{
  "index": "1942-B-5",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "11. Sketch the curve\n\\[\ny=\\frac{x}{1+x^{6} \\sin ^{2} x}\n\\]\nand show that\n\\[\n\\int_{0}^{\\infty} \\frac{x d x}{1+x^{6} \\sin ^{2} x}\n\\]\nexists.",
  "solution": "Solution. Let\n\\[\nf(x)=\\frac{x}{1+x^{6} \\sin ^{2} x}\n\\]\n\nThen \\( f \\) is an odd function, so its graph is symmetric with respect to the origin; hence we need consider only non-negative values of \\( x \\).\n\nClearly\n\\[\nx \\geq \\frac{x}{1+x^{6} \\sin ^{2} x} \\geq \\frac{x}{1+x^{6}} .\n\\]\n\nAt points \\( x=n \\pi, n \\) an integer, the curve is tangent to the line \\( y=x \\). At points \\( x=\\left(n+\\frac{1}{2}\\right) \\pi \\), it is tangent to the graph of \\( y=x /\\left(1+x^{6}\\right) \\).\n\nSo the graph of \\( f(x) \\) oscillates between an upper curve \\( y=x \\) and a lower curve \\( y=x /\\left(1+x^{0}\\right) \\).\n\nFor large values of \\( x \\), except those very near an integral multiple of \\( \\pi \\), \\( f \\) is very small, i.e., near the lower curve. Indeed, the graph is characterized by tall, narrow spikes at integral multiples of \\( \\pi \\). (The spikes are much narrower than the drawing suggests. For example, the function \\( f \\) drops to less than .02 within .1 of \\( 2 \\pi \\).)\nSince \\( f \\) is non-negative on the whole domain of integration, the question of the convergence of\n\\[\n\\int_{0}^{\\infty} f(x) d x\n\\]\nis equivalent to whether or not\n\\[\n\\int_{0}^{t} f(x) d x\n\\]\nis bounded, and in turn to whether or not\n\\[\n\\sum_{n=1}^{\\infty} \\int_{(n-1 / 2) \\pi}^{(n+1 / 2) \\pi} f(x) d x\n\\]\nis convergent.\nNote that \\( |\\sin t| \\geq(2 / \\pi)|t| \\) for \\( |t| \\leq \\pi / 2 \\). Set \\( k_{n}=2 \\pi^{2}\\left(n-\\frac{1}{2}\\right)^{3} \\). Then, for any positive integer \\( n \\), and for\n\\[\n\\left(n-\\frac{1}{2}\\right) \\pi \\leq x \\leq\\left(n+\\frac{1}{2}\\right) \\pi,\n\\]\nwe have\n\\[\n\\begin{aligned}\n1+x^{6} \\sin ^{2} x & =1+x^{6} \\sin ^{2}(x-n \\pi) \\\\\n& \\geq 1+\\left(n-\\frac{1}{2}\\right)^{6} \\pi^{6} \\frac{4}{\\pi^{2}}(x-n \\pi)^{2} \\\\\n& =1+k_{n}^{2}(x-n \\pi)^{2} .\n\\end{aligned}\n\\]\n\nSo we get\n\\[\nf(x) \\leq \\frac{\\left(n+\\frac{1}{2}\\right) \\pi}{1+k_{n}^{2}(x-n \\pi)^{2}}\n\\]\n\nTherefore\n\\[\n\\begin{array}{r}\n\\int_{(n-1 / 2) \\pi}^{(n+1 / 2) \\pi} f(x) d x \\leq\\left(n+\\frac{1}{2}\\right) \\pi \\int_{-\\pi 2}^{\\pi 2} \\frac{d u}{1+k_{n}{ }^{2} u^{2}} \\\\\n\\leq\\left(n+\\frac{1}{2}\\right) \\pi \\int_{-\\infty}^{\\infty} \\frac{d u}{1+k_{n}{ }^{2} u^{2}} \\\\\n=\\frac{\\left(n+\\frac{1}{2}\\right) \\pi^{2}}{k_{n}} \\leq \\frac{A}{n^{2}}\n\\end{array}\n\\]\nfor some number \\( A \\) independent of \\( n \\).\nSince \\( \\sum_{n=1}^{\\infty} 1 / n^{2} \\) is convergent, we conclude that\n\\[\n\\sum_{n=1}^{\\infty} \\int_{(n-1 / 2) \\pi}^{(n+1 / 2) \\pi} f(x) d x\n\\]\nis convergent.\nTherefore \\( \\int_{0}^{\\infty} f(x) d x \\) exists.\nRemark. The problem was first treated by G. H. Hardy, in Messenger of Mathematics, vol. 31, 1902, p. 177. He showed that\n\\[\n\\int_{0}^{\\infty} \\frac{x^{\\mu}}{1+x^{\\nu} \\sin ^{2} x} d x\n\\]\nconverges if and only if \\( \\nu>2 \\mu+2 \\).",
