{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "第１種完全楕円積分\n",
    "$$K(k)=\\int_{0}^{\\frac{\\pi}{2}}{\\frac{1}{\\sqrt{1-k^2\\,\\sin ^2\\varphi}}\\;d\\varphi}$$\n",
    "に成り立つ次の関係式\n",
    "$$K(\\frac{2\\,\\sqrt{x}}{x+1})=(x+1)\\,K(x)$$\n",
    "を示します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{1}$}\\left[ x<1 \\right] \\]"
      ],
      "text/plain": [
       "(%o1)                               [x < 1]"
      ],
      "text/x-maxima": [
       "[x < 1]"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{2}$}\\left[ x>-1 \\right] \\]"
      ],
      "text/plain": [
       "(%o2)                              [x > - 1]"
      ],
      "text/x-maxima": [
       "[x > -1]"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "kill(all)$\n",
    "assume(x<1);\n",
    "assume(-1<x);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "この積分の計算は積分範囲を$0 \\leq \\varphi \\leq \\pi$に広げておく必要があります。被積分関数は$\\varphi=\\frac{\\pi}{2}$で対称ですから積分の値は２倍になるだけです。つまり、\n",
    "$$K(k)=\\frac{1}{2}\\,\\int_{0}^{\\pi}{\\frac{1}{\\sqrt{1-k^2\\,\\sin ^2\\varphi}}\\;d\\varphi}$$\n",
    "以下の計算ではこの$\\frac{1}{2}$を除いた積分の部分だけを計算します。後で忘れずに2で割ることにしましょう。\n",
    "\n",
    "$k=\\frac{2\\,\\sqrt{x}}{x+1}$を積分範囲を広げた積分の$k$に代入すると以下の積分になります。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{3}$}\\int_{0}^{\\pi}{\\frac{1}{\\sqrt{1-\\frac{4\\,\\sin ^2\\varphi\\,x}{\\left(x+1\\right)^2}}}\\;d\\varphi}\\]"
      ],
      "text/plain": [
       "                        %pi\n",
       "                       /\n",
       "                       [               1\n",
       "(%o3)                  I    ----------------------- dphi\n",
       "                       ]                  2\n",
       "                       /             4 sin (phi) x\n",
       "                        0   sqrt(1 - -------------)\n",
       "                                              2\n",
       "                                       (x + 1)"
      ],
      "text/x-maxima": [
       "'integrate(1/sqrt(1-(4*sin(phi)^2*x)/(x+1)^2),phi,0,%pi)"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F1:integrate(1/sqrt(1-4*x/(1+x)^2*sin(phi)^2),phi,0,%pi);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "通分して、$x\\gt0$に注意して$x+1$をくくり出します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{4}$}\\left(x+1\\right)\\,\\int_{0}^{\\pi}{\\frac{1}{\\sqrt{\\left(x+1\\right)^2-4\\,\\sin ^2\\varphi\\,x}}\\;d\\varphi}\\]"
      ],
      "text/plain": [
       "                        %pi\n",
       "                       /\n",
       "                       [                  1\n",
       "(%o4)          (x + 1) I    ------------------------------ dphi\n",
       "                       ]                2        2\n",
       "                       /    sqrt((x + 1)  - 4 sin (phi) x)\n",
       "                        0"
      ],
      "text/x-maxima": [
       "(x+1)*'integrate(1/sqrt((x+1)^2-4*sin(phi)^2*x),phi,0,%pi)"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F2:substpart(xthru(part(F1,1)),F1,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "平方根記号の中を展開します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{5}$}\\left(x+1\\right)\\,\\int_{0}^{\\pi}{\\frac{1}{\\sqrt{x^2-4\\,\\sin ^2\\varphi\\,x+2\\,x+1}}\\;d\\varphi}\\]"
      ],
      "text/plain": [
       "                      %pi\n",
       "                     /\n",
       "                     [                    1\n",
       "(%o5)        (x + 1) I    ---------------------------------- dphi\n",
       "                     ]          2        2\n",
       "                     /    sqrt(x  - 4 sin (phi) x + 2 x + 1)\n",
       "                      0"
      ],
      "text/x-maxima": [
       "(x+1)*'integrate(1/sqrt(x^2-4*sin(phi)^2*x+2*x+1),phi,0,%pi)"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F3:substpart(expand(part(F2,2,1)),F2,2,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "平方根記号の中を$x$の多項式として整理します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{6}$}\\left(x+1\\right)\\,\\int_{0}^{\\pi}{\\frac{1}{\\sqrt{x^2+\\left(2-4\\,\\sin ^2\\varphi\\right)\\,x+1}}\\;d\\varphi}\\]"
