積分一覧
$\BA\D\\ &[~1~]\qquad\int_0^\frac{\pi}{4} \frac{1-\tan x}{\sqrt{\tan x}}\L(\frac{\pi}{4}-x \R)dx =\frac{\pi\ln2}{2\sqrt{2}}\\ &[~2~]\qquad\int_0^\frac{\pi}{2} \frac{x\ln\tan x}{\sqrt{\tan x}}\,dx =\frac{\pi}{8\sqrt{2}}(16\beta(2)-\pi^2-2\pi\ln2)\\ &[~3~]\qquad\int_0^\frac{\pi}{2} \sinh^{-1}\sqrt{\sin x}\,dx=\frac{\pi\ln2}{2}\\ &[~4~]\qquad\int_0^\infty \L(\sinh^{-1}\cosh x-x \R)dx=\frac{3}{8}\zeta(2)\\ &[~5~]\qquad\int_0^\frac{\pi}{2} \frac{\cos x}{\sin^2 x}\L(1-\frac{x^2}{\tan^2 x} \R)dx =-\frac{2}{3}-\zeta(2)+\frac{10}{3}\beta(2)\\ &[~6~]\qquad\int_0^\frac{\pi}{2} \L(1-\frac{x^2}{2}\frac{\cos x}{1-\cos x} \R)\frac{dx}{\sin x} =-\frac{3}{2}+\frac{\ln2}{2}+\frac{\pi}{4}+\frac{3}{8}\zeta(2)+\frac{\pi}{2}\beta(2)-t(3) \\ &[~7~]\qquad\int_0^1 \frac{1}{x\sqrt{1-x}}\L({\rm Li}_1(\alpha x){\rm Li}_3(\alpha x)-\frac{1}{2}{\rm Li}_2(\alpha x)^2 \R)dx =\sum_{0< m< n}\frac{2^{2n}\alpha^n}{m^3 n^2\binom{2n}{n}}\\ &[~8~]\qquad\int_0^\frac{\pi}{2}\L(\frac{x}{\sin x}-1 \R)\ln\frac{1}{\cos x}\,\frac{dx}{\tan\frac{x}{2}} =2\int_0^1 \frac{\sqrt{x}}{1-x^2}\ln\frac{2}{1+x}\,dx\\ &[~9~]\qquad\int_0^\frac{\pi}{4}x^2\L(\frac{1}{\sin x}+\frac{1}{\cos x} \R)dx =\frac{1}{16}\sum_{0< n}\frac{2^{6n}}{n^3\binom{2n}{n}\binom{4n}{2n}}\\ &[10]\qquad\int_0^\frac{\pi}{2}x^2\L(\frac{\pi}{2}-x \R)\frac{\cos x}{\sqrt{\sin x}}\,dx =\frac{(2\pi)^{\frac{3}{2}}}{\Gamma^2\L(\frac{1}{4} \R)}\sum_{0< n}\frac{\binom{4n}{2n}}{2^{4n}n^2}\frac{n!^2}{\L(\frac{5}{4} \R)_n^2}\\ &[11]\qquad\int_0^\frac{\pi}{2}\L(\L(\cos^{-1}\sqrt{\sin x} \R)^4+\L(\sinh^{-1}\sqrt{\sin x} \R)^4 \R)dx =\frac{3\pi}{32}\sum_{0< n}\frac{2^{4n}}{n^4\binom{2n}{n}^2}\\ &[12]\qquad\int_0^1 \frac{(1-x^2)\tanh^{-1}\sqrt{1-x^2}}{(1+x^2)^2-(2x\cos\theta)^2}\,dx=\frac{\pi}{2}\frac{\cos^{-1}\sqrt{\sin\theta}}{\cos\theta} \left(0\le\theta\lt\frac{\pi}{2}\right)\\ &[13]\qquad\int_0^1 x^{-1+\alpha}(1-x)^{-1+\beta}\left(\frac{1+ax}{1-ax}+\frac{1+bx}{1-bx}\right) {_2}F_1\left[\begin{matrix} 1,\alpha\\ \alpha+\beta \end{matrix};abx \right]dx =\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} {_2}F_1\left[\begin{matrix} 1,\alpha\\ \alpha+\beta \end{matrix};a\right] {_2}F_1\left[\begin{matrix} 1,\alpha\\ \alpha+\beta \end{matrix};b\right] \\ &[14]\qquad\int_0^1 \frac{(\sin^{-1}abx)^2}{x}\left(\frac{1+a^2x^2}{1-a^2x^2}+\frac{1+b^2x^2}{1-b^2x^2} \right)\ln\frac{1+\sqrt{1-x^2}}{x}\,dx =(\sin^{-1}a)^2(\sin^{-1}b)^2\\ &[15]\qquad\int_0^\infty \frac{x}{\sinh x}\ln\cosh\ln\tanh\frac{x}{2} \,dx=\pi\beta(2)-\frac{1}{2}\pi^2\ln2\\ &[16]\qquad\int_0^1 \frac{1-x}{(1+x)(1+x+x^2)}\ln\frac{1}{x}\,\ln(1+x)\,dx=\frac{\zeta(3)}{18}\\ &[17]\qquad\int_0^1 \frac{(1-x)(2+x)(1+2x)}{x(1+x)(1+x+x^2)}\ln\frac{(1+x)^2}{1+x+x^2}\,\ln\frac{1}{x}\,dx=\zeta(3)\\ &[18]\qquad\int_0^1 \frac{(1-x)(2+x)(1+2x)}{x(1+x)(1+x+x^2)}\ln^2\frac{(1+x)^2}{1+x+x^2}\,\ln\frac{1}{x}\,dx=\frac{\zeta(4)}{9}\\ &[19]\qquad\int_0^1 \frac{\tan^{-1}\frac{1}{x}}{\sqrt{1-x^4}}dx=\int_0^1 \frac{\tanh^{-1}x}{\sqrt{1-x^4}}dx\\ &[20]\qquad\int_0^1 \frac{x\sinh(t \tanh^{-1}x)}{(1-u^2x^2)\sqrt{1-x^2}}dx =\frac{\pi}{2}\frac{\sinh(t \tanh^{-1}u)}{u\sqrt{1-u^2}\cos\frac{\pi t}{2}}\\ &[21]\qquad\int_0^\frac{\pi}{6}\L(\frac{\sqrt{2}x\sin x}{\sqrt{1-2\sin^2 x}}+\frac{\sqrt{1-4\sin^2 x}}{(1-2\sin^2 x)\tan x}\ln\frac{1}{\cos x}\R)dx=\frac{\zeta(2)}{12}\\ &[22]\qquad\frac{2}{\pi}\int_0^1\int_0^1 \frac{\sqrt{tu}}{(1+t)\sqrt{1-u}}\left(\frac{1}{1-(1-t)u}-\frac{1}{\ln\frac{1}{(1-t)u}} \right) dtdu +\int_0^1 \sqrt\frac{1-x}{1+x}\left(\frac{1}{x}-\frac{1}{\ln\frac{1}{1-x}}\right) dx =\gamma\\ &[23]\qquad\int_0^1 \frac{1-x^2}{x(1+x^2)}\tanh^{-1}x\,\tanh^{-1}\sqrt{1-x^2}\,dx =\frac{\pi}{8}\int_0^1 \frac{\ln\frac{1}{1-{x}^2}}{\sqrt{x(1-x)}}dx =3\zeta(2)\tanh^{-1}\frac{17-8\sqrt{2}}{23}\\ &[24]\qquad\int_0^1 \frac{\sin^{-1}\frac{x^2}{2}}{(1+x^2)\sqrt{2+x^2}}dx=\frac{\zeta(2)}{24}\\ &[25]\qquad \int_0^\frac{\pi}{3} \frac{x(2-\cos x)}{\cos x\,\sqrt{2\cos x-1}}dx=\frac{3}{2}\zeta(2) \EA$
$ \D $
$ \D $
$ \D $
$ \D $
$ \D $
$ \D $
$ \D $
$ \D $
$ \D $