  "vars": [
    "x",
    "y",
    "f",
    "t",
    "u"
  ],
  "params": [
    "n",
    "k_n",
    "A",
    "\\\\mu",
    "\\\\nu"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "f": "functionf",
        "t": "variablet",
        "u": "variableu",
        "n": "indexint",
        "k_n": "largescaling",
        "A": "constanta",
        "\\mu": "exponentmu",
        "\\nu": "exponentnu"
      },
      "question": "11. Sketch the curve\n\\[\nvariabley=\\frac{variablex}{1+variablex^{6} \\sin ^{2} variablex}\n\\]\nand show that\n\\[\n\\int_{0}^{\\infty} \\frac{variablex d variablex}{1+variablex^{6} \\sin ^{2} variablex}\n\\]\nexists.",
      "solution": "Solution. Let\n\\[\nfunctionf(variablex)=\\frac{variablex}{1+variablex^{6} \\sin ^{2} variablex}\n\\]\n\nThen \\( functionf \\) is an odd function, so its graph is symmetric with respect to the origin; hence we need consider only non-negative values of \\( variablex \\).\n\nClearly\n\\[\nvariablex \\geq \\frac{variablex}{1+variablex^{6} \\sin ^{2} variablex} \\geq \\frac{variablex}{1+variablex^{6}} .\n\\]\n\nAt points \\( variablex=indexint \\pi, indexint \\) an integer, the curve is tangent to the line \\( variabley=variablex \\). At points \\( variablex=\\left(indexint+\\frac{1}{2}\\right) \\pi \\), it is tangent to the graph of \\( variabley=variablex /\\left(1+variablex^{6}\\right) \\).\n\nSo the graph of \\( functionf(variablex) \\) oscillates between an upper curve \\( variabley=variablex \\) and a lower curve \\( variabley=variablex /\\left(1+variablex^{6}\\right) \\).\n\nFor large values of \\( variablex \\), except those very near an integral multiple of \\( \\pi \\), \\( functionf \\) is very small, i.e., near the lower curve. Indeed, the graph is characterized by tall, narrow spikes at integral multiples of \\( \\pi \\). (The spikes are much narrower than the drawing suggests. For example, the function \\( functionf \\) drops to less than .02 within .1 of \\( 2 \\pi \\).)\nSince \\( functionf \\) is non-negative on the whole domain of integration, the question of the convergence of\n\\[\n\\int_{0}^{\\infty} functionf(variablex) d variablex\n\\]\nis equivalent to whether or not\n\\[\n\\int_{0}^{variablet} functionf(variablex) d variablex\n\\]\nis bounded, and in turn to whether or not\n\\[\n\\sum_{indexint=1}^{\\infty} \\int_{(indexint-1 / 2) \\pi}^{(indexint+1 / 2) \\pi} functionf(variablex) d variablex\n\\]\nis convergent.\nNote that \\( |\\sin variablet| \\geq(2 / \\pi)|variablet| \\) for \\( |variablet| \\leq \\pi / 2 \\). Set \\( largescaling=2 \\pi^{2}\\left(indexint-\\frac{1}{2}\\right)^{3} \\). Then, for any positive integer \\( indexint \\), and for\n\\[\n\\left(indexint-\\frac{1}{2}\\right) \\pi \\leq variablex \\leq\\left(indexint+\\frac{1}{2}\\right) \\pi,\n\\]\nwe have\n\\[\n\\begin{aligned}\n1+variablex^{6} \\sin ^{2} variablex & =1+variablex^{6} \\sin ^{2}(variablex-indexint \\pi) \\\\\n& \\geq 1+\\left(indexint-\\frac{1}{2}\\right)^{6} \\pi^{6} \\frac{4}{\\pi^{2}}(variablex-indexint \\pi)^{2} \\\\\n& =1+largescaling^{2}(variablex-indexint \\pi)^{2} .