{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Borwein積分\n", "\n", "黒木玄\n", "\n", "2019-06-13\n", "\n", "* Copyright 2019 Gen Kuroki\n", "* License: MIT https://opensource.org/licenses/MIT\n", "* Repository: https://github.com/genkuroki/Calculus\n", "\n", "このファイルは次の場所できれいに閲覧できる:\n", "\n", "* http://nbviewer.jupyter.org/github/genkuroki/Calculus/blob/master/A03%20Borwein%20integral.ipynb\n", "\n", "* https://genkuroki.github.io/documents/Calculus/A03%20Borwein%20integral.pdf\n", "\n", "このファイルは [Free Wolfram Engine](https://www.wolfram.com/engine/) を [Jupyter](https://jupyter.org/) で[使えるようにする](https://github.com/WolframResearch/WolframLanguageForJupyter)と利用できる. 詳しくは次の解説を参照せよ.\n", "\n", "* [Free Wolfram EngineをJupyterで使う方法](https://nbviewer.jupyter.org/github/genkuroki/msfd28/blob/master/Free%20Wolfram%20Engine.ipynb)\n", "\n", "$\n", "\\newcommand\\eps{\\varepsilon}\n", "\\newcommand\\ds{\\displaystyle}\n", "\\newcommand\\Z{{\\mathbb Z}}\n", "\\newcommand\\R{{\\mathbb R}}\n", "\\newcommand\\C{{\\mathbb C}}\n", "\\newcommand\\T{{\\mathbb T}}\n", "\\newcommand\\QED{\\text{□}}\n", "\\newcommand\\root{\\sqrt}\n", "\\newcommand\\bra{\\langle}\n", "\\newcommand\\ket{\\rangle}\n", "\\newcommand\\d{\\partial}\n", "\\newcommand\\sech{\\operatorname{sech}}\n", "\\newcommand\\cosec{\\operatorname{cosec}}\n", "\\newcommand\\sign{\\operatorname{sign}}\n", "\\newcommand\\sinc{\\operatorname{sinc}}\n", "\\newcommand\\real{\\operatorname{Re}}\n", "\\newcommand\\imag{\\operatorname{Im}}\n", "\\newcommand\\Li{\\operatorname{Li}}\n", "\\newcommand\\PROD{\\mathop{\\coprod\\kern-1.35em\\prod}}\n", "\\newcommand\\Si{\\operatorname{Si}}\n", "\\newcommand\\Ci{\\operatorname{Ci}}\n", "\\newcommand\\si{\\operatorname{si}}\n", "\\newcommand\\Cin{\\operatorname{Cin}}\n", "\\newcommand\\Fourier{\\operatorname{\\mathscr{F}}}\n", "$" ] }, { "cell_type": "markdown", "metadata": { "toc": true }, "source": [ "

目次

\n", "
" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "JupyterImageResolution = 84;\n", "JupyterOutTextForm = \"TeX\";\n", "\n", "TeX[x_] := ToString[TeXForm[x]]\n", "TeX[x_, y__] := StringJoin[TeX[x], TeX[y]]\n", "TeXRaw[x__, y_] := StringJoin[x, TeX[y]]\n", "\n", "MappedBy[x_] := x\n", "MappedBy[x_, F___, G_] := MappedBy[x, F] // G\n", "\n", "SetAttributes[TeXEq, HoldFirst]\n", "TeXEq[x_] := TeX[HoldForm[x] == MappedBy[x, ReleaseHold, FullSimplify]]\n", "TeXEq[x_, F__] := TeX[HoldForm[x] == MappedBy[x, ReleaseHold, F]]\n", "TeXEqRaw[x_] := TeX[HoldForm[x] == MappedBy[x, ReleaseHold]]" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Borwein積分の紹介\n", "\n", "以下では $a_0,a_1,\\ldots,a_r$ は正の実数であるとし, $\\sinc$ 函数を\n", "\n", "$$\n", "\\sinc x = \\begin{cases}\n", "\\dfrac{\\sin x}{x} & (x\\ne 0) \\\\\n", "\\;\\;\\,1 & (x=0) \\\\\n", "\\end{cases}\n", "$$\n", "\n", "と定めて利用する.\n", "\n", "次の形の積分を[**Borwein積分**](https://en.wikipedia.org/wiki/Borwein_integral)と呼ぶ:\n", "\n", "$$\n", "\\int_{-\\infty}^\\infty\\prod_{k=0}^r\\sinc(a_k x)\\,dx = \n", "\\int_{-\\infty}^\\infty \\sinc(a_0 x)\\sinc(a_1 x)\\cdots\\sinc(a_r x)\\,dx\n", "$$\n", "\n", "必要ならば条件収束する広義積分でこれを定義しておく. 例えば," ] }, { "cell_type": "code", "execution_count": 15, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \, dx=\pi$$
" ] }, "execution_count": 15, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "上の結果はDirichlet積分の公式\n", "\n", "$$\n", "\\int_{-\\infty}^\\infty \\frac{\\sin(a x)}{x}\\,dx = \\pi\\sign(a)\n", "$$\n", "\n", "の特別な場合として有名である." ] }, { "cell_type": "code", "execution_count": 16, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \text{sinc}\left(\frac{x}{3}\right) \, dx=\pi$$
