{
  "index": "1949-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "2. Answer either (i) or (ii):\n(i) Prove that\n\\[\n\\sum_{n=2}^{\\infty} \\frac{\\cos (\\log \\log n)}{\\log n}\n\\]\ndiverges.\n(page 277)\n(ii) Assume that \\( p>0, a>0 \\), and \\( a c-b^{2}>0 \\), and show that\n\\[\n\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d x d y}{\\left(p+a x^{2}+2 b x y+c y^{2}\\right)^{2}}=\\pi p^{-1}\\left(a c-b^{2}\\right)^{-12}\n\\]",
  "solution": "Solution. A convergent series cannot have blocks of terms whose sum is arbitrarily large. We shall show that the given series has such blocks.\n\nFor a positive integer \\( k \\) consider the set \\( N_{k} \\) of integers \\( n \\) such that\n\\[\n2 \\pi k-\\frac{1}{3} \\pi \\leq \\log \\log n \\leq 2 \\pi k\n\\]\nand let\n\\[\nT_{k}=\\sum_{N_{k}} \\frac{\\cos (\\log \\log n)}{\\log n}\n\\]\n\nWe shall prove that \\( T_{k} \\rightarrow \\infty \\) as \\( k \\rightarrow \\infty \\). Now \\( N_{k}=\\left\\{n: \\exp \\exp \\left(2 \\pi k-\\frac{1}{3} \\pi\\right)\\right. \\) \\( \\leq n \\leq \\exp \\exp (2 \\pi k)\\} \\) and therefore \\( \\left|N_{k}\\right| \\), the number of elements in \\( N_{k} \\), satisfies\n\\[\n\\left|N_{k}\\right| \\geq \\exp \\exp 2 \\pi k-\\exp (\\alpha \\exp 2 \\pi k)-1\n\\]\nwhere \\( \\alpha=\\exp \\left(-\\frac{1}{3} \\pi\\right) \\).\nEach term in the sum (1) is at least\n\\[\n\\frac{\\cos \\left(-\\frac{1}{3} \\pi\\right)}{\\exp 2 \\pi k}=\\frac{1}{2 \\exp 2 \\pi k}\n\\]\nso\n\\[\nT_{k} \\geq \\frac{1}{2 x_{k}}\\left(\\exp \\left(x_{k}\\right)-\\exp \\left(\\alpha x_{k}\\right)-1\\right)\n\\]\nwhere \\( x_{k}=\\exp 2 \\pi k \\).\nNow\n\\[\n\\lim _{x \\rightarrow \\infty} \\frac{1}{x}(\\exp x-\\exp \\alpha x)=\\lim _{x \\rightarrow \\infty}(\\exp x-\\alpha \\exp \\alpha x)=\\infty\n\\]\nby L'Hospital's rule, using the fact that \\( \\alpha<1 \\). Since \\( x_{k} \\rightarrow \\infty \\) as \\( k \\rightarrow \\infty \\), it follows that \\( T_{k} \\rightarrow \\infty \\) as \\( k \\rightarrow \\infty \\), and this proves that the given series diverges.\n\nSolution. By a rotation of axes,\n\\[\n\\begin{array}{c}\nx=u \\cos \\theta+v \\sin \\theta \\\\\ny=-u \\sin \\theta+v \\cos \\theta\n\\end{array}\n\\]\nwith proper choice of the parameter \\( \\theta \\), the quadratic form \\( a x^{2}+2 b x y \\) \\( +c y^{2} \\) becomes\n\\[\nA u^{2}+C v^{2} .\n\\]\n\nThe discriminant \\( a c-b^{\\mathbf{2}} \\) is an invariant under (1), so\n\\[\nA C=a c-b^{2} .\n\\]\n\nSince the original form is positive definite, we also have \\( A>0 \\) and \\( C>0 \\). The transformation is area-preserving, so the required integral in the problem becomes\n\\[\n\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d u d v}{\\left(p+A u^{2}+C v^{2}\\right)^{2}} .\n\\]\n\nUnder the substitutions\n\\[\nu=\\sqrt{\\frac{p}{A}} s, \\quad v=\\sqrt{\\frac{p}{C}} t\n\\]\nthis becomes\n\\[\n\\frac{1}{p \\sqrt{A C}} \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d s d t}{\\left(1+s^{2}+t^{2}\\right)^{2}}\n\\]\n\nWe need only show that the value of the integral in (3) is \\( \\pi \\). Passing to polar coordinates\n\\[\n\\begin{array}{l}\ns=r \\cos \\theta \\\\\nt=r \\sin \\theta,\n\\end{array}\n\\]\nwe get\n\\[\n\\begin{aligned}\n\\int_{0}^{2 \\pi} \\int_{0}^{\\infty} \\frac{r d r d \\theta}{\\left(1+r^{2}\\right)^{2}} & =2 \\pi \\int_{0}^{\\infty} \\frac{r d r}{\\left(1+r^{2}\\right)^{2}} \\\\\n& =2 \\pi\\left[\\frac{-1}{2\\left(1+r^{2}\\right)}\\right]_{0}^{\\infty}=\\pi\n\\end{aligned}\n\\]",