      ],
      "text/plain": [
       "                      %pi\n",
       "                     /\n",
       "                     [                    1\n",
       "(%o6)        (x + 1) I    ---------------------------------- dphi\n",
       "                     ]          2             2\n",
       "                     /    sqrt(x  + (2 - 4 sin (phi)) x + 1)\n",
       "                      0"
      ],
      "text/x-maxima": [
       "(x+1)*'integrate(1/sqrt(x^2+(2-4*sin(phi)^2)*x+1),phi,0,%pi)"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F4:substpart(ratsimp(part(F3,2),x),F3,2);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$x$の係数に三角関数の倍角公式を使います。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{7}$}\\left(x+1\\right)\\,\\int_{0}^{\\pi}{\\frac{1}{\\sqrt{x^2+2\\,\\cos \\left(2\\,\\varphi\\right)\\,x+1}}\\;d\\varphi}\\]"
      ],
      "text/plain": [
       "                         %pi\n",
       "                        /\n",
       "                        [                  1\n",
       "(%o7)           (x + 1) I    ----------------------------- dphi\n",
       "                        ]          2\n",
       "                        /    sqrt(x  + 2 cos(2 phi) x + 1)\n",
       "                         0"
      ],
      "text/x-maxima": [
       "(x+1)*'integrate(1/sqrt(x^2+2*cos(2*phi)*x+1),phi,0,%pi)"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F5:substpart(trigreduce(part(F4,2,1)),F4,2,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "三角関数を指数関数で書き直します。ここで虚数単位が出てきます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{8}$}\\left(x+1\\right)\\,\\int_{0}^{\\pi}{\\frac{1}{\\sqrt{x^2+\\left(e^{2\\,i\\,\\varphi}+e^ {- 2\\,i\\,\\varphi }\\right)\\,x+1}}\\;d\\varphi}\\]"
      ],
      "text/plain": [
       "                 %pi\n",
       "                /\n",
       "                [                         1\n",
       "(%o8)   (x + 1) I    -------------------------------------------- dphi\n",
       "                ]          2      2 %i phi     - 2 %i phi\n",
       "                /    sqrt(x  + (%e         + %e          ) x + 1)\n",
       "                 0"
      ],
      "text/x-maxima": [
       "(x+1)*'integrate(1/sqrt(x^2+(%e^(2*%i*phi)+%e^-(2*%i*phi))*x+1),phi,0,%pi)"
      ]
     },
     "execution_count": 7,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F6:exponentialize(F5);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "分母の平方根記号の中身が因数分解できることを、因数分解の結果を展開することで示します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{9}$}\\left(e^ {- 2\\,i\\,\\varphi }\\,x+1\\right)\\,\\left(e^{2\\,i\\,\\varphi}\\,x+1\\right)\\]"
      ],
      "text/plain": [
       "                       - 2 %i phi           2 %i phi\n",
       "(%o9)               (%e           x + 1) (%e         x + 1)"
      ],
      "text/x-maxima": [
       "(%e^-(2*%i*phi)*x+1)*(%e^(2*%i*phi)*x+1)"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "FT1:(1+x*exp(2*%i*phi))*(1+x*exp(-2*%i*phi));"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{10}$}x^2+e^{2\\,i\\,\\varphi}\\,x+e^ {- 2\\,i\\,\\varphi }\\,x+1\\]"
      ],
      "text/plain": [
       "                     2     2 %i phi       - 2 %i phi\n",
       "(%o10)              x  + %e         x + %e           x + 1"
      ],
      "text/x-maxima": [
       "x^2+%e^(2*%i*phi)*x+%e^-(2*%i*phi)*x+1"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "%,expand;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "というわけでF6の式は以下の式と等しいことが分かります。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{11}$}\\left(x+1\\right)\\,\\int_{0}^{\\pi}{\\frac{1}{\\sqrt{\\left(e^ {- 2\\,i\\,\\varphi }\\,x+1\\right)\\,\\left(e^{2\\,i\\,\\varphi}\\,x+1\\right)}}\\;d\\varphi}\\]"
      ],
      "text/plain": [
       "                 %pi\n",
       "                /\n",
       "                [                          1\n",
       "(%o11)  (x + 1) I    --------------------------------------------- dphi\n",