" ] }, "execution_count": 16, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x]Sinc[x/3], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 17, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \text{sinc}\left(\frac{x}{3}\right) \text{sinc}\left(\frac{x}{5}\right) \, dx=\pi$$
" ] }, "execution_count": 17, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x]Sinc[x/3]Sinc[x/5], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 18, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \text{sinc}\left(\frac{x}{3}\right) \text{sinc}\left(\frac{x}{5}\right) \text{sinc}\left(\frac{x}{7}\right) \, dx=\pi$$
" ] }, "execution_count": 18, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x]Sinc[x/3]Sinc[x/5]Sinc[x/7], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 19, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \text{sinc}\left(\frac{x}{3}\right) \text{sinc}\left(\frac{x}{5}\right) \text{sinc}\left(\frac{x}{7}\right) \text{sinc}\left(\frac{x}{9}\right) \, dx=\pi$$
" ] }, "execution_count": 19, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x]Sinc[x/3]Sinc[x/5]Sinc[x/7]Sinc[x/9], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 20, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \text{sinc}\left(\frac{x}{3}\right) \text{sinc}\left(\frac{x}{5}\right) \text{sinc}\left(\frac{x}{7}\right) \text{sinc}\left(\frac{x}{9}\right) \text{sinc}\left(\frac{x}{11}\right) \, dx=\pi$$
" ] }, "execution_count": 20, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x]Sinc[x/3]Sinc[x/5]Sinc[x/7]Sinc[x/9]Sinc[x/11], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 21, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \text{sinc}\left(\frac{x}{3}\right) \text{sinc}\left(\frac{x}{5}\right) \text{sinc}\left(\frac{x}{7}\right) \text{sinc}\left(\frac{x}{9}\right) \text{sinc}\left(\frac{x}{11}\right) \text{sinc}\left(\frac{x}{13}\right) \, dx=\pi$$
" ] }, "execution_count": 21, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x]Sinc[x/3]Sinc[x/5]Sinc[x/7]Sinc[x/9]Sinc[x/11]Sinc[x/13], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "以上のBorwien積分の値はすべて $\\pi$ になった. しかし, これの次の積分は以下のように $\\pi$ とは異なる $\\pi$ に非常に近い値になる!" ] }, { "cell_type": "code", "execution_count": 22, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \text{sinc}\left(\frac{x}{3}\right) \text{sinc}\left(\frac{x}{5}\right) \text{sinc}\left(\frac{x}{7}\right) \text{sinc}\left(\frac{x}{9}\right) \text{sinc}\left(\frac{x}{11}\right) \text{sinc}\left(\frac{x}{13}\right) \text{sinc}\left(\frac{x}{15}\right) \, dx=\frac{467807924713440738696537864469 \pi }{467807924720320453655260875000}$$
" ] }, "execution_count": 22, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x]Sinc[x/3]Sinc[x/5]Sinc[x/7]Sinc[x/9]Sinc[x/11]Sinc[x/13]Sinc[x/15], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 23, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\frac{467807924713440738696537864469}{467807924720320453655260875000}=0.9999999999852937186$$
" ] }, "execution_count": 23, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "467807924713440738696537864469/467807924720320453655260875000 // TeXEq[#, N[#, 20]&]&" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "このように\n", "\n", "$$\n", "\\frac{467807924713440738696537864469}{467807924720320453655260875000}\\pi\n", "$$\n", "\n", "の値は非常に $\\pi$ に近い. 「これは使用しているWolfram言語のバグなのではないだろうか?」と疑う人がいても不思議ではない結果である." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Borwein積分の公式\n", "\n", "以下で紹介する結果については\n", "\n", "* David Borwein and Jonathan M. Borwein. Some Remarkable Properties of Sinc and Related Integrals. The Ramanujan Journal, March 2001, Volume 5, Issue 1, pp 73–89. [PDF](http://www.thebigquestions.com/borweinintegrals.pdf)\n", "\n", "を参照した.