}\dfrac{_{2n}C_n}{(2n+1)4^n}=\dfrac{\pi}{2}$$
$$\sum_{n=0}^{\infty}\dfrac{_{2n}C_n}{(2n+1)^24^n}=\dfrac{\pi}{2}\log{2}$$
$$\sum_{n=0}^{\infty}\dfrac{_{2n}C_n}{(2n+1)^3 4^n}=\dfrac{\pi^3}{48}+\dfrac{\pi}{4}(\log{2})^2$$
$$\sum_{n=0}^{\infty}\dfrac{1}{(n!)^2}=\dfrac{1}{\pi}\int_{0}^{\pi}e^{2\cos{x}}\,dx$$

$$ \int_{0}^{\pi}e^{2\cos{x}}\,dx=\pi\sum_{n=0}^{\infty}\dfrac{1}{(n!)^2}$$
$$\int_0^1\sqrt[3]{1-x^3}dx=\dfrac{\sqrt{3}{\Gamma\bigl(\frac{1}{3}\bigl)^{3}}}{12\pi}$$
$$\int_0^1\sqrt[4]{1-x^4}dx=\dfrac{{\Gamma\bigl(\frac{1}{4}\bigl)^{2}}}{8\sqrt{\pi}}$$
$$\int_{0}^{\infty}\dfrac{dx}{e^{x}(1+x)}=-e\mathrm{Ei}(-1)$$
$$\int_{0}^{\infty}\dfrac{dx}{e^{x}(1+x^2)}=\mathrm{Ci}(1)\sin{(1)}-\mathrm{Si}(1)\cos{(1)}+\dfrac{\pi}{2}\cos{(1)}$$
$$\int_{0}^{\infty}\dfrac{\ln{x}}{1+x^2}dx=0$$
$$\int_{0}^{\infty}\dfrac{x}{\sinh{x}}dx=\dfrac{\pi^2}{4}$$
$$\int_{0}^{\infty}\dfrac{x}{\cosh{x}}dx=2G$$
$$\int_{0}^{\dfrac{\pi}{2}}\sqrt{\sin{x}}dx=\int_{0}^{\dfrac{\pi}{2}}\sqrt{\cos{x}}dx=\sqrt{\dfrac{2}{\pi}}\Gamma\biggl(\dfrac{3}{4}\biggl)^2$$
$$\int_{0}^{\dfrac{\pi}{2}}\sqrt{\tan{x}}dx=\dfrac{\pi}{\sqrt{2}}$$
$$\int_{0}^{\infty}e^{-x^2}\sin{x^2}dx=\dfrac{\sqrt{\pi(\sqrt{2}-1})}{4}$$
$$\int_{0}^{\infty}e^{-x^2}\cos{x^2}dx=\dfrac{\sqrt{\pi(\sqrt{2}+1})}{4}$$
$$\int_{0}^{\infty}\dfrac{\sin^2{x}}{1+x^2}dx=\dfrac{(e^2-1)\pi}{4e^2}$$
$$\int_{0}^{\infty}\dfrac{\cos^2{x}}{1+x^2}dx=\dfrac{(e^2+1)\pi}{4e^2}$$
$$\int_{0}^{\frac{\pi}{2}}\dfrac{x^2}{\sin{x}}dx=2\pi G-\dfrac{7\zeta(3)}{2}$$
$$\int_{0}^{\frac{\pi}{2}}\dfrac{x^2}{\tan{x}}dx=\dfrac{\pi^2}{4}\ln{2}-\dfrac{7\zeta(3)}{8}$$