\n\\end{aligned}\n\\]\n\nSo we get\n\\[\nfunctionf(variablex) \\leq \\frac{\\left(indexint+\\frac{1}{2}\\right) \\pi}{1+largescaling^{2}(variablex-indexint \\pi)^{2}}\n\\]\n\nTherefore\n\\[\n\\begin{array}{r}\n\\int_{(indexint-1 / 2) \\pi}^{(indexint+1 / 2) \\pi} functionf(variablex) d variablex \\leq\\left(indexint+\\frac{1}{2}\\right) \\pi \\int_{-\\pi 2}^{\\pi 2} \\frac{d variableu}{1+largescaling{ }^{2} variableu^{2}} \\\\\n\\leq\\left(indexint+\\frac{1}{2}\\right) \\pi \\int_{-\\infty}^{\\infty} \\frac{d variableu}{1+largescaling{ }^{2} variableu^{2}} \\\\\n=\\frac{\\left(indexint+\\frac{1}{2}\\right) \\pi^{2}}{largescaling} \\leq \\frac{constanta}{indexint^{2}}\n\\end{array}\n\\]\nfor some number \\( constanta \\) independent of \\( indexint \\).\nSince \\( \\sum_{indexint=1}^{\\infty} 1 / indexint^{2} \\) is convergent, we conclude that\n\\[\n\\sum_{indexint=1}^{\\infty} \\int_{(indexint-1 / 2) \\pi}^{(indexint+1 / 2) \\pi} functionf(variablex) d variablex\n\\]\nis convergent.\nTherefore \\( \\int_{0}^{\\infty} functionf(variablex) d variablex \\) exists.\nRemark. The problem was first treated by G. H. Hardy, in Messenger of Mathematics, vol. 31, 1902, p. 177. He showed that\n\\[\n\\int_{0}^{\\infty} \\frac{variablex^{exponentmu}}{1+variablex^{exponentnu} \\sin ^{2} variablex} d variablex\n\\]\nconverges if and only if \\( exponentnu>2 exponentmu+2 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marigolds",
        "y": "campfires",
        "f": "sugarloaf",
        "t": "ploughman",
        "u": "rainstorm",
        "n": "driftwood",
        "k_n": "lighthouse",
        "A": "waterfall",
        "\\mu": "barleycorn",
        "\\nu": "ivorytower"
      },
      "question": "11. Sketch the curve\n\\[\ncampfires=\\frac{marigolds}{1+marigolds^{6} \\sin ^{2} marigolds}\n\\]\nand show that\n\\[\n\\int_{0}^{\\infty} \\frac{marigolds\\,d marigolds}{1+marigolds^{6} \\sin ^{2} marigolds}\n\\]\nexists.",
      "solution": "Solution. Let\n\\[\nsugarloaf(marigolds)=\\frac{marigolds}{1+marigolds^{6} \\sin ^{2} marigolds}\n\\]\nThen \\( sugarloaf \\) is an odd function, so its graph is symmetric with respect to the origin; hence we need consider only non-negative values of \\( marigolds \\).\n\nClearly\n\\[\nmarigolds \\geq \\frac{marigolds}{1+marigolds^{6} \\sin ^{2} marigolds} \\geq \\frac{marigolds}{1+marigolds^{6}} .\n\\]\n\nAt points \\( marigolds=driftwood \\pi, driftwood \\) an integer, the curve is tangent to the line \\( campfires=marigolds \\). At points \\( marigolds=\\left(driftwood+\\frac{1}{2}\\right) \\pi \\), it is tangent to the graph of \\( campfires=marigolds /\\left(1+marigolds^{6}\\right) \\).\n\nSo the graph of \\( sugarloaf(marigolds) \\) oscillates between an upper curve \\( campfires=marigolds \\) and a lower curve \\( campfires=marigolds /\\left(1+marigolds^{0}\\right) \\).