\n", "\n", "$a_0,a_1,\\ldots,a_r>0$ と仮定する. この節では次の公式を示そう:\n", "\n", "$$\n", "\\begin{aligned}\n", "\\int_{-\\infty}^\\infty \\prod_{k=0}^r \\sinc(a_k x)\\,dx &=\n", "\\frac{\\pi}{r! 2^r a_0 a_1\\cdots a_r}\n", "\\\\ &\\,\\times \n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)^r\n", "\\sign\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right).\n", "\\end{aligned}\n", "\\tag{$*$}\n", "$$\n", "\n", "さらに次の公式も示す:\n", "\n", "$$\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)^k =\n", "\\begin{cases}\n", "0 & (k=0,1,\\ldots,r-1) \\\\\n", "r! 2^r a_1\\cdots a_r & (k=r) \n", "\\\\\n", "(r+1)! 2^r a_0 a_1\\cdots a_r & (k=r+1). \\\\\n", "\\end{cases}\n", "\\tag{$**$}\n", "$$\n", "\n", "特に $a_0 > a_1+\\cdots+a_r$ ならば($*$)の $\\sign$ の因子がすべて $1$ になり, ($**$)の $k=r$ の場合を使うと,\n", "\n", "$$\n", "\\int_{-\\infty}^\\infty \\prod_{k=0}^r \\sinc(a_k x)\\,dx = \\frac{\\pi}{a_0}\n", "\\quad (a_0 > a_1+\\cdots+a_r)\n", "$$\n", "\n", "が得られる. 前節の計算例の最後以外はこの公式の $a_0=1$ の場合になっている.\n", "\n", "さらに, $a_0 < a_1+\\cdots+a_r$ でかつ $\\eps_1=\\cdots=\\eps_r=-1$ 以外のとき $a_0+\\sum_{k=1}^r\\eps_k a_k>0$ となるならば, \n", "\n", "$$\n", "\\int_{-\\infty}^\\infty \\prod_{k=0}^r \\sinc(a_k x)\\,dx = \n", "\\frac{\\pi}{a_0}\\left(\n", "1 - \\frac{\\left(a_1+\\cdots+a_r-a_0\\right)^r}{r! 2^{r-1} a_1\\cdots a_r}\n", "\\right)\n", "$$\n", "\n", "が得られる. 前節の計算例の最後はこの公式の $a_0=1$ の場合になっている." ] }, { "cell_type": "code", "execution_count": 24, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}=\frac{43024}{45045}$$
" ] }, "execution_count": 24, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "1/3+1/5+1/7+1/9+1/11+1/13 // TeXEq" ] }, { "cell_type": "code", "execution_count": 25, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}=\frac{46027}{45045}$$
" ] }, "execution_count": 25, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "1/3+1/5+1/7+1/9+1/11+1/13+1/15 // TeXEq" ] }, { "cell_type": "code", "execution_count": 26, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \text{sinc}\left(\frac{x}{3}\right) \text{sinc}\left(\frac{x}{5}\right) \text{sinc}\left(\frac{x}{7}\right) \text{sinc}\left(\frac{x}{9}\right) \text{sinc}\left(\frac{x}{11}\right) \text{sinc}\left(\frac{x}{13}\right) \text{sinc}\left(\frac{x}{15}\right) \, dx=\frac{467807924713440738696537864469 \pi }{467807924720320453655260875000}$$
" ] }, "execution_count": 26, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x]Sinc[x/3]Sinc[x/5]Sinc[x/7]Sinc[x/9]Sinc[x/11]Sinc[x/13]Sinc[x/15], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 27, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\pi \left(1-\frac{\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}-1\right)^7}{\frac{7! 2^6}{3\ 5\ 7\ 9\ 11\ 13\ 15}}\right)=\frac{467807924713440738696537864469 \pi }{467807924720320453655260875000}$$
" ] }, "execution_count": 27, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Pi(1 - (1/3+1/5+1/7+1/9+1/11+1/13+1/15-1)^7/(7! 2^6 (1/3)(1/5)(1/7)(1/9)(1/11)(1/13)(1/15))) // TeXEq" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Borwein積分の公式の証明\n", "\n", "三角函数に関する簡単な計算によって以下が成立していることを示せる.\n", "\n", "$r$ が偶数のとき,\n", "\n", "$$\n", "\\prod_{k=0}^r \\sin(a_k x) =\n", "\\frac{(-1)^{r/2}}{2^r}\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r \\sin\\left(\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)x\\right).\n", "$$\n", "\n", "$r$ が奇数のとき, \n", "\n", "$$\n", "\\prod_{k=0}^r \\sin(a_k x) =\n", "\\frac{(-1)^{(r+1)/2}}{2^r}\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r \\cos\\left(\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)x\\right).\n", "$$\n", "\n", "これらの公式の両辺を $r$ 回 $x$ で微分すると, $r$ の偶奇によらず, 結果は次のようになることがすぐにわかる:\n", "\n", "$$\n", "\\left(\\frac{\\d}{\\d x}\\right)^r\n", "\\prod_{k=0}^r \\sin(a_k x) =\n", "\\frac{1}{2^r}\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)^r\n", "\\sin\\left(\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)x\\right).