       "                ]            - 2 %i phi           2 %i phi\n",
       "                /    sqrt((%e           x + 1) (%e         x + 1))\n",
       "                 0"
      ],
      "text/x-maxima": [
       "(x+1)*'integrate(1/sqrt((%e^-(2*%i*phi)*x+1)*(%e^(2*%i*phi)*x+1)),phi,0,%pi)"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F7:substpart(FT1,F6,2,1,2,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "一旦$(x+1)$の項を忘れて積分の変形に集中します。後で忘れずに$(x+1)$をかけることにします。\n",
    "\n",
    "被積分関数を$\\frac{1}{\\sqrt{e^{-2\\,i\\,\\varphi}\\,x+1}}\\,\\frac{1}{\\sqrt{e^{2\\,i\\,\\varphi}\\,x+1}}$と見ると$\\frac{1}{\\sqrt{1+t}}$の形をした２つの関数の積になっています。一旦それぞれを別々に冪級数に展開します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{12}$}\\sqrt{\\pi}\\,\\sum_{n=1}^{\\infty }{\\frac{t^{n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}+1\\]"
      ],
      "text/plain": [
       "                           inf\n",
       "                           ====              n\n",
       "                           \\                t\n",
       "(%o12)           sqrt(%pi)  >    ------------------------- + 1\n",
       "                           /           1\n",
       "                           ====  gamma(- - n) gamma(n + 1)\n",
       "                           n = 1       2"
      ],
      "text/x-maxima": [
       "sqrt(%pi)*'sum(t^n/(gamma(1/2-n)*gamma(n+1)),n,1,inf)+1"
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "niceindices(makegamma(powerseries(1/sqrt(1+t),t,0))),niceindicespref:[n];"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "最初の項は$t=e^{-2\\,i\\,\\varphi}\\,x$なのでこの代入を行います。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{13}$}\\sqrt{\\pi}\\,\\sum_{n=1}^{\\infty }{\\frac{e^ {- 2\\,i\\,n\\,\\varphi }\\,x^{n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}+1\\]"
      ],
      "text/plain": [
       "                           inf\n",
       "                           ====        - 2 %i n phi  n\n",
       "                           \\         %e             x\n",
       "(%o13)           sqrt(%pi)  >    ------------------------- + 1\n",
       "                           /           1\n",
       "                           ====  gamma(- - n) gamma(n + 1)\n",
       "                           n = 1       2"
      ],
      "text/x-maxima": [
       "sqrt(%pi)*'sum((%e^-(2*%i*n*phi)*x^n)/(gamma(1/2-n)*gamma(n+1)),n,1,inf)+1"
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "%,t:exp(-2*%i*phi)*x;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "第１項で$n=0$の場合1になることから第２項の1を総和の中に取り込みます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{14}$}\\sqrt{\\pi}\\,\\sum_{n=0}^{\\infty }{\\frac{e^ {- 2\\,i\\,n\\,\\varphi }\\,x^{n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}\\]"
      ],
      "text/plain": [
       "                             inf\n",
       "                             ====        - 2 %i n phi  n\n",
       "                             \\         %e             x\n",
       "(%o14)             sqrt(%pi)  >    -------------------------\n",
       "                             /           1\n",
       "                             ====  gamma(- - n) gamma(n + 1)\n",
       "                             n = 0       2"
      ],
      "text/x-maxima": [
       "sqrt(%pi)*'sum((%e^-(2*%i*n*phi)*x^n)/(gamma(1/2-n)*gamma(n+1)),n,0,inf)"
      ]
     },
     "execution_count": 13,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F8:substpart(0,part(%,1),2,3);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "総和のインデックス変数を$m$に変えて冪級数展開を求めます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{15}$}\\sqrt{\\pi}\\,\\sum_{m=1}^{\\infty }{\\frac{t^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}+1\\]"
      ],
      "text/plain": [
       "                           inf\n",
       "                           ====              m\n",
       "                           \\                t\n",
       "(%o15)           sqrt(%pi)  >    ------------------------- + 1\n",
       "                           /           1\n",