\n\nFor large values of \\( marigolds \\), except those very near an integral multiple of \\( \\pi \\), \\( sugarloaf \\) is very small, i.e., near the lower curve. Indeed, the graph is characterized by tall, narrow spikes at integral multiples of \\( \\pi \\). (The spikes are much narrower than the drawing suggests. For example, the function \\( sugarloaf \\) drops to less than .02 within .1 of \\( 2 \\pi \\).)\n\nSince \\( sugarloaf \\) is non-negative on the whole domain of integration, the question of the convergence of\n\\[\n\\int_{0}^{\\infty} sugarloaf(marigolds) \\, d marigolds\n\\]\nis equivalent to whether or not\n\\[\n\\int_{0}^{ploughman} sugarloaf(marigolds) \\, d marigolds\n\\]\nis bounded, and in turn to whether or not\n\\[\n\\sum_{driftwood=1}^{\\infty} \\int_{(driftwood-1 / 2) \\pi}^{(driftwood+1 / 2) \\pi} sugarloaf(marigolds) \\, d marigolds\n\\]\nis convergent.\n\nNote that \\( |\\sin ploughman| \\geq(2 / \\pi)|ploughman| \\) for \\( |ploughman| \\leq \\pi / 2 \\). Set \\( lighthouse_{driftwood}=2 \\pi^{2}\\left(driftwood-\\frac{1}{2}\\right)^{3} \\). Then, for any positive integer \\( driftwood \\), and for\n\\[\n\\left(driftwood-\\frac{1}{2}\\right) \\pi \\leq marigolds \\leq\\left(driftwood+\\frac{1}{2}\\right) \\pi,\n\\]\nwe have\n\\[\n\\begin{aligned}\n1+marigolds^{6} \\sin ^{2} marigolds & =1+marigolds^{6} \\sin ^{2}(marigolds-driftwood \\pi) \\\\\n& \\geq 1+\\left(driftwood-\\frac{1}{2}\\right)^{6} \\pi^{6} \\frac{4}{\\pi^{2}}(marigolds-driftwood \\pi)^{2} \\\\\n& =1+lighthouse_{driftwood}^{2}(marigolds-driftwood \\pi)^{2} .\n\\end{aligned}\n\\]\n\nSo we get\n\\[\nsugarloaf(marigolds) \\leq \\frac{\\left(driftwood+\\frac{1}{2}\\right) \\pi}{1+lighthouse_{driftwood}^{2}(marigolds-driftwood \\pi)^{2}}\n\\]\n\nTherefore\n\\[\n\\begin{array}{r}\n\\int_{(driftwood-1 / 2) \\pi}^{(driftwood+1 / 2) \\pi} sugarloaf(marigolds) \\, d marigolds \\leq\\left(driftwood+\\frac{1}{2}\\right) \\pi \\int_{-\\pi 2}^{\\pi 2} \\frac{d rainstorm}{1+lighthouse_{driftwood}{ }^{2} rainstorm^{2}} \\\\\n\\leq\\left(driftwood+\\frac{1}{2}\\right) \\pi \\int_{-\\infty}^{\\infty} \\frac{d rainstorm}{1+lighthouse_{driftwood}{ }^{2} rainstorm^{2}} \\\\\n=\\frac{\\left(driftwood+\\frac{1}{2}\\right) \\pi^{2}}{lighthouse_{driftwood}} \\leq \\frac{waterfall}{driftwood^{2}}\n\\end{array}\n\\]\nfor some number \\( waterfall \\) independent of \\( driftwood \\).\n\nSince \\( \\sum_{driftwood=1}^{\\infty} 1 / driftwood^{2} \\) is convergent, we conclude that\n\\[\n\\sum_{driftwood=1}^{\\infty} \\int_{(driftwood-1 / 2) \\pi}^{(driftwood+1 / 2) \\pi} sugarloaf(marigolds) \\, d marigolds\n\\]\nis convergent.\n\nTherefore \\( \\int_{0}^{\\infty} sugarloaf(marigolds) \\, d marigolds \\) exists.\n\nRemark. The problem was first treated by G. H. Hardy, in Messenger of Mathematics, vol. 31, 1902, p. 177. He showed that\n\\[\n\\int_{0}^{\\infty} \\frac{marigolds^{barleycorn}}{1+marigolds^{ivorytower} \\sin ^{2} marigolds} \\, d marigolds\n\\]\nconverges if and only if \\( ivorytower>2\\,barleycorn+2 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "dependentvalue",
        "y": "independentvalue",
        "f": "constantvalue",
        "t": "anglelessquantity",
        "u": "discreteindex",
        "n": "continuousparameter",