\n", "$$\n", "\n", "この公式を使うと, Borwein積分の公式($*$)は部分積分によって容易に証明される.\n", "\n", "$$\n", "\\frac{1}{r!}\\left(-\\frac{\\d}{\\d x}\\right)^r\\frac{1}{x} = \\frac{1}{x^{r+1}}\n", "$$\n", "\n", "を使って, 部分積分を $r$ 回繰り返すと, \n", "\n", "$$\n", "\\begin{aligned}\n", "\\int_{-\\infty}^\\infty \\prod_{k=0}^r \\sinc(a_k x)\\,dx &=\n", "\\frac{1}{a_0a_1\\cdots a_r}\n", "\\int_{-\\infty}^\\infty \\frac{1}{x^{r+1}} \\prod_{k=0}^r \\sin(a_k x)\\,dx\n", "\\\\ &=\n", "\\frac{1}{r! a_0a_1\\cdots a_r}\n", "\\int_{-\\infty}^\\infty \\frac{1}{x}\\;\\left(\\frac{\\d}{\\d x}\\right)^r\\prod_{k=0}^r \\sin(a_k x)\\,dx\n", "\\\\ &=\n", "\\frac{1}{r! 2^r a_0a_1\\cdots a_r}\n", "\\\\ &\\,\\times\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)^r\n", "\\int_{-\\infty}^\\infty\\frac{1}{x}\n", "\\sin\\left(\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)x\\right)\n", "\\,dx\n", "\\\\ &=\n", "\\frac{1}{r! 2^r a_0a_1\\cdots a_r}\n", "\\\\ &\\,\\times\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)^r\n", "\\sign\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right).\n", "\\end{aligned}\n", "$$\n", "\n", "これが示したかったBorwein積分の公式($*$)である:\n", "\n", "$$\n", "\\begin{aligned}\n", "\\int_{-\\infty}^\\infty \\prod_{k=0}^r \\sinc(a_k x)\\,dx &=\n", "\\frac{\\pi}{r! 2^r a_0 a_1\\cdots a_r}\n", "\\\\ &\\,\\times \n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)^r\n", "\\sign\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right).\n", "\\end{aligned}\n", "\\tag{$*$}\n", "$$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### 公式($\\boldsymbol{**}$)の証明\n", "\n", "$$\n", "\\begin{aligned}\n", "e^{a_0 t}\\prod_{k=1}^r(e^{a_k t}-e^{-a_k t}) &=\n", "e^{a_0 t}\\prod_{k=1}^r\\sum_{\\eps_k=\\pm1}\\eps_k e^{\\eps_k a_k t}\n", "\\\\ &=\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\\eps_1\\cdots\\eps_r \n", "\\exp\\left(\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)t\\right).\n", "\\end{aligned}\n", "$$\n", "\n", "左辺の $t$ に関するべき級数の展開は\n", "\n", "$$\n", "(1+a_0 t + O(t^2))(2^r a_1\\cdots a_r t^r + O(t^{r+2})) =\n", "2^r a_1\\cdots a_r t^r + 2^r a_0 a_1\\cdots a_r t^{r+1} + O(t^{r+2})\n", "$$\n", "\n", "となり, 右辺は\n", "\n", "$$\n", "\\sum_{k=0}^\\infty \\frac{t^k}{k!}\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\\eps_1\\cdots\\eps_r\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)^k\n", "$$\n", "\n", "となる. $k=0,1,\\ldots,r,r-1$ に関する $t^k$ の係数を比較すると公式 ($**$) が得られる:\n", "\n", "$$\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r\n", "\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)^k =\n", "\\begin{cases}\n", "0 & (k=0,1,\\ldots,r-1) \\\\\n", "r! 2^r a_1\\cdots a_r & (k=r) \n", "\\\\\n", "(r+1)! 2^r a_0 a_1\\cdots a_r & (k=r+1). \\\\\n", "\\end{cases}\n", "\\tag{$**$}\n", "$$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### sinの積をsinまたはcosの和で表す公式の証明\n", "\n", "sinの積に関する公式も証明しておこう.\n", "\n", "$$\n", "\\sin z = \\dfrac{e^{iz}-e^{-iz}}{2i} =\n", "\\frac{1}{2i}\\sum_{\\eps=\\pm1}\\eps e^{i\\eps z}\n", "$$\n", "\n", "を使うと, \n", "\n", "$$\n", "\\begin{aligned}\n", "\\prod_{k=0}^r\\sin(a_k x) &=\n", "\\frac{1}{(2i)^{r+1}}(e^{ia_0x}-e^{-ia_0x})\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\\eps_1\\cdots\\eps_r e^{i\\left(\\sum_{k=1}^r \\eps_k a_k\\right) x}\n", "\\\\ &=\n", "\\frac{1}{(2i)^{r+1}}\\left(\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\\eps_1\\cdots\\eps_r e^{i\\left(a_0+\\sum_{k=1}^r \\eps_k a_k\\right) x} -\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\\eps_1\\cdots\\eps_r e^{i\\left(-a_0+\\sum_{k=1}^r \\eps_k a_k\\right) x}\n", "\\right)\n", "\\\\ &=\n", "\\frac{1}{(2i)^{r+1}}\\left(\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\\eps_1\\cdots\\eps_r e^{i\\left(a_0+\\sum_{k=1}^r \\eps_k a_k\\right) x} -\n", "(-1)^r\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\\eps_1\\cdots\\eps_r e^{-i\\left(a_0+\\sum_{k=1}^r \\eps_k a_k\\right) x}\n", "\\right)\n", "\\\\ &=\n", "\\frac{1}{(2i)^{r+1}}\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\\eps_1\\cdots\\eps_r\n", "\\left(\n", "e^{i\\left(a_0+\\sum_{k=1}^r \\eps_k a_k\\right) x} -\n", "(-1)^r\n", "e^{-i\\left(a_0+\\sum_{k=1}^r \\eps_k a_k\\right) x}\n", "\\right).