積分問題集２

Xであげた問題を載せます.問題数は50問です.

定義

\begin{align} \beta_n\coloneqq\binom{2n}{n}2^{-2n},A\coloneqq\sum_{n\geq0}\beta_n^3,\kappa(x)\coloneqq\sum_{n\geq0}\beta_n^2 x^{n},K(x)\coloneqq\int_0^1\frac{dt}{\sqrt{1-t^2}\sqrt{1-x^2t^2}},K'(x)\coloneqq K(\sqrt{1-x^2}) \\ \end{align}
\begin{align} \zeta(s)\coloneqq\sum_{n\gt 0}\frac{1}{n^s},\beta(s)\coloneqq\sum_{n\geq0}\frac{(-1)^n}{(2n+1)^s},\mathrm{Li}_s(x)\coloneqq\sum_{n\gt0}\frac{x^n}{n^s},\Gamma(z)\coloneqq\int_0^\infty x^{z-1}e^{-x}dx,\psi(z)\coloneqq\frac{d}{dz}\ln\Gamma(z),\gamma\coloneqq-\psi(1) \end{align}
$Ai(x)$:Airy function of the first kind

$\mathrm{Proposition.1}$
\begin{eqnarray} \int_0^1\frac{\ln(1+x+x^2)}{x}dx=\frac{\pi^2}{9} \end{eqnarray}
$\mathrm{Proposition.2}$
\begin{eqnarray} \int_{-\infty}^{\infty}\ln\left(1+\frac{4\pi^2}{\left(e^x-x+\ln \frac{\pi}{2}\right)^2}\right)dx &=&2\pi^2 \\ \end{eqnarray}
$\mathrm{Proposition.3} $
\begin{eqnarray} \int_{-\infty}^{\infty}e^{aix}\ln\left(\frac{1}{1-\frac{1}{\cosh^2 x}}\right)dx &=&\frac{2\pi}{a}\tanh\left(\frac{a\pi}{4}\right)\\ \end{eqnarray}
$\mathrm{Proposition.4}$
\begin{eqnarray} \int_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac{\pi}{8}\ln2 \end{eqnarray}
$\mathrm{Proposition.5}$
\begin{eqnarray} \int_0^{\pi/2}\ln \left(1+\frac{\sin x}{2}\right)dx=\frac{4}{3}\beta(2)-\pi\ln2+\frac{\pi}{6}\ln\frac{2+\sqrt3}{2-\sqrt3} \end{eqnarray}
$\mathrm{Proposition.6}$
\begin{eqnarray} \int_0^\infty\frac{\cos 2ax}{\cosh^{2n} x}dx=\frac{2^{2n}\Gamma(n)^2}{4\Gamma(2n)}\prod_{m\geq0}\left(1+\frac{a^2}{(n+m)^2}\right)^{-1} \end{eqnarray}
$\mathrm{Proposition.7}$
\begin{eqnarray} \int_0^{\frac{\pi}{4}}\ln\left(\frac{\pi^2}{4}+\ln^2\tan x\right)dx&=&\frac{\pi}{2}\ln2\\ \end{eqnarray}
$\mathrm{Proposition.8}$
\begin{eqnarray} \int_0^1K(x)^2\tanh^{-1}x\ dx&=&\frac{\pi^4}{32}\sum_{n=0}^{\infty}\beta_n^4\\ \int_0^1K\left(\sqrt\frac{1-\sqrt{1-x^2}}{2}\right)^2\frac{\tanh^{-1}x}{\sqrt{1-x^2}}dx&=&\frac{\pi^4}{16}\sum_{n=0}^{\infty}\beta_n^4-\frac{\Gamma\left(\frac{3}{4}\right)^4}{32}\sum_{n=1}^{\infty}\frac{1}{n^3\beta_n^3}\sum_{k=0}^{n-1}\beta_k^4\\ \end{eqnarray}
$\mathrm{Proposition.9}$
\begin{eqnarray} \int_0^{\pi/2}\frac{x}{\sin x+\cos x+1}dx=\frac{\pi}{4}\ln2 \end{eqnarray}
$\mathrm{Proposition.10}$ $0\leq\alpha\leq1$
\begin{eqnarray} \int_0^\pi\ln\tanh^{-1}(\alpha e^{ix})dx=\pi\ln\alpha+\frac{\pi^2i}{2} \end{eqnarray}
$\mathrm{Proposition.11}$
\begin{eqnarray} \int_0^1\left( _2F_1\left[\begin{matrix}\frac{1}{4},\frac{1}{4}\\1\end{matrix};x\right]\right)^2dx=A-\frac{4}{\pi^2}A^{-1} \end{eqnarray}