        "k_n": "flexiblescalar",
        "A": "varyinglimit",
        "\\mu": "radicalvalue",
        "\\nu": "logarithmorder"
      },
      "question": "11. Sketch the curve\n\\[\nindependentvalue=\\frac{dependentvalue}{1+dependentvalue^{6} \\sin ^{2} dependentvalue}\n\\]\nand show that\n\\[\n\\int_{0}^{\\infty} \\frac{dependentvalue d dependentvalue}{1+dependentvalue^{6} \\sin ^{2} dependentvalue}\n\\]\nexists.",
      "solution": "Solution. Let\n\\[\nconstantvalue(dependentvalue)=\\frac{dependentvalue}{1+dependentvalue^{6} \\sin ^{2} dependentvalue}\n\\]\n\nThen \\( constantvalue \\) is an odd function, so its graph is symmetric with respect to the origin; hence we need consider only non-negative values of \\( dependentvalue \\).\n\nClearly\n\\[\ndependentvalue \\geq \\frac{dependentvalue}{1+dependentvalue^{6} \\sin ^{2} dependentvalue} \\geq \\frac{dependentvalue}{1+dependentvalue^{6}} .\n\\]\n\nAt points \\( dependentvalue=continuousparameter \\pi, continuousparameter \\) an integer, the curve is tangent to the line \\( independentvalue=dependentvalue \\). At points \\( dependentvalue=\\left(continuousparameter+\\frac{1}{2}\\right) \\pi \\), it is tangent to the graph of \\( independentvalue=dependentvalue /\\left(1+dependentvalue^{6}\\right) \\).\n\nSo the graph of \\( constantvalue(dependentvalue) \\) oscillates between an upper curve \\( independentvalue=dependentvalue \\) and a lower curve \\( independentvalue=dependentvalue /\\left(1+dependentvalue^{0}\\right) \\).\n\nFor large values of \\( dependentvalue \\), except those very near an integral multiple of \\( \\pi \\), \\( constantvalue \\) is very small, i.e., near the lower curve. Indeed, the graph is characterized by tall, narrow spikes at integral multiples of \\( \\pi \\). (The spikes are much narrower than the drawing suggests. For example, the function \\( constantvalue \\) drops to less than .02 within .1 of \\( 2 \\pi \\).)\nSince \\( constantvalue \\) is non-negative on the whole domain of integration, the question of the convergence of\n\\[\n\\int_{0}^{\\infty} constantvalue(dependentvalue) d dependentvalue\n\\]\nis equivalent to whether or not\n\\[\n\\int_{0}^{anglelessquantity} constantvalue(dependentvalue) d dependentvalue\n\\]\nis bounded, and in turn to whether or not\n\\[\n\\sum_{continuousparameter=1}^{\\infty} \\int_{(continuousparameter-1 / 2) \\pi}^{(continuousparameter+1 / 2) \\pi} constantvalue(dependentvalue) d dependentvalue\n\\]\nis convergent.\nNote that \\( |\\sin anglelessquantity| \\geq(2 / \\pi)|anglelessquantity| \\) for \\( |anglelessquantity| \\leq \\pi / 2 \\). Set \\( flexiblescalar_{continuousparameter}=2 \\pi^{2}\\left(continuousparameter-\\frac{1}{2}\\right)^{3} \\). Then, for any positive integer \\( continuousparameter \\), and for\n\\[\n\\left(continuousparameter-\\frac{1}{2}\\right) \\pi \\leq dependentvalue \\leq\\left(continuousparameter+\\frac{1}{2}\\right) \\pi,\n\\]\nwe have\n\\[\n\\begin{aligned}\n1+dependentvalue^{6} \\sin ^{2} dependentvalue & =1+dependentvalue^{6} \\sin ^{2}(dependentvalue-continuousparameter \\pi) \\\\\n& \\geq 1+\\left(continuousparameter-\\frac{1}{2}\\right)^{6} \\pi^{6} \\frac{4}{\\pi^{2}}(dependentvalue-continuousparameter \\pi)^{2} \\\\\n& =1+flexiblescalar_{continuousparameter}^{2}(dependentvalue-continuousparameter \\pi)^{2} .