\n", "\\end{aligned}\n", "$$\n", "\n", "3番目の等号で括弧の中の2つ目の和の $\\eps_k$ 達をすべて $-1$ 倍にした. この結果は, $r$ が偶数のとき,\n", "\n", "$$\n", "\\prod_{k=0}^r \\sin(a_k x) =\n", "\\frac{(-1)^{r/2}}{2^r}\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r \\sin\\left(\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)x\\right)\n", "$$\n", "\n", "になり, $r$ が奇数のとき, \n", "\n", "$$\n", "\\prod_{k=0}^r \\sin(a_k x) =\n", "\\frac{(-1)^{(r+1)/2}}{2^r}\n", "\\sum_{\\eps_1,\\ldots,\\eps_r=\\pm1}\n", "\\eps_1\\cdots\\eps_r \\cos\\left(\\left(a_0+\\sum_{k=1}^r\\eps_k a_k\\right)x\\right)\n", "$$\n", "\n", "になる. これが示したい公式であった. \n", "\n", "このように奇数個の $\\sin$ の積は $\\sin$ の和に展開され, 偶数個の $\\sin$ の積は $\\cos$ の和に展開される." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Borwein積分とFourier変換の関係" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Fourier解析からの準備\n", "\n", "Fourier変換 $f(x)\\mapsto\\Fourier[f(x)](p)$ とその逆変換 $\\phi(p)\\mapsto\\Fourier^{-1}[\\phi(p)](x)$ が次のように定義される:\n", "\n", "$$\n", "\\Fourier[f(x)](p) = \\int_{-\\infty}^\\infty e^{-ipx}f(x)\\,dx, \\quad\n", "\\Fourier^{-1}[\\phi(p)](x) = \\int_{-\\infty}^\\infty e^{ipx}\\phi(p)\\,\\frac{dp}{2\\pi}.\n", "$$\n", "\n", "さらに, 函数 $\\phi(p),\\psi(p)$ のたたみ込み積が\n", "\n", "$$\n", "\\phi*\\psi(p)\\, = \\int_{-\\infty}^\\infty \\phi(q)\\psi(p-q)\\,\\frac{dq}{2\\pi}\n", "$$\n", "\n", "と定義される. このとき, \n", "\n", "$$\n", "\\int_{-\\infty}^\\infty \\phi*\\psi(p)\\,\\frac{dp}{2\\pi} =\n", "\\left(\\int_{-\\infty}^\\infty \\phi(p)\\,\\frac{dp}{2\\pi}\\right)\n", "\\left(\\int_{-\\infty}^\\infty \\psi(p)\\,\\frac{dp}{2\\pi}\\right).\n", "$$\n", "\n", "さらに, 函数 $f(x),g(x)$ に関する適切な仮定のもとで次が成立することを示せる:\n", "\n", "$$\n", "\\Fourier[f(x)g(x)] = \\Fourier[f(x)]*\\Fourier[g(x)].\n", "$$\n", "\n", "**例:** $a>0$ に対して, $\\chi_a(p)$ を\n", "\n", "$$\n", "\\chi_a(p) = \n", "\\begin{cases}\n", "\\;1 & (-a$$\left(\frac{1}{2}+\frac{1}{3}>1\right)=\text{False}$$" ] }, "execution_count": 28, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "1/2+1/3 > 1 // TeXEq" ] }, { "cell_type": "code", "execution_count": 29, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \text{sinc}\left(\frac{x}{2}\right) \text{sinc}\left(\frac{x}{3}\right) \, dx=\pi$$
" ] }, "execution_count": 29, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x]Sinc[x/2]Sinc[x/3], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 30, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>1\right)=\text{True}$$
" ] }, "execution_count": 30, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "1/2+1/3+1/4 > 1 // TeXEq" ] }, { "cell_type": "code", "execution_count": 31, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \text{sinc}(x) \text{sinc}\left(\frac{x}{2}\right) \text{sinc}\left(\frac{x}{3}\right) \text{sinc}\left(\frac{x}{4}\right) \, dx=\frac{1727 \pi }{1728}$$
" ] }, "execution_count": 31, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[Sinc[x]Sinc[x/2]Sinc[x/3]Sinc[x/4], {x,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 32, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$g(a,p)=\frac{\pi \text{If}[-a\leq p\leq a,1,0]}{a}$$