       "                           ====  gamma(- - m) gamma(m + 1)\n",
       "                           m = 1       2"
      ],
      "text/x-maxima": [
       "sqrt(%pi)*'sum(t^m/(gamma(1/2-m)*gamma(m+1)),m,1,inf)+1"
      ]
     },
     "execution_count": 14,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "niceindices(makegamma(powerseries(1/sqrt(1+t),t,0))),niceindicespref:[m];"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "F7の被積分関数の第２項なので$t=e^{2\\,i\\,\\varphi}\\,x$の代入を行います。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{16}$}\\sqrt{\\pi}\\,\\sum_{m=1}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi}\\,x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}+1\\]"
      ],
      "text/plain": [
       "                           inf\n",
       "                           ====         2 %i m phi  m\n",
       "                           \\          %e           x\n",
       "(%o16)           sqrt(%pi)  >    ------------------------- + 1\n",
       "                           /           1\n",
       "                           ====  gamma(- - m) gamma(m + 1)\n",
       "                           m = 1       2"
      ],
      "text/x-maxima": [
       "sqrt(%pi)*'sum((%e^(2*%i*m*phi)*x^m)/(gamma(1/2-m)*gamma(m+1)),m,1,inf)+1"
      ]
     },
     "execution_count": 15,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "%,t:exp(2*%i*phi)*x;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "第２項の1を総和の$m=0$として取り込みます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{17}$}\\sqrt{\\pi}\\,\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi}\\,x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}\\]"
      ],
      "text/plain": [
       "                             inf\n",
       "                             ====         2 %i m phi  m\n",
       "                             \\          %e           x\n",
       "(%o17)             sqrt(%pi)  >    -------------------------\n",
       "                             /           1\n",
       "                             ====  gamma(- - m) gamma(m + 1)\n",
       "                             m = 0       2"
      ],
      "text/x-maxima": [
       "sqrt(%pi)*'sum((%e^(2*%i*m*phi)*x^m)/(gamma(1/2-m)*gamma(m+1)),m,0,inf)"
      ]
     },
     "execution_count": 16,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F9:substpart(0,part(%,1),2,3);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "従ってF7の被積分関数は次の式に等しくなります。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{18}$}\\pi\\,\\left(\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi}\\,x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}\\right)\\,\\sum_{n=0}^{\\infty }{\\frac{e^ {- 2\\,i\\,n\\,\\varphi }\\,x^{n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}\\]"
      ],
      "text/plain": [
       "            inf                              inf\n",
       "            ====         2 %i m phi  m       ====        - 2 %i n phi  n\n",
       "            \\          %e           x        \\         %e             x\n",
       "(%o18) %pi ( >    -------------------------)  >    -------------------------\n",
       "            /           1                    /           1\n",
       "            ====  gamma(- - m) gamma(m + 1)  ====  gamma(- - n) gamma(n + 1)\n",
       "            m = 0       2                    n = 0       2"
      ],
      "text/x-maxima": [
       "%pi*('sum((%e^(2*%i*m*phi)*x^m)/(gamma(1/2-m)*gamma(m+1)),m,0,inf))\n",
       "   *'sum((%e^-(2*%i*n*phi)*x^n)/(gamma(1/2-n)*gamma(n+1)),n,0,inf)"
      ]
     },
     "execution_count": 17,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F8*F9;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "この式は２つの無限級数の積になっており、展開することで$0\\le n, m\\lt \\infty$の全ての組み合わせについて積を取ることになります。和の順番を変更し入れ子にすることで次の式と等しいことが分かります。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{19}$}\\pi\\,\\sum_{n=0}^{\\infty }{\\frac{\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi-2\\,i\\,n\\,\\varphi}\\,x^{n+m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}\\]"