\n\\end{aligned}\n\\]\n\nSo we get\n\\[\nconstantvalue(dependentvalue) \\leq \\frac{\\left(continuousparameter+\\frac{1}{2}\\right) \\pi}{1+flexiblescalar_{continuousparameter}^{2}(dependentvalue-continuousparameter \\pi)^{2}}\n\\]\n\nTherefore\n\\[\n\\begin{array}{r}\n\\int_{(continuousparameter-1 / 2) \\pi}^{(continuousparameter+1 / 2) \\pi} constantvalue(dependentvalue) d dependentvalue \\leq\\left(continuousparameter+\\frac{1}{2}\\right) \\pi \\int_{-\\pi 2}^{\\pi 2} \\frac{d discreteindex}{1+flexiblescalar_{continuousparameter}^{2} discreteindex^{2}} \\\\\n\\leq\\left(continuousparameter+\\frac{1}{2}\\right) \\pi \\int_{-\\infty}^{\\infty} \\frac{d discreteindex}{1+flexiblescalar_{continuousparameter}^{2} discreteindex^{2}} \\\\\n=\\frac{\\left(continuousparameter+\\frac{1}{2}\\right) \\pi^{2}}{flexiblescalar_{continuousparameter}} \\leq \\frac{varyinglimit}{continuousparameter^{2}}\n\\end{array}\n\\]\nfor some number \\( varyinglimit \\) independent of \\( continuousparameter \\).\nSince \\( \\sum_{continuousparameter=1}^{\\infty} 1 / continuousparameter^{2} \\) is convergent, we conclude that\n\\[\n\\sum_{continuousparameter=1}^{\\infty} \\int_{(continuousparameter-1 / 2) \\pi}^{(continuousparameter+1 / 2) \\pi} constantvalue(dependentvalue) d dependentvalue\n\\]\nis convergent.\nTherefore \\( \\int_{0}^{\\infty} constantvalue(dependentvalue) d dependentvalue \\) exists.\nRemark. The problem was first treated by G. H. Hardy, in Messenger of Mathematics, vol. 31, 1902, p. 177. He showed that\n\\[\n\\int_{0}^{\\infty} \\frac{dependentvalue^{radicalvalue}}{1+dependentvalue^{logarithmorder} \\sin ^{2} dependentvalue} d dependentvalue\n\\]\nconverges if and only if \\( logarithmorder>2 radicalvalue+2 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qxzpmrtd",
        "y": "ljcbryun",
        "f": "spqmdnva",
        "t": "krzyploq",
        "u": "vdawerui",
        "n": "hsgdplma",
        "k_n": "nvtqzxea",
        "A": "brfiklqw",
        "\\mu": "pdnzqray",
        "\\nu": "ctargwle"
      },
      "question": "11. Sketch the curve\n\\[\nljcbryun=\\frac{qxzpmrtd}{1+qxzpmrtd^{6} \\sin ^{2} qxzpmrtd}\n\\]\nand show that\n\\[\n\\int_{0}^{\\infty} \\frac{qxzpmrtd \\, d qxzpmrtd}{1+qxzpmrtd^{6} \\sin ^{2} qxzpmrtd}\n\\]\nexists.",
      "solution": "Solution. Let\n\\[\nspqmdnva(qxzpmrtd)=\\frac{qxzpmrtd}{1+qxzpmrtd^{6} \\sin ^{2} qxzpmrtd}\n\\]\n\nThen \\( spqmdnva \\) is an odd function, so its graph is symmetric with respect to the origin; hence we need consider only non-negative values of \\( qxzpmrtd \\).\n\nClearly\n\\[\nqxzpmrtd \\geq \\frac{qxzpmrtd}{1+qxzpmrtd^{6} \\sin ^{2} qxzpmrtd} \\geq \\frac{qxzpmrtd}{1+qxzpmrtd^{6}} .\n\\]\n\nAt points \\( qxzpmrtd=hsgdplma \\pi, hsgdplma \\) an integer, the curve is tangent to the line \\( ljcbryun=qxzpmrtd \\). At points \\( qxzpmrtd=\\left(hsgdplma+\\frac{1}{2}\\right) \\pi \\), it is tangent to the graph of \\( ljcbryun=qxzpmrtd /\\left(1+qxzpmrtd^{6}\\right) \\).