" ] }, "execution_count": 33, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "g[a_, p_] := Pi/a If[-a <= p <= a, 1, 0]\n", "g[a,p] // TeXEq" ] }, { "cell_type": "code", "execution_count": 34, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \frac{g(a,p)}{2 \pi } \, dp=1$$
" ] }, "execution_count": 34, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Assuming[a > 0, Integrate[g[a,p]/(2Pi), {p,-Infinity,Infinity}] // TeXEq]" ] }, { "cell_type": "code", "execution_count": 35, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{G1}(p)=2 \pi \text{If}\left[-\frac{1}{2}\leq p\leq \frac{1}{2},1,0\right]$$
" ] }, "execution_count": 36, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "G1[p_] = g[1/2, p];\n", "G1[p] // TeX[HoldForm[G1[p]], \"=\", #]&" ] }, { "cell_type": "code", "execution_count": 37, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\"Output\"
" ] }, "execution_count": 37, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Plot[G1[p]/(2Pi), {p,-1.5,1.5}]" ] }, { "cell_type": "code", "execution_count": 38, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \frac{\text{G1}(p)}{2 \pi } \, dp=1$$
" ] }, "execution_count": 38, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[G1[p]/(2Pi), {p,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 39, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{G2}(p)=\begin{cases} 2 \pi & -\frac{1}{6}<p\leq \frac{1}{6} \\ -\frac{1}{2} \pi (6 p-5) & \frac{1}{6}<p<\frac{5}{6} \\ \frac{1}{2} \pi (6 p+5) & -\frac{5}{6}<p\leq -\frac{1}{6} \end{cases}$$
" ] }, "execution_count": 40, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "G2[p_] = Integrate[g[1/2,q]g[1/3,p-q]/(2Pi), {q,-Infinity,Infinity}];\n", "G2[p] // TeX[HoldForm[G2[p]], \"=\", #]&" ] }, { "cell_type": "code", "execution_count": 41, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\"Output\"
" ] }, "execution_count": 41, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Plot[G2[p]/(2Pi), {p,-1.5,1.5}]" ] }, { "cell_type": "code", "execution_count": 42, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \frac{\text{G2}(p)}{2 \pi } \, dp=1$$
" ] }, "execution_count": 42, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[G2[p]/(2Pi), {p,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 43, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{G3}(p)=\begin{cases} \frac{1}{2} (5 \pi -6 \pi p) & \frac{5}{12}<p<\frac{7}{12} \\ \frac{1}{2} (6 \pi p+5 \pi ) & -\frac{7}{12}<p\leq -\frac{5}{12} \\ \frac{1}{24} \left(47 \pi -144 \pi p^2\right) & -\frac{1}{12}<p<\frac{1}{12} \\ \frac{1}{48} \left(-144 \pi p^2-24 \pi p+95 \pi \right) & \frac{1}{12}<p<\frac{5}{12} \\ \frac{1}{48} \left(-144 \pi p^2+24 \pi p+95 \pi \right) & -\frac{5}{12}<p\leq -\frac{1}{12} \\ \frac{1}{48} \left(-144 \pi p^2+168 \pi p+79 \pi \right) & p=\frac{1}{12} \\ \frac{1}{16} \left(-48 \pi p^2+56 \pi p+5 \pi \right) & p=\frac{5}{12} \\ \frac{1}{48} \left(144 \pi p^2-312 \pi p+169 \pi \right) & \frac{7}{12}\leq p<\frac{13}{12} \\ \frac{1}{48} \left(144 \pi p^2+312 \pi p+169 \pi \right) & -\frac{13}{12}<p\leq -\frac{7}{12} \end{cases}$$
" ] }, "execution_count": 44, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "G3[p_] = Integrate[G2[q]g[1/4,p-q]/(2Pi), {q,-Infinity,Infinity}];\n", "G3[p] // TeX[HoldForm[G3[p]], \"=\", #]&" ] }, { "cell_type": "code", "execution_count": 45, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\"Output\"
" ] }, "execution_count": 45, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Plot[G3[p]/(2Pi), {p,-1.5,1.5}]" ] }, { "cell_type": "code", "execution_count": 46, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\int_{-\infty }^{\infty } \frac{\text{G3}(p)}{2 \pi } \, dp=1$$
" ] }, "execution_count": 46, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Integrate[G3[p]/(2Pi), {p,-Infinity,Infinity}] // TeXEq" ] }, { "cell_type": "code", "execution_count": 47, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\pi \int_{-1}^1 \frac{\text{G3}(p)}{2 \pi } \, dp=\frac{1727 \pi }{1728}$$