$\mathrm{Proposition.12}$
\begin{eqnarray} \int_0^1\frac{K'(x)}{K(x)^2+K'(x)^2}dx=\frac{1}{4} \end{eqnarray}
$\mathrm{Proposition.13}$
\begin{eqnarray} \int_0^{\pi/2}\frac{\cos x}{\sin^2x}\left(1-\frac{x}{\tan x}\right)dx=\beta(2)-\frac{1}{2} \end{eqnarray}
$\mathrm{Proposition.14}$
\begin{eqnarray} \int_0^1\frac{1}{(x-\tanh^{-1}x)^2+\pi^2}\frac{dx}{1-x^2}=\frac{3}{5} \end{eqnarray}
$\mathrm{Proposition.15}$ $pqr=1,\bar{x}\coloneqq\sqrt{x^2+1}$
\begin{eqnarray} \int_0^1\frac{\tan^{-1}q\sqrt{p^2x^2+1}}{q\sqrt{p^2x^2+1}}\frac{dx}{(r^2+1)p^2x^2+1}=\frac{\pi^2}{8}-\frac{1}{2}(\tan^{-1}\bar{p}q)^2+\frac{1}{2}\left(\tan^{-1}\frac{1}{\bar{q}r}\right)^2-\frac{1}{2}(\tan^{-1}\bar{r}p)^2 \end{eqnarray}
$\mathrm{Proposition.16}$
\begin{eqnarray} \int_0^1\frac{\ln x}{1+x^2}\ln(1-x^4)dx=\frac{\pi^3}{16}-3\beta(2)\ln2 \end{eqnarray}
$\mathrm{Proposition.17}$
\begin{eqnarray} \int_0^\infty\frac{\sin(\tan x)}{x}dx&=&\frac{\pi}{2}(1-e^{-1})\\ \int_0^\infty\left(\frac{\sin(\tan x)}{x}\right)^2dx&=&\frac{\pi}{2}\\ \int_0^\infty\left(\frac{\sin(\tan x)}{x}\right)^3dx&=&\frac{3\pi}{8} \end{eqnarray}
$\mathrm{Proposition.18}$
\begin{eqnarray} \int_0^1x^{2n}\kappa\left(\frac{1-\sqrt{1-x^2}}{2}\right)^2dx&=&\beta_n^3\left(\int_0^1\kappa\left(\frac{1-\sqrt{1-x^2}}{2}\right)^2dx +\sum_{k=1}^n\frac{1}{8k^3\beta_k^3}\left(\left(2k-\frac{1}{2}\right)A-\frac{2}{\pi^2A}\right)\right)\\ \int_0^1x^{2n-1}\kappa\left(\frac{1-\sqrt{1-x^2}}{2}\right)^2dx&=&\frac{1}{8n^3\beta_n^3}\sum_{k=0}^{n-1}\beta_k^3\left(\left(2k+\frac{1}{2}\right)A-\frac{2}{\pi^2A}\right) \end{eqnarray}
$\mathrm{Proposition.19}$
\begin{eqnarray} \int_0^1\frac{\mathrm{Li}_2(1-x^2)}{1-x^2}dx=\frac{\pi^2}{2}\ln2-\frac{7}{4}\zeta(3) \end{eqnarray}
$\mathrm{Proposition.20}$
\begin{eqnarray} \int_0^1\frac{\tan^{-1}x\ln(1+x^2)}{1+x^2}dx=\frac{21}{64}\zeta(3)-\frac{\pi}{4}\beta(2)+\frac{\pi^2}{16}\ln2 \end{eqnarray}
$\mathrm{Proposition.21}$
\begin{eqnarray} \int_0^\infty\frac{\sin4\pi x^2}{\sinh^2\pi x}dx=\frac{3-\sqrt2}{4} \end{eqnarray}
$\mathrm{Proposition.22}$
\begin{eqnarray} \int_0^\infty\frac{1}{1+(x+\tan x)^2}dx=\frac{\pi}{2} \end{eqnarray}
$\mathrm{Proposition.23}$
\begin{eqnarray} \int_0^\infty\frac{\sin\frac{\pi x^2}{2}}{\sinh^2\pi x}dx=\frac{1}{4\pi} \end{eqnarray}
$\mathrm{Proposition.24}$
\begin{eqnarray} \int_0^1x^{3/2}\frac{K'(x)}{\sqrt{K(x)+\sqrt{K(x)^2+K'(x)^2}}}dx=\frac{\pi^{3/2}}{16} \end{eqnarray}
$\mathrm{Proposition.25}$