\n\nSo the graph of \\( spqmdnva(qxzpmrtd) \\) oscillates between an upper curve \\( ljcbryun=qxzpmrtd \\) and a lower curve \\( ljcbryun=qxzpmrtd /\\left(1+qxzpmrtd^{0}\\right) \\).\n\nFor large values of \\( qxzpmrtd \\), except those very near an integral multiple of \\( \\pi \\), \\( spqmdnva \\) is very small, i.e., near the lower curve. Indeed, the graph is characterized by tall, narrow spikes at integral multiples of \\( \\pi \\). (The spikes are much narrower than the drawing suggests. For example, the function \\( spqmdnva \\) drops to less than .02 within .1 of \\( 2 \\pi \\).)\n\nSince \\( spqmdnva \\) is non-negative on the whole domain of integration, the question of the convergence of\n\\[\n\\int_{0}^{\\infty} spqmdnva(qxzpmrtd) d qxzpmrtd\n\\]\nis equivalent to whether or not\n\\[\n\\int_{0}^{krzyploq} spqmdnva(qxzpmrtd) d qxzpmrtd\n\\]\nis bounded, and in turn to whether or not\n\\[\n\\sum_{hsgdplma=1}^{\\infty} \\int_{(hsgdplma-1 / 2) \\pi}^{(hsgdplma+1 / 2) \\pi} spqmdnva(qxzpmrtd) d qxzpmrtd\n\\]\nis convergent.\n\nNote that \\( |\\sin krzyploq| \\geq(2 / \\pi)|krzyploq| \\) for \\( |krzyploq| \\leq \\pi / 2 \\). Set \\( nvtqzxea=2 \\pi^{2}\\left(hsgdplma-\\frac{1}{2}\\right)^{3} \\). Then, for any positive integer \\( hsgdplma \\), and for\n\\[\n\\left(hsgdplma-\\frac{1}{2}\\right) \\pi \\leq qxzpmrtd \\leq\\left(hsgdplma+\\frac{1}{2}\\right) \\pi,\n\\]\nwe have\n\\[\n\\begin{aligned}\n1+qxzpmrtd^{6} \\sin ^{2} qxzpmrtd & =1+qxzpmrtd^{6} \\sin ^{2}(qxzpmrtd-hsgdplma \\pi) \\\\\n& \\geq 1+\\left(hsgdplma-\\frac{1}{2}\\right)^{6} \\pi^{6} \\frac{4}{\\pi^{2}}(qxzpmrtd-hsgdplma \\pi)^{2} \\\\\n& =1+nvtqzxea^{2}(qxzpmrtd-hsgdplma \\pi)^{2} .\n\\end{aligned}\n\\]\n\nSo we get\n\\[\nspqmdnva(qxzpmrtd) \\leq \\frac{\\left(hsgdplma+\\frac{1}{2}\\right) \\pi}{1+nvtqzxea^{2}(qxzpmrtd-hsgdplma \\pi)^{2}}\n\\]\n\nTherefore\n\\[\n\\begin{array}{r}\n\\int_{(hsgdplma-1 / 2) \\pi}^{(hsgdplma+1 / 2) \\pi} spqmdnva(qxzpmrtd) d qxzpmrtd \\leq\\left(hsgdplma+\\frac{1}{2}\\right) \\pi \\int_{-\\pi 2}^{\\pi 2} \\frac{d vdawerui}{1+nvtqzxea^{2} vdawerui^{2}} \\\\\n\\leq\\left(hsgdplma+\\frac{1}{2}\\right) \\pi \\int_{-\\infty}^{\\infty} \\frac{d vdawerui}{1+nvtqzxea^{2} vdawerui^{2}} \\\\\n=\\frac{\\left(hsgdplma+\\frac{1}{2}\\right) \\pi^{2}}{nvtqzxea} \\leq \\frac{brfiklqw}{hsgdplma^{2}}\n\\end{array}\n\\]\nfor some number \\( brfiklqw \\) independent of \\( hsgdplma \\).\n\nSince \\( \\sum_{hsgdplma=1}^{\\infty} 1 / hsgdplma^{2} \\) is convergent, we conclude that\n\\[\n\\sum_{hsgdplma=1}^{\\infty} \\int_{(hsgdplma-1 / 2) \\pi}^{(hsgdplma+1 / 2) \\pi} spqmdnva(qxzpmrtd) d qxzpmrtd\n\\]\nis convergent.\nTherefore \\( \\int_{0}^{\\infty} spqmdnva(qxzpmrtd) d qxzpmrtd \\) exists.\n\nRemark. The problem was first treated by G. H. Hardy, in Messenger of Mathematics, vol. 31, 1902, p. 177. He showed that\n\\[\n\\int_{0}^{\\infty} \\frac{qxzpmrtd^{pdnzqray}}{1+qxzpmrtd^{ctargwle} \\sin ^{2} qxzpmrtd} d qxzpmrtd\n\\]\nconverges if and only if \\( ctargwle>2 pdnzqray+2 \\)."
