Landen变换与Gauss变换

青青子衿

\begin{align*}
\frac{K\big(\sqrt{1-k^{2}}\,\big)}{K(k)}&=2\cdot\frac{K\big({\small{\frac{1-k}{1+k}}}\,\big)}{K({\small{\frac{2\sqrt{k}}{1+k}}})}=\dfrac{1}{2}\cdot\frac{K\big({\small{\frac{2\sqrt{\sqrt{1-k^{2}}}}{1+\sqrt{1-k^{2}}}}}\,\big)}{K\big({\small{\frac{1-\sqrt{1-k^{2}}}{1+\sqrt{1-k^{2}}}}}\big)}\\
\end{align*}

\begin{align*}
\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-k^2t^2\right)}}&=
\frac{1}{1+k}\int_{0}^{\frac{(1+k)x}{1+kx^2}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\big({\small{\frac{2\sqrt{k}}{1+k}}}t\big)^2\big)}}\\
\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-k^2t^2\right)}}&=
\frac{2}{1+k}\int_{0}^{\frac{\sqrt{(1+x)(1+kx)}-\sqrt{(1-x)(1-kx)}}{2}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\big({\small{\frac{2\sqrt{k}}{1+k}}}t\big)^2\big)}}\\


\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-(1-k^2)t^2\right)}}&=
\frac{2}{1+\sqrt{1-k^2}}\int_{0}^{\frac{\sqrt{(1+x)(1+{\scriptsize{\sqrt{1-k^2}}}x)}-\sqrt{(1-x)(1-{\scriptsize{\sqrt{1-k^2}}}x)}}{2}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\big({\small{{\small{\frac{2\sqrt{\sqrt{1-k^{2}}}}{1+\sqrt{1-k^{2}}}}}}}t\big)^2\big)}}\\
\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-k^2t^2\right)}}&=
\frac{1}{1+\sqrt{1-k^2}}\int_{0}^{\frac{\big(1+\sqrt{1-k^2}\big)x\sqrt{1-x^2}}{\sqrt{1-k^2x^2\,}}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\big({\small{\frac{1-\sqrt{1-k^{2}}}{1+\sqrt{1-k^{2}}}}}t\big)^2\big)}}\\
\int_{0}^{x}\sqrt{\dfrac{1-k^2t^2}{1-t^{2}}}\mathrm{d}t+kx&=
(1+k)\int_{0}^{\frac{\sqrt{(1+x)(1+kx)}-\sqrt{(1-x)(1-kx)}}{2}}\sqrt{\dfrac{1-\big({\small{\frac{2\sqrt{k}}{1+k}}}t\big)^2}{1-t^{2}}}\mathrm{d}t\\
&\qquad+(1-k)\int_{0}^{\frac{\sqrt{(1+x)(1+kx)}-\sqrt{(1-x)(1-kx)}}{2}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\big({\small{\frac{2\sqrt{k}}{1+k}}}t\big)^2\big)}}\\

\int_{0}^{x}\sqrt{\frac{1-k^2t^{2}}{1-t^{2}}}{\mathrm{d}}t&=
-\frac{kx\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}{1+kx^{2}}\\
&\qquad+\frac{1+k}{2}\int_{0}^{\frac{(1+k)x}{1+kx^{2}}}\sqrt{\dfrac{1-\big({\small{\frac{2\sqrt{k}}{1+k}}}t\big)^2}{1-t^{2}}}\mathrm{d}t\\
&\qquad\qquad+\frac{1-k}{2}\int_{0}^{\frac{(1+k)x}{1+kx^{2}}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\big({\small{\frac{2\sqrt{k}}{1+k}}}t\big)^2\big)}}\\
\\
\int_{0}^{\frac{2x\sqrt{1-x^{2}}}{\left(1+k\right)\sqrt{1-(\left.2\sqrt{k}\>\!x \middle/(1+k)\right.)^{2}}}}\sqrt{\frac{1-k^2t^{2}}{1-t^{2}}}dt&=-\frac{2kx\sqrt{1-x^{2}}}{\left(1+k\right)\sqrt{1-\big({\small{\frac{2\sqrt{k}}{1+k}}}t\big)^2}}\\
&\qquad+(1+k)\int_{0}^{x}\sqrt{\dfrac{1-\big({\small{\frac{2\sqrt{k}}{1+k}}}t\big)^2}{1-t^{2}}}\mathrm{d}t\\
&\qquad\qquad+(1-k)\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\big({\small{\frac{2\sqrt{k}}{1+k}}}t\big)^2\big)}}\\