\begin{eqnarray} \int_0^{\pi/2}\frac{x^2}{x^2+\ln^2(2\cos x)}dx=\frac{\pi}{8}(1-\gamma+\ln(2\pi)) \end{eqnarray}
$\mathrm{Proposition.26}$
\begin{eqnarray} \int_0^\infty\frac{\ln(1+x)}{\ln^2x+\pi^2}\frac{dx}{x^2}=\gamma \end{eqnarray}
$\mathrm{Proposition.27}$
\begin{eqnarray} \int_0^\pi x\ln^2(2\sin x)dx=\frac{\pi^4}{24} \end{eqnarray}
$\mathrm{Proposition.28}$ $\phi=\frac{1+\sqrt5}{2},\bar{\phi}=\frac{-1+\sqrt5}{2}$
\begin{eqnarray} \int_0^{\pi/2}\frac{\ln(x^2+\ln^2\cos x)}{1+\frac{\bar{\phi}^2-1}{\bar{\phi}^2+1}\cos2x}dx=\frac{\bar{\phi}+\phi}{2}\ln\ln\phi \end{eqnarray}
$\mathrm{Proposition.29}$
\begin{eqnarray} \int_{-\infty}^\infty\frac{\cos\frac{\pi x^2}{2}}{\cosh\pi x}dx=\sqrt{2+\sqrt2}-1 \end{eqnarray}
$\mathrm{Proposition.30}$
\begin{eqnarray} \int_{-\infty}^\infty\frac{\cos2\pi x^2}{\cosh\pi x}dx=\sqrt{1-\frac{1}{\sqrt2}} \end{eqnarray}
$\mathrm{Proposition.31}$
\begin{eqnarray} \int_0^\infty x^{3n}Ai(x)^2dx=\frac{1}{3^{2/3}\Gamma(\frac{1}{3})^2}\frac{(3n)!\Gamma(\frac{7}{6})}{\Gamma(n+\frac{7}{6})12^n} \end{eqnarray}
$\mathrm{Proposition.32}$
\begin{eqnarray} \int_{-\infty}^\infty Ai(x)e^{sx}dx&=&e^{s^3/3}\\ \int_{-\infty}^\infty Ai(x)^2e^{sx}dx&=&\frac{1}{2\sqrt{s\pi}}e^{s^3/12} \end{eqnarray}
$\mathrm{Proposition.33}$ $0\lt X\leq\pi,f(x)$は$\lim\limits_{x\to0}\frac{f(x)}{x}$が存在し$[0,X]$で連続な複素関数
\begin{eqnarray} \int_0^X\frac{f(t)}{1-e^{it}}dt=\sum_{n\geq0}\int_0^Xe^{int}f(t)dt \end{eqnarray}
$\mathrm{Proposition.34}$
\begin{eqnarray} \int_0^\infty\frac{2x}{x^2+1}\ln\frac{x^2+2x\cos a+1}{x^2+2x\cos b+1}dx=b^2-a^2 \end{eqnarray}
$\mathrm{Proposition.35}$
\begin{eqnarray} \int_0^1\ln\left(\frac{1+x+\frac{x^2}{4}}{1+x-2x^2}\right)\frac{dx}{x(2+x)}&=&\frac{\pi^2}{24}\\ \int_0^1\ln\left(\frac{1+2x+x^2}{1+2x-3x^2}\right)\frac{dx}{x(1+x)}&=&\frac{\pi^2}{12} \end{eqnarray}
$\mathrm{Proposition.36}$
\begin{eqnarray} \int_0^1\frac{1-\sqrt{1-s^2x^2}}{x\sqrt{1-x^2}\sqrt{1-s^2x^2}}\ln\frac{1}{1-x^2}dx=(\tanh^{-1}s)^2 \end{eqnarray}
$\mathrm{Proposition.37}$
\begin{eqnarray} \int_0^1\frac{\ln(1-sx)}{x}\ln\frac{1-x}{\sqrt{1-sx}}dx=\mathrm{Li}_3(s) \end{eqnarray}
$\mathrm{Proposition.38}$
\begin{eqnarray} \int_0^1\frac{\tanh^{-1}x}{1-x^2}(\tan^{-1}sx-s\tan^{-1}x)dx=\frac{1}{6}(\tan^{-1}s)^3 \end{eqnarray}
$\mathrm{Proposition.39}$ $a\gt0,a\notin2\mathbb{N}-1$
\begin{eqnarray} \int_0^\infty\sin\left(\frac{a\pi x}{2}\right)\left(\psi\left(\frac{3+x}{4}\right)-\psi\left(\frac{1+x}{4}\right)\right)dx=\frac{\pi}{\cos(\frac{a\pi}{2})}+\frac{1}{2}\left(\psi\left(\frac{1-a}{4}\right)+\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{1+a}{4}\right)-\psi\left(\frac{3-a}{4}\right)\right) \end{eqnarray}
