\u305d\u308c\u3092\u4f7f\u3063\u3066sinc\u95a2\u6570\u306e\u7a4d\u5206\u3092\u8907\u7d20\u7a4d\u5206\u3092\u4f7f\u308f\u306a\u3044\u3067\u8a08\u7b97\u3067\u304d\u306a\u3044\u304b\u3068\u601d\u3044\u3001\u90e8\u5206\u7a4d\u5206\u3092\u4f7f\u3063\u3066\u8a08\u7b97\u3057\u3066\u307f\u305f\u3068\u3053\u308d\u30014\u4e57\u3042\u305f\u308a\u307e\u3067\u306a\u3089\u4f55\u3068\u304b\u8a08\u7b97\u3067\u304d\u305d\u3046\u306a\u306e\u3067\u3001\u30e1\u30e2\u3059\u308b\u3053\u3068\u306b\u3057\u307e\u3057\u305f\u3002<\/p>\n
\u306a\u304a\u3001\u3053\u306e\u8a18\u4e8b\u3067\u306f\u300csinc\u95a2\u6570\u300d\u306f(\\ref{eq:sincdef})\u5f0f\u3067\u8868\u3055\u308c\u308b\u95a2\u6570$f(x)$\u306e\u3053\u3068\u3092\u6307\u3059\u3082\u306e\u3068\u3057\u307e\u3059\u3002
\n\\begin{align}
\nf(x) &= \\frac{\\sin x}{x} \\label{eq:sincdef}
\n\\end{align}<\/p>\n
\u9806\u756a\u306b\u8a08\u7b97\u3057\u3066\u3044\u304d\u307e\u3059\u3002<\/h2>\n
$[0,\\infty )$\u306e\u5e83\u7fa9\u7a4d\u5206\u3092\u8a08\u7b97\u3057\u307e\u3059\u304c\u3001sinc\u95a2\u6570\u306e2,3,4\u4e57\u306f\u3044\u305a\u308c\u3082\u5076\u95a2\u6570\u3068\u306a\u308a\u307e\u3059\u306e\u3067\u3001\u7a4d\u5206\u533a\u9593\u3092$(-\\infty , \\infty)$\u3068\u3057\u305f\u3068\u304d\u306e\u8a08\u7b97\u7d50\u679c\u306f\u7a4d\u5206\u533a\u9593\u3092$[0,\\infty )$\u3068\u3057\u305f\u3068\u304d\u306e2\u500d\u306b\u306a\u308a\u307e\u3059\u3002<\/p>\n
\u307e\u305f\u3001\u3053\u306e\u8a18\u4e8b\u3067\u306f
\n\\begin{align}
\n\\int_{0}^{\\infty} \\frac{\\sin x}{x} dx &= \\frac{\\pi}{2} \\label{eq:dirchletintegral}
\n\\end{align}
\n\u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u306f\u65e2\u77e5\u3068\u3057\u307e\u3059\u3002<\/p>\n
sinc\u95a2\u6570\u306e2\u4e57($+\\alpha$)<\/h3>\n
$\\displaystyle\\int_{0}^{\\infty} \\displaystyle\\frac{\\sin^2 x}{x^2}dx$\u306f\u90e8\u5206\u7a4d\u5206\u3092\u7528\u3044\u3066\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u5909\u5f62\u53ca\u3073\u8a08\u7b97\u3067\u304d\u307e\u3059\u3002
\n\\begin{align}
\n\\int_{0}^{\\infty} \\frac{\\sin^2 x}{x^2}dx &= \\int_{0}^{\\infty}\\left(-\\frac{1}{x}\\right)^{\\prime} \\sin^2 x dx \\nonumber \\cr
\n&= \\left[ -\\frac{\\sin^2 x}{x} \\right]_{0}^{\\infty} + \\int_{0}^{\\infty}\\frac{2\\cos x\\sin x}{x} dx \\nonumber \\cr
\n&= \\int_{0}^{\\infty}\\frac{\\sin 2x}{x} dx \\nonumber \\cr
\n&= \\displaystyle\\int_{0}^{\\infty}\\frac{\\sin t}{\\displaystyle\\frac{t}{2}}\\cdot\\displaystyle\\frac{1}{2} dt \\nonumber \\cr &= \\frac{\\pi}{2} \\label{eq:sincsquare}
\n\\end{align}
\n\u306a\u304a\u3001(\\ref{eq:sincsquare})\u5f0f\u306e\u7d50\u679c\u3092\u5229\u7528\u3059\u308b\u3068\u3001$a > 0$\u306e\u3068\u304d\u3001
\n\\begin{align}