\end{align*}

2. \int_{0}^{x}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-\left(kt\right)^{2}\right)}}dt
3. \frac{2}{1+k}\int_{0}^{\frac{\sqrt{(1+x)(1+kx)}-\sqrt{(1-x)(1-kx)}}{2}}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-\left(\frac{2\sqrt{k}t}{1+k}\right)^{2}\right)}}dt
4. \int_{0}^{x}\sqrt{\frac{1-\left(kt\right)^{2}}{1-t^{2}}}dt+kx
5. \left(1+k\right)\int_{0}^{\frac{\sqrt{(1+x)(1+kx)}-\sqrt{(1-x)(1-kx)}}{2}}\sqrt{\frac{1-\left(\frac{2\sqrt{k}t}{1+k}\right)^{2}}{1-t^{2}}}dt+\left(1-k\right)\int_{0}^{\frac{\sqrt{(1+x)(1+kx)}-\sqrt{(1-x)(1-kx)}}{2}}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-\left(\frac{2\sqrt{k}t}{1+k}\right)^{2}\right)}}dt

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青青子衿

\begin{align*}
\Pi(x,\beta,k)
&=\int_{0}^{x}\frac{{\mathrm{d}t}}{\left(1-\beta^{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}\\
\\
&=\dfrac{2k}{\left(1+k\right){\small{\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}}}\int_{0}^{\frac{\sqrt{\left(1+x\right)\left(1+kx\right)}-\sqrt{\left(1-x\right)\left(1-kx\right)}}{2}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)(1-({\scriptsize{\frac{2\sqrt{k}}{1+k}}}t)^{2})}}\\
&\qquad-\int_{0}^{\frac{\sqrt{\left(1+x\right)\left(1+kx\right)}-\sqrt{\left(1-x\right)\left(1-kx\right)}}{2}}\frac{\tfrac{\scriptsize{\big(\sqrt{(1-\beta)(k+\beta)}-\sqrt{(1+\beta)(k-\beta)}\>\!\>\!\big)^{2}}}{{\scriptsize{\left(1+k\right)\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}}}}{\left(1-({\scriptsize{\frac{\sqrt{(1-\beta)(k-\beta)}+\sqrt{(1+\beta)(k+\beta)}}{1+k}}}t)^{2}\right)\sqrt{\left(1-t^{2}\right)(1-({\scriptsize{\frac{2\sqrt{k}}{1+k}}}t)^{2})}}{\mathrm{d}}t\\
&\qquad\qquad-\dfrac{\beta^{2}x}{{\small{\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}}}\int_{0}^{1}\frac{{\mathrm{d}}s}{\sqrt{s^{2}+\left(1-s^{2}\right)\left(1-\beta^{2}x^{2}\right)}}
\end{align*}

1. \int_{0}^{x}\frac{1}{\left(1-\beta^{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}dt
2. \int_{0}^{\frac{\sqrt{\left(1+x\right)\left(1+kx\right)}-\sqrt{\left(1-x\right)\left(1-kx\right)}}{2}}\frac{\frac{2k}{\left(1+k\right)\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}}{\sqrt{\left(1-t^{2}\right)\left(1-\left(\frac{2\sqrt{k}}{1+k}t\right)^{2}\right)}}dt-\int_{0}^{\frac{\sqrt{\left(1+x\right)\left(1+kx\right)}-\sqrt{\left(1-x\right)\left(1-kx\right)}}{2}}\frac{\frac{\left(\sqrt{\left(1-\beta\right)\left(k+\beta\right)}-\sqrt{\left(1+\beta\right)\left(k-\beta\right)}\right)^{2}}{\left(1+k\right)\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}}{\left(1-\left(\frac{\sqrt{\left(1-\beta\right)\left(k-\beta\right)}+\sqrt{\left(1+\beta\right)\left(k+\beta\right)}}{1+k}\right)^{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-\left(\frac{2\sqrt{k}}{1+k}t\right)^{2}\right)}}dt-\int_{0}^{1}\frac{\frac{\beta^{2}x}{\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}}{\sqrt{s^{2}+\left(1-s^{2}\right)\left(1-\beta^{2}x^{2}\right)}}ds