    },
    "kernel_variant": {
      "question": "Fix the two real parameters  \n\n  m ≥ 4  and  α > 3.  \n\nFor x ∈ ℝ set  \n  F_{m,α}(x)=\\dfrac{x^{2}}{1+x^{2m}\\sin^{2}x}+\\dfrac{\\sin x}{1+x^{α}}.  \n\n(a)  Give a qualitative sketch of y = F_{m,α}(x), pointing out its parity, its behaviour near the origin, and the form of the alternating spikes that occur close to the points x = nπ, n ∈ ℤ.  \n\n(b)  Show that the improper integral  \n\n  I_{m,α}=∫_{0}^{∞}|F_{m,α}(x)|\\,dx  \n\nexists for every m ≥ 4 and α > 3.  \n\n(c)  Prove that, for fixed m, the map α↦I_{m,α} is continuous on (3,∞).",
      "solution": "(a)  Preliminaries.  Because sin is odd and the quotients involve only even powers of x, one checks immediately that F_{m,α}(−x)=−F_{m,α}(x); hence the graph is centrally symmetric and it suffices to discuss x ≥ 0.  Near the origin we have sin x≈x, so  \n\n F_{m,α}(x)=x^{2}+O(x^{4})+O(x),  \n\nand therefore the curve leaves the origin tangentially to y = x².  \nAt x=nπ the sine term vanishes; the first summand equals (nπ)², the second vanishes, so tall positive or negative spikes of height ~(nπ)² appear.  Writing x=nπ+t with |t|≪1 and using sin t≈t,   \n\n F_{m,α}(x)≈\\dfrac{(nπ)²}{1+(nπ)^{2m}t²}+O\\!\\bigl((nπ)^{−(α−1)}\\bigr),  \n\nwe see that the spikes have width comparable to (nπ)^{−m}.  Between spikes, |sin x|≥sin(π/4)=1/√2, so the first summand is O(x^{2−2m}) and the second is O(x^{−α}).  Since 2−2m≤−6 and −α≤−4, the mid-level of the graph decays at least like x^{−4}.  \n\n(b)  Absolute convergence.  We split the half-line into three disjoint sets:\n\nE₀=[0,1],  A_n=[nπ−π/4,nπ+π/4] (n≥1),  and  M=ℝ_{≥0}\\setminus(E₀∪⋃A_n).\n\nStep 1: the origin.  On E₀ we have |F_{m,α}(x)|≤x²+x, so ∫_{E₀}|F_{m,α}|≤2/3.  \n\nStep 2: the spikes.  For x∈A_n write x=nπ+t, |t|≤π/4.  By |sin t|≥(2/π)|t| we get  \n\n 1+x^{2m}\\sin²x ≥ 1+c_n²t², c_n=\\dfrac{2(nπ−π/4)^{m}}{π}.  \n\nMoreover x²≤(nπ+π/4)².  Hence  \n\n |F_{m,α}(x)|≤\\dfrac{(nπ+π/4)²}{1+c_n²t²}+C(nπ)^{−α}, C constant.  \n\nIntegrating first term over t gives  \n\n ∫_{A_n}≤π²(nπ+π/4)²/(2c_n)=O(n^{2−m}),  \n\nwhile the second term contributes O(n^{1−α}).  Because m ≥ 4 and α > 3, the series Σ_n O(n^{2−m})+O(n^{1−α}) converges; hence Σ_n∫_{A_n}|F_{m,α}| converges.  \n\nStep 3: the mid-region.  On M we have |sin x|≥1/√2, so  \n\n |F_{m,α}(x)|≤2x^{2−2m}+x^{−α}.  \n\nBoth integrands are dominated by Cx^{−4}, and ∫_{1}^{∞}x^{−4}dx=1/3.  \n\nCombining the three steps we conclude I_{m,α}<∞.  \n\n(c)  Continuity in α.  Fix m.  From the estimate in Step 3 we already know |F_{m,α}(x)|≤Cx^{−4} for x≥1, C independent of α.  On [0,1] the integrand is bounded by 2.  Hence we may apply the dominated convergence theorem: if α_k→α, then F_{m,α_k}(x)→F_{m,α}(x) pointwise and the uniform bound above serves as a global integrable majorant.  It follows that I_{m,α_k}→I_{m,α}, establishing continuity. ∎",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.136285",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}