\n\\int_{0}^{\\infty} \\frac{\\sin^2 ax}{x^2}dx &= \\int_{0}^{\\infty} \\frac{\\sin^2 t}{\\displaystyle\\frac{t^2}{a^2}}\\cdot\\frac{dt}{a} \\nonumber \\cr
\n&= a \\int_{0}^{\\infty} \\frac{\\sin^2 t}{t}dt \\nonumber \\cr
\n&= \\frac{a\\pi}{2} \\label{eq:sincsquareatimes}
\n\\end{align}
\n\u3067\u3042\u308b\u3053\u3068\u3082\u308f\u304b\u308a\u307e\u3059\u3002<\/p>\n
sinc\u95a2\u6570\u306e3\u4e57<\/h3>\n
2\u4e57\u306e\u5834\u5408\u3068\u540c\u69d8\u306b\u3001$\\displaystyle\\int_{0}^{\\infty} \\displaystyle\\frac{\\sin^3 x}{x^3}dx$\u306f\u90e8\u5206\u7a4d\u5206\u3092\u7528\u3044\u3066\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u5909\u5f62\u53ca\u3073\u8a08\u7b97\u3067\u304d\u307e\u3059\u3002
\n\\begin{align}
\n\\int_{0}^{\\infty} \\frac{\\sin^3 x}{x^3}dx &= \\int_{0}^{\\infty}\\left(-\\frac{1}{2x^2}\\right)^{\\prime} \\sin^3 x dx \\nonumber \\cr
\n&= \\left[ -\\frac{\\sin^3 x}{2x^2} \\right]_{0}^{\\infty} + \\int_{0}^{\\infty}\\frac{3\\cos x\\sin^2 x}{2x^2} dx \\nonumber \\cr
\n&= \\int_{0}^{\\infty}\\frac{3\\sin 2x\\sin x}{4x^2} dx \\nonumber \\cr
\n&= \\int_{0}^{\\infty}\\left(-\\frac{1}{x}\\right)^{\\prime}\\frac{3\\sin 2x\\sin x}{4} dx \\nonumber \\cr
\n&= \\left[ -\\frac{3\\sin 2x\\sin x}{4} \\right]_{0}^{\\infty} + \\int_{0}^{\\infty}\\frac{3}{4x}(2\\cos 2x\\sin x + \\sin 2x\\cos x) dx \\nonumber \\cr
\n&= \\int_{0}^{\\infty}\\frac{3}{4x}(\\cos 2x\\sin x + \\sin 3x) dx \\nonumber \\cr
\n&= \\int_{0}^{\\infty}\\frac{3}{4x}\\left[\\frac{1}{2}(\\sin 3x-\\sin x) + \\sin 3x\\right] dx \\nonumber \\cr
\n&= \\int_{0}^{\\infty}\\frac{3}{4x}\\left(\\frac{3}{2}\\sin 3x-\\frac{1}{2}\\sin x\\right) dx \\nonumber \\cr
\n&= \\frac{9}{16}\\pi – \\frac{3}{16}\\pi \\nonumber \\cr
\n&= \\frac{3}{8}\\pi \\label{eq:sinccube}
\n\\end{align}<\/p>\n
sinc\u95a2\u6570\u306e4\u4e57<\/h3>\n
[2020\/06\/12\u88dc\u8db3] \u9014\u4e2d\u306e\u5f0f\u5909\u5f62\u306b\u8aa4\u308a\u304c\u3042\u3063\u305f\u306e\u3067\u3001\u4fee\u6b63\u3057\u307e\u3057\u305f\u3002
\n2\u4e57\u53ca\u30733\u4e57\u306e\u5834\u5408\u3068\u540c\u69d8\u306b\u3001$\\displaystyle\\int_{0}^{\\infty} \\displaystyle\\frac{\\sin^4 x}{x^4}dx$\u306f\u90e8\u5206\u7a4d\u5206\u3092\u7528\u3044\u3066(\\ref{eq:sincfourthpowerfirst})\u5f0f\u306e\u3088\u3046\u306b\u3072\u3068\u307e\u305a\u5909\u5f62\u3057\u307e\u3059\u3002
\n\\begin{align}
\n\\int_{0}^{\\infty} \\frac{\\sin^4 x}{x^4}dx &= \\int_{0}^{\\infty}\\left(-\\frac{1}{3x^3}\\right)^{\\prime} \\sin^4 x dx \\nonumber \\cr
\n&= \\left[ -\\frac{\\sin^4 x}{3x^3} \\right]_{0}^{\\infty} + \\int_{0}^{\\infty}\\frac{4\\cos x\\sin^3 x}{3x^3} dx \\nonumber \\cr
\n&= \\int_{0}^{\\infty}\\frac{2\\sin 2x\\sin^2 x}{3x^3} dx \\nonumber \\cr