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青青子衿

\begin{align*}
\left(1+k\right)\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}
&=\int_{0}^{\frac{(1+k)x^{2}}{1+kx^2}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\left(\scriptsize\frac{2\sqrt{k}}{1+k}t\right)^{2}\big)}}\\

&=\int_{\frac{1+kx^{2}}{2\sqrt{k}x}}^{+\infty}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\left(\scriptsize\frac{2\sqrt{k}}{1+k}t\right)^{2}\big)}}
\end{align*}

\begin{align*}
\Pi(1,\beta,k)
&=
\int_{0}^{1}\frac{\mathrm{d}t}{\left(1-\beta^{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}
\\
&=\frac{k-\beta^{2}}{2\left(1+k\right)\left(k+\beta^{2}\right)}\int_{0}^{1}\frac{\mathrm{d}t}{\big(1-({\scriptsize{\frac{2\sqrt{k}\beta}{k+\beta^{2}}}}t)^{2}\big)\sqrt{(1-t^{2})\big(1-\big({\scriptsize{\frac{2\sqrt{k}}{1+k}}}t\big)^{2}\big)}}\\
&\qquad+\frac{1}{2}\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}\\

\\
\Pi(x,\beta,k)&=\int_{0}^{x}\frac{\mathrm{d}t}{\left(1-\beta^{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}\\
\\
&=\frac{k}{\left(1+k\right)\left(k+\beta^{2}\right)}\int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\big({\scriptsize{\frac{2\sqrt{k}}{1+k}}}t\big)^{2}\big)}}\\

&\qquad-\frac{k-\beta^{2}}{2\left(1+k\right)\left(k+\beta^{2}\right)}\int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{\mathrm{d}t}{\big(1-({\scriptsize{\frac{k+\beta^{2}}{\left(1+k\right)\beta}}}t)^{2}\big)\sqrt{\left(1-t^{2}\right)\big(1-\big({\scriptsize{\frac{2\sqrt{k}}{1+k}}}t\big)^{2}\big)}}\\

&\qquad\qquad+\tfrac{\beta}{2\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}\operatorname{artanh}\left(\tfrac{x\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}{\beta\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}\right)\\
\\
&=\frac{1}{2\left(1+k\right)}\int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\big(1-\big({\scriptsize{\frac{2\sqrt{k}}{1+k}}}t\big)^{2}\big)}}\\
&\qquad\>\>+\frac{k-\beta^{2}}{2\left(1+k\right)\left(k+\beta^{2}\right)}\int_{\frac{1+kx^{2}}{2\sqrt{k}x}}^{+\infty}\frac{\mathrm{d}t}{\big(1-({\scriptsize{\frac{2\sqrt{k}\beta}{k+\beta^{2}}}}t)^{2}\big)\sqrt{\left(1-t^{2}\right)\big(1-\big({\scriptsize{\frac{2\sqrt{k}}{1+k}}}t\big)^{2}\big)}}\\
&\qquad\qquad\>\>+\tfrac{\beta}{2\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}\operatorname{artanh}\left(\tfrac{x\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}{\beta\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}\right)\\
\\
&=\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}\\
&\qquad-\int_{0}^{x}\frac{\mathrm{d}t}{\left(1-\frac{k^{2}}{\beta^{2}}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}\\
&\qquad\qquad+\tfrac{\beta}{\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}\operatorname{artanh}\left(\tfrac{x\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}{\beta\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}\right)\\
\end{align*}