      ],
      "text/plain": [
       "                         inf\n",
       "                         ====    2 %i m phi - 2 %i n phi  n + m\n",
       "                         \\     %e                        x\n",
       "                          >    --------------------------------\n",
       "                   inf   /              1\n",
       "                   ====  ====     gamma(- - m) gamma(m + 1)\n",
       "                   \\     m = 0          2\n",
       "(%o19)         %pi  >    --------------------------------------\n",
       "                   /                 1\n",
       "                   ====        gamma(- - n) gamma(n + 1)\n",
       "                   n = 0             2"
      ],
      "text/x-maxima": [
       "%pi*'sum(('sum((%e^(2*%i*m*phi-2*%i*n*phi)*x^(n+m))/(gamma(1/2-m)*gamma(m+1)),\n",
       "               m,0,inf))\n",
       "          /(gamma(1/2-n)*gamma(n+1)),n,0,inf)"
      ]
     },
     "execution_count": 18,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F10:%pi*sum(sum(part(F8,2,1)*part(F9,2,1),m,0,inf),n,0,inf);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "２重総和の内側の総和を取り出します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{20}$}\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi-2\\,i\\,n\\,\\varphi}\\,x^{n+m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}\\]"
      ],
      "text/plain": [
       "                    inf\n",
       "                    ====    2 %i m phi - 2 %i n phi  n + m\n",
       "                    \\     %e                        x\n",
       "(%o20)               >    --------------------------------\n",
       "                    /              1\n",
       "                    ====     gamma(- - m) gamma(m + 1)\n",
       "                    m = 0          2"
      ],
      "text/x-maxima": [
       "'sum((%e^(2*%i*m*phi-2*%i*n*phi)*x^(n+m))/(gamma(1/2-m)*gamma(m+1)),m,0,inf)"
      ]
     },
     "execution_count": 19,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "G1:part(F10,2,1,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$x^n$は$m$と無関係なので総和の外にくくり出すことが来ます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{21}$}x^{n}\\,\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi-2\\,i\\,n\\,\\varphi}\\,x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}\\]"
      ],
      "text/plain": [
       "                        inf\n",
       "                        ====    2 %i m phi - 2 %i n phi  m\n",
       "                      n \\     %e                        x\n",
       "(%o21)               x   >    ----------------------------\n",
       "                        /            1\n",
       "                        ====   gamma(- - m) gamma(m + 1)\n",
       "                        m = 0        2"
      ],
      "text/x-maxima": [
       "x^n*'sum((%e^(2*%i*m*phi-2*%i*n*phi)*x^m)/(gamma(1/2-m)*gamma(m+1)),m,0,inf)"
      ]
     },
     "execution_count": 20,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "G2:x^n*substpart(part(G1,1)/x^n,G1,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "従ってF10は次の式と等しくなります。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{22}$}\\pi\\,\\sum_{n=0}^{\\infty }{\\frac{x^{n}\\,\\sum_{m=0}^{\\infty }{\\frac{e^{2\\,i\\,m\\,\\varphi-2\\,i\\,n\\,\\varphi}\\,x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}}\\]"
      ],
      "text/plain": [
       "                             inf\n",
       "                             ====    2 %i m phi - 2 %i n phi  m\n",
       "                           n \\     %e                        x\n",
       "                          x   >    ----------------------------\n",
       "                    inf      /            1\n",
       "                    ====     ====   gamma(- - m) gamma(m + 1)\n",
       "                    \\        m = 0        2\n",
       "(%o22)          %pi  >    -------------------------------------\n",
       "                    /                 1\n",
