f(x)/(x+t)の積分3

$\frac{f(\kappa(x,1-x))}{1-xt}$

\begin{eqnarray} \int_0^1\frac{\kappa(1-x)}{\sqrt{x}}\frac{dx}{1-tx}&=&\pi\kappa(t)\\ \int_0^1\left(\kappa(1-x)\ln\left(\frac{1-x}{x}\right)-\pi\kappa(x)\right)\frac{1}{\sqrt{x}}\frac{dx}{1-tx}&=&\pi\kappa(t)\ln(1-t) \\ \int_0^1\kappa(x)\kappa(1-x)\frac{dx}{1-tx}&=&\frac{\pi}{2}\kappa(t)^2\\ \int_0^1\kappa(1-x)^2\left(\frac{1}{1-(1-t)x}+\frac{1}{1-tx}\right)dx&=&\pi\kappa(t)\kappa(1-t)\\ \int_0^1(3\kappa(x)^2\kappa(1-x)-\kappa(1-x)^3)\frac{\sqrt{x}dx}{1-tx}&=&\pi\kappa(t)^3 \\ \int_0^1\frac{2\sqrt x\kappa(x)\kappa(1-x)^2}{1-tx}+\frac{\sqrt{x}\kappa(1-x)^3}{1-(1-t)x}dx&=&\pi\kappa(t)^2\kappa(1-t)\\ \int_0^1(\kappa(x)^3\kappa(1-x)-\kappa(x)\kappa(1-x)^3)\frac{x}{1-tx}dx&=&\frac{\pi}{4}\kappa(t)^4\\ \end{eqnarray}

\begin{eqnarray} \int_0^1\frac{\kappa(1-x)}{\kappa(x)^2+\kappa(1-x)^2}\frac{dx}{(1-tx)\sqrt{x}}&=&\frac{\pi}{t}\left(1-\frac{1}{\kappa(t)}\right)\\ \int_0^1\frac{\kappa(1-x)^2}{\kappa(x)^2+\kappa(1-x)^2}\frac{dx}{1-tx}&=&\frac{4\ln2-\ln t}{t}-\frac{\pi\kappa(1-t)}{t\kappa(t)} \\ \int_0^1\frac{\kappa(x)}{\sqrt{x}}\frac{2}{1-(1-t)x}-\frac{1}{\sqrt{x}}\frac{\kappa(1-x)^3}{\kappa(x)^2+\kappa(1-x)^2}\frac{1}{1-tx}dx&=&\pi\frac{\kappa(1-t)^2}{\kappa(t)}\\ \int_0^1\frac{3\kappa(x)^2-\kappa(1-x)^2}{1-(1-t)x}-\frac{\kappa(1-x)^4}{\kappa(x)^2+\kappa(1-x)^2}\frac{1}{1-xt}dx&=&\pi\frac{\kappa(1-t)^3}{\kappa(t)} \end{eqnarray}

$\frac{1}{K^2+K'^2}$

\begin{align} \int_0^1\frac{K(x)K'(x)^2}{K(x)^2+K'(x)^2}dx=\frac{7}{8}\zeta(3)\left(=\int_0^1\frac{1}{x}\left(\frac{1}{\sqrt{1-x^2}}-1\right)\frac{K'(x)^2}{K(x)}dx=\frac{1}{4\pi}\int_0^{\infty}t^2(\theta_3^4(e^{-t})\theta_4(e^{-t})-\theta_3^2(e^{-t})\theta_4^4(e^{-t}))dt\right) \end{align}
\begin{align} \int_0^1\frac{K(x)^3}{K(x)^2+K'(x)^2}\frac{dx}{x\sqrt{1-x^2}}=4\beta(2) \end{align}
\begin{align} \int_0^1\frac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}dx=\int_0^1\frac{K'(x)^3}{K(x)\sqrt{1-x^2}}dx-2\int_0^1K'(x)^2dx \end{align}
\begin{align} \int_0^1\frac{K(x)K'(x)^3}{K(x)^2+K'(x)^2}dx=2\int_0^1\frac{K(x)^2}{\sqrt{1-x^2}}dx-\int_0^1\frac{K'(x)^3}{K(x)}dx,\int_0^1\frac{K(x)^3K'(x)}{K(x)^2+K'(x)^2}dx=\int_0^1\frac{K'(x)^3}{K(x)}dx-3\int_0^1K(x)K'(x)dx \end{align}
\begin{eqnarray} \int_0^1\frac{K(x)^2K'(x)^3}{K(x)^2+K'(x)^2}dx&=&\frac{\Gamma(\frac{1}{4})^8}{640\pi^2}, \int_0^1\frac{K(x)^4K'(x)}{K(x)^2+K'(x)^2}dx&=&\frac{\Gamma(\frac{1}{4})^8}{960\pi^2} \end{eqnarray}
\begin{align} \int_0^1x\frac{K(x)^2K'(x)^4}{K(x)^2+K'(x)^2}dx=\int_0^1xK(x)^4-xK(x)^2K'(x)^2dx,\int_0^13xK(x)^2K'(x)^2dx=\int_0^12xK(x)^4dx \end{align}