1. NumberForm[
2. N[2 EllipticPi[\[Beta]^2, \[Pi]/2, k^2] -
3.     EllipticK[k^2] /. {\[Beta] -> 2/10, k -> 3/10}, 50], 50]
4. NumberForm[
5. N[(k - \[Beta]^2)/((1 + k) (k + \[Beta]^2))
6.      EllipticPi[((2 Sqrt[k] \[Beta])/(k + \[Beta]^2))^2, \[Pi]/
7.      2, ((2 Sqrt[k])/(1 + k))^2] /. {\[Beta] -> 2/10, k -> 3/10},
8.   50], 50]
9. (D[((1 + k) x)/(1 + k*x^2), x]/
10.    Sqrt[(1 - (((1 + k) x)/(
11.        1 + k*x^2))^2) (1 - ((2 Sqrt[k])/(1 + k)*((1 + k) x)/(
12.         1 + k*x^2))^2)]  -
13.    D[((1 + k) x)/(1 + k*x^2),
14.     x]/((1 - (\[Beta]*((1 + k) x)/(
15.         1 + k*x^2))^2) Sqrt[(1 - (((1 + k) x)/(
16.         1 + k*x^2))^2) (1 - ((2 Sqrt[k])/(1 + k)*((1 + k) x)/(
17.          1 + k*x^2))^2)]))^2 // Factor
18. (D[1/((2 Sqrt[k])/(1 + k)*((1 + k) x)/(1 + k*x^2)),
19.    x]/((1 - ((2 Sqrt[k])/(
20.        1 + k)/\[Beta]*1/((2 Sqrt[k])/(1 + k)*((1 + k) x)/(
21.         1 + k*x^2)))^2) Sqrt[(1 - (1/((2 Sqrt[k])/(
22.         1 + k)*((1 + k) x)/(1 + k*x^2)))^2) (1 - ((2 Sqrt[k])/(
23.         1 + k)*1/((2 Sqrt[k])/(1 + k)*((1 + k) x)/(
24.          1 + k*x^2)))^2)] ))^2 // Factor

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（次参倒数关系）
\begin{align*}
\Pi_a+\Pi_b&=\int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{\mathrm{d}t}{\big(1-\big({\scriptsize{\frac{k+\beta^{2}}{(1+k)\beta}}}t\big)^{2}\big)\sqrt{(1-t^{2})(1-\big({\scriptsize{\frac{2\sqrt{k}}{1+k}}}t\big)^{2})}}\\
&\qquad\qquad+\int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{\mathrm{d}t}{\big(1-\big({\scriptsize{\frac{2\sqrt{k}\beta}{k+\beta^{2}}}}t\big)^{2}\big)\sqrt{(1-t^{2})\big(1-\big({\scriptsize{\frac{2\sqrt{k}}{1+k}}}t\big)^{2}\big)}}\\
\\
&=\int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})\big(1-\big({\scriptsize{\frac{2\sqrt{k}}{1+k}}}t\big)^{2}\big)}}
\\
&\qquad\quad+\tfrac{\beta\left(1+k\right)\left(k+\beta^{2}\right)}{\left(k-\beta^{2}\right)\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}\operatorname{artanh}\left(\tfrac{x\left(k-\beta^{2}\right)\left(1+kx^{2}\right)\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}{\beta\left(k+\beta^{2}\right)\left(1-kx^{2}\right)\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}\right)

\end{align*}

\begin{align*}
G&=\int_{0}^{x}\frac{\mathrm{d}t}{\left(1-\beta^{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}
-\frac{1}{2}\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^2\right)}}\\
&\qquad\quad-\int_{0}^{\frac{(1+k)x}{1+kx^{2}}}\frac{\frac{k-\beta^{2}}{2(1+k)(k+\beta^{2})}}{\big(1-\big(\frac{2\sqrt{k}\beta}{k+\beta^{2}}t\big)^{2}\big)\sqrt{(1-t^{2})\big(1-\big(\frac{2\sqrt{k}}{1+k}t\big)^{2}\big)}}\mathrm{d}t\\
\\
&={\scriptsize{\frac{\beta}{2\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}}}\operatorname{artanh}\left({\scriptsize{\frac{2\beta x\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}{k+\beta^{2}+k^{2}x^{2}-2\beta^{2}x^{2}-2k\beta^{2}x^{2}-2k^{2}\beta^{2}x^{2}+\beta^{4}x^{2}+k\beta^{4}x^{4}+k^{2}\beta^{2}x^{4}}}}\right)
\end{align*}