       "                    ====        gamma(- - n) gamma(n + 1)\n",
       "                    n = 0             2"
      ],
      "text/x-maxima": [
       "%pi*'sum((x^n*'sum((%e^(2*%i*m*phi-2*%i*n*phi)*x^m)/(gamma(1/2-m)*gamma(m+1)),\n",
       "                   m,0,inf))\n",
       "          /(gamma(1/2-n)*gamma(n+1)),n,0,inf)"
      ]
     },
     "execution_count": 21,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F11:substpart(G2,F10,2,1,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "F7の被積分関数がこの2重総和と等しいことが分かりました。積分変数$\\varphi$でこの式を積分をすると、積分変数に関係する部分は以下の部分式だけです。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{23}$}e^{2\\,i\\,m\\,\\varphi-2\\,i\\,n\\,\\varphi}\\]"
      ],
      "text/plain": [
       "                             2 %i m phi - 2 %i n phi\n",
       "(%o23)                     %e"
      ],
      "text/x-maxima": [
       "%e^(2*%i*m*phi-2*%i*n*phi)"
      ]
     },
     "execution_count": 22,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "G3:part(F11,2,1,1,2,1,1,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$n,m$はどちらも0以上の整数であることに注意します。すると積分の値はm=nの場合にはその値に関わらず$\\pi$になります。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{24}$}\\pi\\]"
      ],
      "text/plain": [
       "(%o24)                                %pi"
      ],
      "text/x-maxima": [
       "%pi"
      ]
     },
     "execution_count": 23,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "integrate(G3,phi,0,%pi),m=k,n=k;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "また$n\\neq m$の場合にはそれらの値に関わらず積分の値は0になります。例えば$m=4, n=9$で試しに計算してみます（これが成り立つのは積分範囲が$0\\le\\varphi\\le\\pi$の場合です。$0\\le\\varphi\\le\\frac{\\pi}{2}$では上手くいきません）。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{25}$}0\\]"
      ],
      "text/plain": [
       "(%o25)                                 0"
      ],
      "text/x-maxima": [
       "0"
      ]
     },
     "execution_count": 24,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "integrate(G3,phi,0,%pi),m=4,n=9;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "これらのことからF11を$\\varphi$で積分すると内側の総和は$m=n$の場合だけ積分の値が$\\pi$として残り、それ以外の$m,n$の組み合わせは積分値が0になることから消えてしまいます。従ってF11の2重総和の内側の総和は次の式と等しいことが分かります。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 25,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{26}$}\\pi\\,x^{n}\\,\\sum_{m=n}^{n}{\\frac{x^{m}}{\\Gamma\\left(\\frac{1}{2}-m\\right)\\,\\Gamma\\left(m+1\\right)}}\\]"
      ],
      "text/plain": [
       "                            n\n",
       "                           ====              m\n",
       "                         n \\                x\n",
       "(%o26)              %pi x   >    -------------------------\n",
       "                           /           1\n",
       "                           ====  gamma(- - m) gamma(m + 1)\n",
       "                           m = n       2"
      ],
      "text/x-maxima": [
       "%pi*x^n*'sum(x^m/(gamma(1/2-m)*gamma(m+1)),m,n,n)"
      ]
     },
     "execution_count": 25,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "G4:substpart(%pi,substpart(n,substpart(n,G2,2,3),2,4),2,1,1,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "この総和は自明に簡約することが出来ます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 26,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{27}$}\\frac{\\pi\\,x^{2\\,n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)\\,\\Gamma\\left(n+1\\right)}\\]"
      ],
      "text/plain": [
       "                                        2 n\n",
       "                                   %pi x\n",
       "(%o27)                     -------------------------\n",
       "                                 1\n",
       "                           gamma(- - n) gamma(n + 1)\n",
       "                                 2"
      ],
      "text/x-maxima": [
       "(%pi*x^(2*n))/(gamma(1/2-n)*gamma(n+1))"
      ]
     },