$\mathrm{Proposition.40}$
\begin{eqnarray} \int_{-\infty}^\infty\frac{\cos\left(s\tan^{-1}\frac{x}{a}\right)}{(a^2+x^2)^{s/2}}\frac{dx}{1+x^2}=\int_{-\infty}^\infty\frac{\sin\left(s\tan^{-1}\frac{x}{a}\right)}{(a^2+x^2)^{s/2}}\frac{x}{1+x^2}dx=\frac{\pi}{(1+a)^s} \end{eqnarray}
$\mathrm{Proposition.41}$
\begin{eqnarray} \int_0^1\frac{\kappa(x)^2}{\kappa(x)^2+\kappa(1-x)^2}\frac{dx}{1-xt}=\frac{\pi\kappa(1-t)}{t\kappa(t)}-\frac{4\ln2-\ln\frac{t}{1-t}}{t} \end{eqnarray}
$\mathrm{Proposition.42}$
\begin{eqnarray} \int_0^\infty\left(\frac{x}{e^x-1}-1+\frac{x}{2}\right)\frac{e^{-sx}}{x^2}dx=\ln\frac{\Gamma(s)e^s}{s^{s-1/2}\sqrt{2\pi}} \end{eqnarray}
$\mathrm{Proposition.43}$
\begin{eqnarray} \int_0^1\frac{\tan^{-1}\frac{1}{\sqrt{x^2+2a^2}}}{\sqrt{x^2+2a^2}(x^2+a^2)}dx=\frac{1}{2a^2}\left(\tan^{-1}\frac{1}{a}\right)^2 \end{eqnarray}
$\mathrm{Proposition.44}$
\begin{eqnarray} \int_0^1\mathrm{Li}_2(x)\ln\frac{1}{1-x}dx=\zeta(2)+2\zeta(3)-3 \end{eqnarray}
$\mathrm{Proposition.45}$
\begin{eqnarray} \int_0^\infty\frac{\tan^{-1}\frac{1}{x}\ln^2(1+x^2)}{x}dx&=&\frac{9\pi}{4}\zeta(3)\\ \int_0^\infty\frac{\tan^{-1}\frac{1}{x}\left(\tan^{-1}x\right)^2}{x}dx&=&\frac{7\pi}{16}\zeta(3) \end{eqnarray}
$\mathrm{Proposition.46}$
\begin{eqnarray} \int_0^1 x^n \ _2F_1\left[\begin{matrix} \frac{1}{3},\frac{1}{3}\\ \frac{2}{3}\end{matrix};1-x \right]\frac{dx}{x^{\frac{1}{3}}(1-x)^{\frac{1}{3}}}=\Gamma\left(\frac{2}{3}\right)^3\frac{\left(\frac{2}{3}\right)_n^2}{n!^2} \end{eqnarray}
$\mathrm{Proposition.47}$
\begin{eqnarray} \int_0^1\frac{dx}{\sqrt{1+x^4}}=\frac{\pi^{\frac{3}{2}}}{4\Gamma\left(\frac{3}{4}\right)^2} \end{eqnarray}
$\mathrm{Proposition.48}$
\begin{eqnarray} \int_0^{\pi/2}\left(\sinh^{-1}\sin x\right)^2dx&=&\frac{\pi^3}{48}\\ \int_0^{\pi/2}\left(\sinh^{-1}\sqrt{\sin x}\right)^3dx&=&\frac{\pi^3}{16}\ln 2-\frac{3\pi}{16}\zeta(3) \end{eqnarray}
$\mathrm{Proposition.49}$
\begin{eqnarray} \int_0^{\pi/2}\ln\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)\ln\left(\ln\sqrt{\frac{1+\cos x}{1-\cos x}}\right)\frac{dx}{1+\sin x}=2\ln2-\ln^2 2 \end{eqnarray}
$\mathrm{Proposition.50}$
\begin{eqnarray} \int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}dx&=&\sum_{n=-\infty}^{\infty}\frac{\sin^2n}{n^2}\\ \int_{-\infty}^{\infty}\binom{n}{x}\binom{m}{x}\binom{n+m+x}{x}dx&=&\sum_{k=0}^{\min(n,m)}\binom{n}{k}\binom{m}{k}\binom{n+m+k}{k} \end{eqnarray}