" ] }, "execution_count": 47, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Pi Integrate[G3[p]/(2Pi), {p,-1,1}] // TeXEq" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "次に $I_k(p) = g_0*g_1*\\cdots*g_k(p)$ を計算して, $I_k(0)$ を計算し, $I_k(p)/\\pi$ プロットしてみよう." ] }, { "cell_type": "code", "execution_count": 48, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{I0}(p)=\pi \text{If}[-1\leq p\leq 1,1,0]$$
" ] }, "execution_count": 49, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "I0[p_] = g[1,p];\n", "I0[p] // TeX[HoldForm[I0[p]], \"=\", #]&" ] }, { "cell_type": "code", "execution_count": 50, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{I0}(0)=\pi$$
" ] }, "execution_count": 50, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "I0[0] // TeXEq" ] }, { "cell_type": "code", "execution_count": 51, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\"Output\"
" ] }, "execution_count": 51, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Plot[I0[p]/Pi, {p,-2,2}]" ] }, { "cell_type": "code", "execution_count": 52, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{I1}(p)=\begin{cases} \pi & -\frac{1}{2}<p\leq \frac{1}{2} \\ -\frac{1}{2} \pi (2 p-3) & \frac{1}{2}<p<\frac{3}{2} \\ \frac{1}{2} \pi (2 p+3) & -\frac{3}{2}<p\leq -\frac{1}{2} \end{cases}$$
" ] }, "execution_count": 53, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "I1[p_] = Integrate[I0[q]g[1/2,p-q]/(2Pi), {q,-Infinity,Infinity}];\n", "I1[p] // TeX[HoldForm[I1[p]], \"=\", #]&" ] }, { "cell_type": "code", "execution_count": 54, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{I1}(0)=\pi$$
" ] }, "execution_count": 54, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "I1[0] // TeXEq" ] }, { "cell_type": "code", "execution_count": 55, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\"Output\"
" ] }, "execution_count": 55, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Plot[I1[p]/Pi, {p,-2,2}]" ] }, { "cell_type": "code", "execution_count": 56, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{I2}(p)=\begin{cases} \pi & -\frac{1}{6}<p\leq \frac{1}{6} \\ \frac{1}{4} \pi (6 p+5) & p=-\frac{1}{6} \\ \frac{1}{2} (3 \pi -2 \pi p) & \frac{5}{6}<p<\frac{7}{6} \\ \frac{1}{2} (2 \pi p+3 \pi ) & -\frac{7}{6}<p\leq -\frac{5}{6} \\ \frac{1}{48} \left(-36 \pi p^2-12 \pi p+47 \pi \right) & -\frac{5}{6}<p<-\frac{1}{6} \\ \frac{1}{48} \left(-36 \pi p^2+12 \pi p+47 \pi \right) & \frac{1}{6}<p<\frac{5}{6} \\ \frac{1}{48} \left(-36 \pi p^2+84 \pi p-13 \pi \right) & p=\frac{5}{6} \\ \frac{1}{48} \left(36 \pi p^2-132 \pi p+121 \pi \right) & \frac{7}{6}\leq p<\frac{11}{6} \\ \frac{1}{48} \left(36 \pi p^2+132 \pi p+121 \pi \right) & -\frac{11}{6}<p\leq -\frac{7}{6} \end{cases}$$
" ] }, "execution_count": 57, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "I2[p_] = Integrate[I1[q]g[1/3,p-q]/(2Pi), {q,-Infinity,Infinity}];\n", "I2[p] // TeX[HoldForm[I2[p]], \"=\", #]&" ] }, { "cell_type": "code", "execution_count": 58, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{I2}(0)=\pi$$
" ] }, "execution_count": 58, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "I2[0] // TeXEq" ] }, { "cell_type": "code", "execution_count": 59, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\"Output\"