     "execution_count": 26,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "G5:G4,simpsum:true;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "上記のG5の式で内側の総和を置き換えることでF7の積分は以下の式に変形できます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 27,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{28}$}\\pi^2\\,\\sum_{n=0}^{\\infty }{\\frac{x^{2\\,n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)^2\\,\\Gamma\\left(n+1\\right)^2}}\\]"
      ],
      "text/plain": [
       "                         inf\n",
       "                         ====              2 n\n",
       "                       2 \\                x\n",
       "(%o28)              %pi   >    ---------------------------\n",
       "                         /          2 1           2\n",
       "                         ====  gamma (- - n) gamma (n + 1)\n",
       "                         n = 0        2"
      ],
      "text/x-maxima": [
       "%pi^2*'sum(x^(2*n)/(gamma(1/2-n)^2*gamma(n+1)^2),n,0,inf)"
      ]
     },
     "execution_count": 27,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F12:substpart(G5,F11,2,1,1);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "以下はこの式を超幾何関数の形に変形していきます。\n",
    "まずポッホハマー記号$\\left(\\frac{1}{2}\\right)_n$を思い出します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{29}$}\\left(\\left(\\frac{1}{2}\\right)\\right)_{n}\\]"
      ],
      "text/plain": [
       "                                      1\n",
       "(%o29)                               (-)\n",
       "                                      2 n"
      ],
      "text/x-maxima": [
       "pochhammer(1/2,n)"
      ]
     },
     "execution_count": 28,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::SIMP-UNIT-STEP in DEFUN\n",
      "SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::SIMP-POCHHAMMER in DEFUN\n"
     ]
    }
   ],
   "source": [
    "pochhammer(1/2,n);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$\\left(\\frac{1}{2}\\right)_n$を単純にガンマ関数で表示すると以下のようになりますが、これではちょっと上手くいきません。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{30}$}\\frac{\\Gamma\\left(n+\\frac{1}{2}\\right)}{\\sqrt{\\pi}}\\]"
      ],
      "text/plain": [
       "                                           1\n",
       "                                 gamma(n + -)\n",
       "                                           2\n",
       "(%o30)                           ------------\n",
       "                                  sqrt(%pi)"
      ],
      "text/x-maxima": [
       "gamma(n+1/2)/sqrt(%pi)"
      ]
     },
     "execution_count": 29,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "makegamma(%);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "ポッホハマー記号について次の等式が成り立ちます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{31}$}\\left(\\left(1-x\\right)\\right)_{n}=\\frac{\\left(-1\\right)^{n}\\,\\Gamma\\left(x\\right)}{\\Gamma\\left(x-n\\right)}\\]"
      ],
      "text/plain": [
       "                                          n\n",
       "                                     (- 1)  gamma(x)\n",
       "(%o31)                    (1 - x)  = ---------------\n",
       "                                 n    gamma(x - n)"
      ],
      "text/x-maxima": [
       "pochhammer(1-x,n) = ((-1)^n*gamma(x))/gamma(x-n)"
      ]
     },
     "execution_count": 30,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "PoId:pochhammer(1-x,n)=gamma(x)*(-1)^n/gamma(x-n);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "この等式で$x=\\frac{1}{2}$を代入すると次の式を得ます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{32}$}\\left(\\left(\\frac{1}{2}\\right)\\right)_{n}=\\frac{\\sqrt{\\pi}\\,\\left(-1\\right)^{n}}{\\Gamma\\left(\\frac{1}{2}-n\\right)}\\]"
      ],
      "text/plain": [
       "                                                  n\n",
       "                             1     sqrt(%pi) (- 1)\n",
       "(%o32)                      (-)  = ----------------\n",
       "                             2 n           1\n",
       "                                     gamma(- - n)\n",
       "                                           2"
      ],
      "text/x-maxima": [
       "pochhammer(1/2,n) = (sqrt(%pi)*(-1)^n)/gamma(1/2-n)"
      ]
     },