2. \beta=0.56
3. k=0.63
4. \int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{1}{\left(1-\left(\frac{k+\beta^{2}}{\left(1+k\right)\beta}t\right)^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-\left(\frac{2\sqrt{k}}{1+k}t\right)^{2}\right)}}dt
6. \int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-\left(\frac{2\sqrt{k}}{1+k}t\right)^{2}\right)}}dt-\int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{1}{\left(1-\left(\frac{2\sqrt{k}\beta}{k+\beta^{2}}t\right)^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-\left(\frac{2\sqrt{k}}{1+k}t\right)^{2}\right)}}dt+\frac{\beta\left(1+k\right)\left(k+\beta^{2}\right)}{\left(k-\beta^{2}\right)\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}\operatorname{artanh}\left(\frac{x\left(k-\beta^{2}\right)\left(1+kx^{2}\right)\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}{\beta\left(k+\beta^{2}\right)\left(1-kx^{2}\right)\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}\right)
8. \int_{0}^{x}\frac{1}{\left(1-\beta^{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}dt-\frac{1}{2}\int_{0}^{x}\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}dt-\int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{\frac{k-\beta^{2}}{2\left(1+k\right)\left(k+\beta^{2}\right)}}{\left(1-\left(\frac{2\sqrt{k}\beta}{k+\beta^{2}}t\right)^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-\left(\frac{2\sqrt{k}}{1+k}t\right)^{2}\right)}}dt
9. \frac{\beta}{2\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}\operatorname{artanh}\left(\frac{2\beta x\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}{k+\beta^{2}+k^{2}x^{2}-2\beta^{2}x^{2}-2k\beta^{2}x^{2}-2k^{2}\beta^{2}x^{2}+\beta^{4}x^{2}+k\beta^{4}x^{4}+k^{2}\beta^{2}x^{4}}\right)

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青青子衿

青青子衿 发表于 2023-10-13 09:40
\begin{align*}
\frac{\beta x}{2}\int_{0}^{1}\frac{\mathrm{d}s}{\sqrt{s^{2}\beta^{2}\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)
+(1-s^{2})(\beta^{2}-k^{2}x^{2})(1-\beta^{2}x^{2})}}
\end{align*}

\begin{align*}
\int_{0}^{x}\frac{1-k^{2}t^{4}}{\left(1-\beta^{2}t^{2}\right)\left(\beta^{2}-k^{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}\mathrm{d}t=\frac{\operatorname{artanh}\left(\frac{x\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}{\beta\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}\right)}{\beta\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}
\end{align*}

\begin{align*}
\Pi_c+\Pi_d&=\int_{0}^{x}\frac{1}{\left(1-\beta^{2}t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}dt\\
&\qquad\quad+\tfrac{k-\beta^{2}}{2\left(1+k\right)\left(k+\beta^{2}\right)}\int_{0}^{\frac{\left(1+k\right)x}{1+kx^{2}}}\frac{\mathrm{d}t}{\big(1-\big({\scriptsize\frac{k+\beta^{2}}{(1+k)\beta}}t\big)^{2}\big)\sqrt{(1-t^{2})\big(1-\big({\scriptsize\frac{2\sqrt{k}}{1+k}}t\big)^{2}\big)}}\\
&=\frac{k}{k+\beta^{2}}\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})\left(1-k^2t^{2}\right)}}\\
&\qquad\quad+\tfrac{\beta}{2\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}\operatorname{artanh}\left(\tfrac{x\sqrt{\left(1-\beta^{2}\right)\left(k^{2}-\beta^{2}\right)}}{\beta\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}}\right)
\end{align*}

\begin{align*}
\left\{\begin{split}
k&=s^2\\
l&=\frac{2s}{1+s^2}
\end{split}\right.\\
\\
l^2=\frac{4k}{(1+k)^2}
\end{align*}


\begin{align*}
&\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-k^{2}t^{2})}}=\frac{1}{(1+\sqrt{k}\>\!)^{2}}
\int_{0}^{\frac{(1+\sqrt{k}\>\!)^{2}x(1+k\ x^{2})}{1+2(1+\sqrt{k}+k)\sqrt{k}x^{2}+k^{2}x^{4}}}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-{\scriptsize\frac{8(1+k)\sqrt{k}}{(1+\sqrt{k}\>\!)^{4}}}t^{2})}}
\end{align*}

1. \int_{0}^{x}\frac{1}{\sqrt{(1-t^{2})(1-k^{2}t^{2})}}dt
2. \frac{1}{(1+\sqrt{k})^{2}}\int_{0}^{\frac{(1+\sqrt{k})^{2}x(1+k\ x^{2})}{1+2\left(1+\sqrt{k}+k\right)\sqrt{k}x^{2}+k^{2}x^{4}}}\frac{1}{\sqrt{(1-t^{2})(1-\frac{8(1+k)\sqrt{k}}{(1+\sqrt{k})^{4}}t^{2})}}dt