\n&= \\int_{0}^{\\infty}\\left(-\\frac{1}{3x^2}\\right)^{\\prime}\\sin 2x\\sin^2 x dx \\nonumber \\cr
\n&= \\left[ -\\frac{1}{3x^2}\\sin 2x\\sin^2 x \\right]_{0}^{\\infty} + \\int_{0}^{\\infty}\\frac{1}{3x^2}(2\\cos 2x\\sin^2 x + 2\\sin 2x\\cos x\\sin x) dx \\nonumber \\cr
\n&= \\int_{0}^{\\infty}\\frac{1}{3x^2}\\left[\\sin^2 2x + \\cos 2x(1- \\cos 2x)\\right] dx
\n \\label{eq:sincfourthpowerfirst}
\n\\end{align}
\n\u3053\u3053\u3067\u3001(\\ref{eq:sincfourthpowerfirst})\u5f0f\u306e\u53f3\u8fba\u304b\u3089$\\cos 2x$\u3092\u6d88\u53bb\u3067\u304d\u306a\u3044\u304b\u8003\u3048\u307e\u3059\u3002<\/p>\n
$\\cos^2 2x = 1 – \\sin^2 2x$\u3067\u3042\u308b\u3053\u3068\u3092\u5229\u7528\u3059\u308b\u3068\u2026
\n\\begin{align}
\n\\int_{0}^{\\infty} \\frac{\\sin^4 x}{x^4}dx &= \\int_{0}^{\\infty}\\frac{1}{3x^2}(2\\sin^2 2x + \\cos 2x – 1) dx \\label{eq:sincfourthpowersecond}
\n\\end{align}
\n\u3068\u5909\u5f62\u3067\u304d\u3066\u3001\u3055\u3089\u306b$\\cos 2x – 1 = -2\\sin^2 x$\u3067\u3042\u308b\u3053\u3068\u3092\u5229\u7528\u3059\u308b\u3068\u2026
\n\\begin{align}
\n\\int_{0}^{\\infty} \\frac{\\sin^4 x}{x^4}dx &= \\int_{0}^{\\infty}\\frac{1}{3x^2}(2\\sin^2 2x – 2\\sin^2 x) dx \\nonumber \\cr
\n&= \\frac{2}{3}\\pi – \\frac{\\pi}{3} \\nonumber \\cr
\n&= \\frac{\\pi}{3} \\label{eq:sincfourthpower}
\n\\end{align}
\n\u3068\u8a08\u7b97\u3067\u304d\u307e\u3059\u3002\u306a\u304a\u3001\u6700\u5f8c\u306e\u5909\u5f62\u3067(\\ref{eq:sincsquareatimes})\u5f0f\u3092\u5229\u7528\u3057\u3066\u3044\u307e\u3059\u3002<\/p>\n
\u304a\u307e\u3051: sinc^2, sinc^3, sinc^4\u306e\u30b0\u30e9\u30d5<\/h2>\n
sinc\u95a2\u6570\u306e5\u4e57\u306e\u8a08\u7b97\u306f\u3067\u304d\u307e\u305b\u3093\u3067\u3057\u305f\u3002orz<\/p>\n
\u305d\u3053\u3067\u3001\u533a\u9593$\\left[-2\\pi,2\\pi\\right]$\u306b\u304a\u3051\u308b$f(x) = \\displaystyle\\frac{\\sin^n x}{x^n}, (n = 2,3,4)$\u306e\u30b0\u30e9\u30d5\u3092Inkscape\u3067\u66f8\u3044\u3066\u307f\u307e\u3057\u305f\u3002 \u4e0a\u8a18\u30b0\u30e9\u30d5\u4e2d\u3001\u8d64\u8272\u306e\u5b9f\u7dda\u304c$f(x) = \\displaystyle\\frac{\\sin^2 x}{x^2}$\u3001\u9752\u8272\u306e\u70b9\u7dda\u304c$f(x) = \\displaystyle\\frac{\\sin^3 x}{x^3}$\u3001\u7dd1\u8272\u306e\u4e00\u70b9\u9396\u7dda\u304c$f(x) = \\displaystyle\\frac{\\sin^4 x}{x^4}$\u306e\u30b0\u30e9\u30d5\u3067\u3059\u3002<\/p>\n \u3053\u306e\u8a18\u4e8b\u3067\u66f8\u3044\u305f2,3,4\u4e57\u306e\u8a08\u7b97\u53ca\u3073\u9014\u4e2d\u7d4c\u904e\u306b\u3064\u3044\u3066\u306f\u6b63\u3057\u304f\u306a\u3044\u90e8\u5206\u304c\u3042\u308b\u304b\u3082\u3057\u308c\u307e\u305b\u3093\u306e\u3067\u3001\u5225\u9014\u8a08\u7b97\u306a\u3069\u3092\u884c\u3063\u305f\u969b\u306e\u7d50\u679c\u306e\u78ba\u8a8d\u7528\u306a\u3069\u306b\u5229\u7528\u3057\u3066\u3044\u305f\u3060\u3051\u308b\u3068\u5e78\u3044\u3067\u3059\u3002\ud83d\ude47\u200d\u2642\ufe0f<\/p>\n \u3053\u306e\u8a18\u4e8b\u306f\u4ee5\u4e0a\u3067\u3059\u3002<\/p>\n \u306f\u3058\u3081\u306b \u3061\u3087\u3063\u3068\u524d\u306e\u8a18\u4e8b\u3067\u30c7\u30a3\u30ea\u30af\u30ec\u7a4d\u5206\u306e\u8a08\u7b97\u3092\u884c\u3044\u307e\u3057\u305f\u3002 \u305d\u308c\u3092\u4f7f\u3063\u3066sinc\u95a2\u6570\u306e\u7a4d\u5206\u3092\u8907\u7d20\u7a4d\u5206\u3092\u4f7f\u308f\u306a\u3044\u3067\u8a08\u7b97\u3067\u304d\u306a\u3044\u304b\u3068\u601d\u3044\u3001\u90e8\u5206\u7a4d\u5206\u3092\u4f7f\u3063\u3066\u8a08\u7b97\u3057\u3066\u307f\u305f\u3068\u3053\u308d\u30014\u4e57\u3042\u305f\u308a\u307e\u3067\u306a\u3089\u4f55\u3068\u304b\u8a08\u7b97\u3067\u304d\u305d\u3046\u306a\u306e\u3067\u3001\u30e1\u30e2\u3059\u308b\u3053\u3068\u306b\u3057\u307e\u3057\u305f\u3002 \u306a\u304a\u3001\u3053\u306e\u8a18\u4e8b\u3067\u306f\u300csinc\u95a2\u6570\u300d\u306f(\\ref{eq:sincdef})\u5f0f\u3067\u8868\u3055\u308c\u308b\u95a2\u6570$f(x)$\u306e\u3053\u3068\u3092\u6307\u3059\u3082\u306e\u3068\u3057\u307e\u3059\u3002 \\begin{align} \u2026 Read More »<\/a><\/span><\/p>\n","protected":false},"author":1,"featured_media":4416,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[31,5,13],"tags":[],"class_list":["post-4400","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-inkscape","category-latex","category-13"],"_links":{"self":[{"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/posts\/4400","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/pandanote.info\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=4400"}],"version-history":[{"count":20,"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/posts\/4400\/revisions"}],"predecessor-version":[{"id":9369,"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/posts\/4400\/revisions\/9369"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/pandanote.info\/index.php?rest_route=\/wp\/v2\/media\/4416"}],"wp:attachment":[{"href":"https:\/\/pandanote.info\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4400"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/pandanote.info\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4400"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/pandanote.info\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4400"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n![\"\"](\"https:\/\/pandanote.info\/wordpress\/wp-content\/uploads\/2019\/05\/dirchlet_integral_scene3-300x171.png\")
<\/a><\/p>\n\u307e\u3068\u3081<\/h2>\n
References \/ \u53c2\u8003\u6587\u732e<\/h2>\n
\n