     "execution_count": 31,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "PoId,x:1/2;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$\\Gamma\\left(\\frac{1}{2}-n\\right)$を作り出すことが出来ました。この式をこのガンマ関数について解いてみます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{33}$}\\left[ \\Gamma\\left(\\frac{1}{2}-n\\right)=\\frac{\\sqrt{\\pi}\\,\\left(-1\\right)^{n}}{\\left(\\left(\\frac{1}{2}\\right)\\right)_{n}} \\right] \\]"
      ],
      "text/plain": [
       "                                                      n\n",
       "                              1        sqrt(%pi) (- 1)\n",
       "(%o33)                 [gamma(- - n) = ----------------]\n",
       "                              2               1\n",
       "                                             (-)\n",
       "                                              2 n"
      ],
      "text/x-maxima": [
       "[gamma(1/2-n) = (sqrt(%pi)*(-1)^n)/pochhammer(1/2,n)]"
      ]
     },
     "execution_count": 32,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "PoId2:solve(%,gamma(1/2-n));"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "この式をF13に代入します。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 33,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{34}$}\\pi\\,\\sum_{n=0}^{\\infty }{\\frac{\\left(\\left(\\frac{1}{2}\\right)\\right)_{n}^2\\,x^{2\\,n}}{\\Gamma\\left(n+1\\right)^2}}\\]"
      ],
      "text/plain": [
       "                                           2\n",
       "                                        1     2 n\n",
       "                                inf    (-)   x\n",
       "                                ====    2 n\n",
       "                                \\\n",
       "(%o34)                      %pi  >    -------------\n",
       "                                /          2\n",
       "                                ====  gamma (n + 1)\n",
       "                                n = 0"
      ],
      "text/x-maxima": [
       "%pi*'sum((pochhammer(1/2,n)^2*x^(2*n))/gamma(n+1)^2,n,0,inf)"
      ]
     },
     "execution_count": 33,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "F13:F12,PoId2;"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "ガンマ関数を階乗に直してみます。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 34,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/latex": [
       "\\[\\tag{${\\it \\%o}_{35}$}\\pi\\,\\sum_{n=0}^{\\infty }{\\frac{\\left(\\left(\\frac{1}{2}\\right)\\right)_{n}^2\\,x^{2\\,n}}{n!^2}}\\]"
      ],
      "text/plain": [
       "                                           2\n",
       "                                        1     2 n\n",
       "                                 inf   (-)   x\n",
       "                                 ====   2 n\n",
       "                                 \\\n",
       "(%o35)                       %pi  >    ----------\n",
       "                                 /          2\n",
       "                                 ====     n!\n",
       "                                 n = 0"
      ],
      "text/x-maxima": [
       "%pi*'sum((pochhammer(1/2,n)^2*x^(2*n))/n!^2,n,0,inf)"
      ]
     },
     "execution_count": 34,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "makefact(F13);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "F7の積分を変形して上記の式が得られました。この式を超幾何関数の定義と見比べれば\n",
    "$$\\pi\\,F\\left(\\frac{1}{2},\\frac{1}{2};1,x^2\\right)$$\n",
    "と等しいことが分かります。\n",
    " \n",
    "忘れずに$(x+1)$を掛けて2で割り\n",
    "$$(x+1)\\,\\frac{\\pi}{2}\\,F\\left(\\frac{1}{2},\\frac{1}{2};1,x^2\\right)$$\n",
    "を得ます。\n",
    "\n",
    "既に証明した\n",
    "$$\\int_{0}^{\\frac{\\pi}{2}}{\\frac{1}{\\sqrt{1-k^2\\,\\sin ^2\\vartheta}}\\;d\\vartheta}=\\frac{\\pi\\,F\\left( \\left. \\begin{array}{c}\\frac{1}{2},\\;\\frac{1}{2}\\\\1\\end{array} \\right |,k^2\\right)}{2}$$\n",
    "と合わせると、\n",
    "$$(x+1)\\,K(x)$$\n",
    "となります。というわけで、\n",
    "$$K(\\frac{2\\,\\sqrt{x}}{x+1})=(x+1)\\,K(x)$$\n",
    "が証明できました。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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