" ] }, "execution_count": 59, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Plot[I2[p]/Pi, {p,-2,2}]" ] }, { "cell_type": "code", "execution_count": 60, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{I3}(p)=\begin{cases} \frac{1727 \pi -432 \pi p^2}{1728} & -\frac{1}{12}<p<\frac{1}{12} \\ \frac{1}{192} \left(-144 \pi p^2-48 \pi p+185 \pi \right) & -\frac{7}{12}<p<-\frac{5}{12} \\ \frac{1}{192} \left(-144 \pi p^2+48 \pi p+185 \pi \right) & \frac{5}{12}<p<\frac{7}{12} \\ \frac{1}{192} \left(144 \pi p^2-528 \pi p+487 \pi \right) & \frac{17}{12}<p<\frac{19}{12} \\ \frac{1}{192} \left(144 \pi p^2+528 \pi p+487 \pi \right) & -\frac{19}{12}<p\leq -\frac{17}{12} \\ \frac{-1728 \pi p^3-5616 \pi p^2-2628 \pi p+2987 \pi }{3456} & -\frac{11}{12}<p<-\frac{7}{12} \\ \frac{-1728 \pi p^3-4752 \pi p^2-900 \pi p+3853 \pi }{3456} & -\frac{17}{12}<p\leq -\frac{13}{12} \\ \frac{-1728 \pi p^3-2160 \pi p^2+6012 \pi p+6851 \pi }{3456} & p=-\frac{7}{12} \\ \frac{-1728 \pi p^3-2160 \pi p^2+6012 \pi p+8003 \pi }{3456} & p=-\frac{11}{12} \\ \frac{-1728 \pi p^3-432 \pi p^2-36 \pi p+3455 \pi }{3456} & \frac{1}{12}<p<\frac{5}{12} \\ \frac{-1728 \pi p^3-432 \pi p^2+6876 \pi p+575 \pi }{3456} & p=\frac{5}{12} \\ \frac{-1728 \pi p^3-432 \pi p^2+6876 \pi p+2879 \pi }{3456} & p=\frac{1}{12} \\ \frac{-1728 \pi p^3+10800 \pi p^2-22500 \pi p+15625 \pi }{3456} & \frac{19}{12}\leq p<\frac{25}{12} \\ \frac{1}{48} \left(-48 \pi p^3-144 \pi p^2-97 \pi p+23 \pi \right) & -\frac{13}{12}<p<-\frac{11}{12} \\ \frac{1}{48} \left(48 \pi p^3-144 \pi p^2+97 \pi p+23 \pi \right) & \frac{11}{12}<p<\frac{13}{12} \\ \frac{1728 \pi p^3-8208 \pi p^2+12996 \pi p-6347 \pi }{3456} & p=\frac{17}{12} \\ \frac{1728 \pi p^3-8208 \pi p^2+12996 \pi p-5195 \pi }{3456} & p=\frac{13}{12} \\ \frac{1728 \pi p^3-5616 \pi p^2+2628 \pi p+2987 \pi }{3456} & \frac{7}{12}<p<\frac{11}{12} \\ \frac{1728 \pi p^3-4752 \pi p^2+900 \pi p+3853 \pi }{3456} & \frac{13}{12}<p<\frac{17}{12} \\ \frac{1728 \pi p^3-2160 \pi p^2-6012 \pi p+6851 \pi }{3456} & p=\frac{7}{12} \\ \frac{1728 \pi p^3-2160 \pi p^2-6012 \pi p+8003 \pi }{3456} & p=\frac{11}{12} \\ \frac{1728 \pi p^3-432 \pi p^2-6876 \pi p+575 \pi }{3456} & p=-\frac{5}{12} \\ \frac{1728 \pi p^3-432 \pi p^2+36 \pi p+3455 \pi }{3456} & -\frac{5}{12}<p\leq -\frac{1}{12} \\ \frac{1728 \pi p^3+10800 \pi p^2+22500 \pi p+15625 \pi }{3456} & -\frac{25}{12}<p\leq -\frac{19}{12} \end{cases}$$
" ] }, "execution_count": 61, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "I3[p_] = Integrate[I2[q]g[1/4,p-q]/(2Pi), {q,-Infinity,Infinity}];\n", "I3[p] // TeX[HoldForm[I3[p]], \"=\", #]&" ] }, { "cell_type": "code", "execution_count": 62, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
$$\text{I3}(0)=\frac{1727 \pi }{1728}$$
" ] }, "execution_count": 62, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "I3[0] // TeXEq" ] }, { "cell_type": "code", "execution_count": 63, "metadata": {}, "outputs": [ { "data": { "text/html": [ "
\"Output\"
" ] }, "execution_count": 63, "metadata": { "text/html": [] }, "output_type": "execute_result" } ], "source": [ "Plot[I3[p]/Pi, {p,-2,2}]" ] } ], "metadata": { "@webio": { "lastCommId": null, "lastKernelId": null }, "_draft": { "nbviewer_url": "https://gist.github.com/b28ca309a828672d2b70ba7aaad296ba" }, "gist": { "data": { "description": "A03 Borwein integral.ipynb", "public": true }, "id": "b28ca309a828672d2b70ba7aaad296ba" }, "jupytext": { "formats": "ipynb,md" }, "kernelspec": { "display_name": "Wolfram Language 13.1", "language": "Wolfram Language", "name": "wolframlanguage13.1" }, "language_info": { "codemirror_mode": "mathematica", "file_extension": ".m", "mimetype": "application/vnd.wolfram.m", "name": "Wolfram Language", "pygments_lexer": "python", "version": "12.0" }, "toc": { "base_numbering": 1, "nav_menu": {}, "number_sections": true, "sideBar": true, "skip_h1_title": true, "title_cell": "目次", "title_sidebar": "目次", "toc_cell": true, "toc_position": {}, "toc_section_display": true, "toc_window_display": false } }, "nbformat": 4, "nbformat_minor": 2 }