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青青子衿

青青子衿 发表于 2024-1-20 13:12
\begin{align*}
&\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-k^{2}t^{2})}}=\frac{1}{(1+\sqrt{k}\>\!)^{2}}
\int_{0}^{\frac{(1+\sqrt{k}\>\!)^{2}x(1+k\ x^{2})}{1+2(1+\sqrt{k}+k)\sqrt{k}x^{2}+k^{2}x^{4}}}\frac{\mathrm{d}t}{\sqrt{(1-t^{2})(1-{\scriptsize\frac{8(1+k)\sqrt{k}}{(1+\sqrt{k}\>\!)^{4}}}t^{2})}}
\end{align*}

【虚模】


\begin{gather*}
\int_{0}^{x}\frac{{\mathrm{d}}t}{\sqrt{\left(1-t^{2}\right)\left(1-m^{2}t^{2}\right)}}\\
\\
=\int_{0}^{\frac{\sqrt{1-m^{2}}x}{\sqrt{1-m^{2}x^{2}}}}\frac{\frac{1}{\sqrt{1-m^{2}}}}{\sqrt{(1-t^{2})(1-(\raise{1pt}{\scriptsize\frac{m}{i\sqrt{1-m^2}}}t)^{2})}}{\mathrm{d}}t\\
=\int_{0}^{\frac{(m-i\sqrt{1-m^{2}})x\sqrt{1-m^{2}x^{2}}}{\sqrt{1-x^{2}}}}\frac{\frac{i}{\sqrt{1-m^{2}}+im}}{\sqrt{(1-t^{2})(1-(\raise{0.6pt}{\scriptsize\frac{\sqrt{1-m^{2}}-im}{\sqrt{1-m^{2}}+im}}t)^{2})}}{\mathrm{d}}t\\

=\int_{0}^{\frac{(\sqrt{1-m^{2}}+im)x\sqrt{1-m^{2}x^{2}}}{1+({\scriptsize\frac{\sqrt{2m(1-m)}+i\sqrt{2m(m+1)}}{2}}x)^{2}}}\frac{\frac{1}{\sqrt{1-m^{2}}+im}}{\sqrt{(1-t^{2})(1-(\raise{0.6pt}{\scriptsize\frac{2\sqrt{im\sqrt{1-m^{2}}}}{\sqrt{1-m^{2}}+im}}t)^{2})}}{\mathrm{d}}t


\end{gather*}

1. (I*EllipticK[
2.    1 - ((2 Sqrt[I (Sqrt[2] - 1) Sqrt[1 - (Sqrt[2] - 1)^2]])/(
3.      Sqrt[1 - (Sqrt[2] - 1)^2] + I (Sqrt[2] - 1)))^2])/
4. EllipticK[((2 Sqrt[I (Sqrt[2] - 1) Sqrt[1 - (Sqrt[2] - 1)^2]])/(
5.    Sqrt[1 - (Sqrt[2] - 1)^2] + I (Sqrt[2] - 1)))^2] // N
6. (1 + Sqrt[2] I)/2 // N
7. (I*EllipticK[
8.    1 - ((Sqrt[1 - (Sqrt[2] - 1)^2] - I (Sqrt[2] - 1))/(
9.      Sqrt[1 - (Sqrt[2] - 1)^2] + I (Sqrt[2] - 1)))^2])/
10. EllipticK[((Sqrt[1 - (Sqrt[2] - 1)^2] - I (Sqrt[2] - 1))/(
11.    Sqrt[1 - (Sqrt[2] - 1)^2] + I (Sqrt[2] - 1)))^2] // N
12. (-2 + 2 Sqrt[2] I)/3 // N

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\begin{gather*}
K(\>{\scriptsize(1+i)(2(\sqrt{2}-1)-i\ (2-\sqrt{2})\sqrt{\tiny\sqrt{2}-1})(2(\sqrt{2}-1))^{1/4}}\>)\\
=({\scriptsize\sqrt{2}}+i({\scriptsize\sqrt{2}-1})\sqrt{\tiny1+\sqrt{2}}\,)\frac{\Gamma\left(\frac{1}{8}\right)\Gamma\left(\frac{3}{8}\right)}{2^{13/4}\sqrt{\pi}}
\end{gather*}

1. I EllipticK[
2.    1 - ((1 + I) (2 (Sqrt[2] - 1) -
3.         I (2 - Sqrt[2]) Sqrt[Sqrt[2] - 1]) (2 Sqrt[2] - 2)^(1/4))^2]/
4.   EllipticK[((1 + I) (2 (Sqrt[2] - 1) -
5.        I (2 - Sqrt[2]) Sqrt[Sqrt[2] - 1]) (2 Sqrt[2] - 2)^(
6.      1/4))^2] // N
7. (1 + I Sqrt[2])/2 // N

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\begin{gather*}
\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2}t^{2}\right)}}=\frac{1}{1+2\sqrt{\scriptsize{k\sqrt{\left.k\middle/l\right.\,}\,}}}\int_{0}^{y_{3,1}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-l^{2}t^{2}\right)}}\\
\\
y_{3,1}=\frac{x\left(1+2\sqrt{\scriptsize{k\sqrt{\left.k\middle/l\right.\,}\,}}+{\small{k\sqrt{\left.k\middle/l\right.\,}\,}}x^{2}\right)}{1+2\sqrt{\scriptsize{k\sqrt{\left.k\middle/l\right.\,}\,}}x^{2}+{\small{k\sqrt{\left.k\middle/l\right.\,}\,}}x^{2}}\\
\\
\\
1+2i\sqrt{2}=i\frac{K\big(\sqrt{1-k^{2}}\,\big)}{K(k)}=3i\frac{K\big(\sqrt{1-l^{2}}\,\big)}{K(l)}=3\cdot\tfrac{1+2i\sqrt{2}}{3}\\
\\
\left\{\begin{split}
k=\left(\tfrac{s^{3}(2+s)}{1+2s}\right)^{\frac{1}{2}}\\
l=\left(\tfrac{s(2+s)^{3}}{(1+2s)^{3}}\right)^{\frac{1}{2}}
\end{split}\right.\\
\\
s=\sqrt{\scriptsize{k\sqrt{\left.k\middle/l\right.\,}\,}}
\end{gather*}

\begin{gather*}
K\left(\tau=\tfrac{1+2i\sqrt{2}}{3}\right)\\
=
K\left(k=\tfrac{(1+i)(\sqrt{2}-1)(2+\sqrt{2}-i\sqrt{\scriptsize1+\sqrt{2}}\,)}{4}(2+2\sqrt{2})^{1/4}\right)\\
=(1+i)(\sqrt{2}-i)(2+\sqrt{2}-2\sqrt{\scriptsize1+\sqrt{2}})(2+2\sqrt{2})^{1/4}K\left(k=5+4\sqrt{2}-2(2+\sqrt{2})\sqrt{\scriptsize1+\sqrt{2}}\right)\\
=(1+i)(\sqrt{2}-i)(2+\sqrt{2}-2\sqrt{\scriptsize1+\sqrt{2}})(2+2\sqrt{2})^{1/4}\frac{(\sqrt{2}+\sqrt{\scriptsize1+\sqrt{2}})\Gamma\left(\frac{1}{8}\right)\Gamma\left(\frac{3}{8}\right)}{2^{17/4}\sqrt{\pi}}\\
=\frac{(1+i)(\sqrt{2}-i)(\sqrt{2}-1)(2+2\sqrt{2})^{3/4}\Gamma\left(\frac{1}{8}\right)\Gamma\left(\frac{3}{8}\right)}{2^{17/4}\sqrt{\pi}}
\end{gather*}

1. N[EllipticK[ModularLambda[1 + I Sqrt[17]]], 20]
2. N[Sqrt[1 - ModularLambda[I Sqrt[17]]]
3.    EllipticK[ModularLambda[I Sqrt[17]]], 20]
4. N[EllipticK[ModularLambda[(1 + I Sqrt[17])/3]], 20]
5. N[(1 + 2 Sqrt[
6.      Sqrt[ModularLambda[1 + I Sqrt[17]]] Sqrt[Sqrt[
7.       ModularLambda[1 + I Sqrt[17]]]/Sqrt[
8.       ModularLambda[(1 + I Sqrt[17])/3]]]]) EllipticK[
9.    ModularLambda[1 + I Sqrt[17]]], 20]

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