  "vars": [
    "n",
    "k",
    "x",
    "y",
    "u",
    "v",
    "s",
    "t",
    "r",
    "\\\\theta",
    "N_k",
    "T_k",
    "x_k"
  ],
  "params": [
    "p",
    "a",
    "b",
    "c",
    "A",
    "C",
    "\\\\alpha"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "integer",
        "k": "blockidx",
        "x": "coordx",
        "y": "coordy",
        "u": "rotuvar",
        "v": "rotvvar",
        "s": "varscale",
        "t": "varscalt",
        "r": "radiusr",
        "\\theta": "angleth",
        "N_k": "blockset",
        "T_k": "blocksum",
        "x_k": "expvalue",
        "p": "paramp",
        "a": "parama",
        "b": "paramb",
        "c": "paramc",
        "A": "rotcoefu",
        "C": "rotcoefv",
        "\\alpha": "alphaval"
      },
      "question": "2. Answer either (i) or (ii):\n(i) Prove that\n\\[\n\\sum_{integer=2}^{\\infty} \\frac{\\cos (\\log \\log integer)}{\\log integer}\n\\]\ndiverges.\n(page 277)\n(ii) Assume that paramp>0, parama>0, and parama paramc-paramb^{2}>0, and show that\n\\[\n\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d coordx d coordy}{\\left(paramp+parama coordx^{2}+2 paramb coordx coordy+paramc coordy^{2}\\right)^{2}}=\\pi paramp^{-1}\\left(parama paramc-paramb^{2}\\right)^{-12}\n\\]",
      "solution": "A convergent series cannot have blocks of terms whose sum is arbitrarily large. We shall show that the given series has such blocks.\n\nFor a positive integer blockidx consider the set blockset of integers integer such that\n\\[\n2 \\pi blockidx-\\frac{1}{3} \\pi \\leq \\log \\log integer \\leq 2 \\pi blockidx\n\\]\nand let\n\\[\nblocksum=\\sum_{blockset} \\frac{\\cos (\\log \\log integer)}{\\log integer}\n\\]\n\nWe shall prove that blocksum \\rightarrow \\infty as blockidx \\rightarrow \\infty. Now blockset=\\{integer: \\exp \\exp (2 \\pi blockidx-\\frac{1}{3} \\pi) \\leq integer \\leq \\exp \\exp (2 \\pi blockidx)\\} and therefore |blockset|, the number of elements in blockset, satisfies\n\\[\n|blockset| \\geq \\exp \\exp 2 \\pi blockidx-\\exp (alphaval \\exp 2 \\pi blockidx)-1\n\\]\nwhere alphaval=\\exp (-\\frac{1}{3} \\pi).\nEach term in the sum (1) is at least\n\\[\n\\frac{\\cos (-\\frac{1}{3} \\pi)}{\\exp 2 \\pi blockidx}=\\frac{1}{2 \\exp 2 \\pi blockidx}\n\\]\nso\n\\[\nblocksum \\geq \\frac{1}{2\\,expvalue}\\bigl(\\exp (expvalue)-\\exp (alphaval\\,expvalue)-1\\bigr)\n\\]\nwhere expvalue=\\exp 2 \\pi blockidx.\nNow\n\\[\n\\lim _{coordx \\rightarrow \\infty} \\frac{1}{coordx}(\\exp coordx-\\exp alphaval\\,coordx)=\\lim _{coordx \\rightarrow \\infty}(\\exp coordx-alphaval \\exp alphaval\\,coordx)=\\infty\n\\]\nby L'Hospital's rule, using the fact that alphaval<1. Since expvalue \\rightarrow \\infty as blockidx \\rightarrow \\infty, it follows that blocksum \\rightarrow \\infty as blockidx \\rightarrow \\infty, and this proves that the given series diverges.\n\nBy a rotation of axes,\n\\[\n\\begin{array}{c}\ncoordx=rotuvar \\cos angleth+rotvvar \\sin angleth \\\\\ncoordy=-rotuvar \\sin angleth+rotvvar \\cos angleth\n\\end{array}\n\\]\nwith proper choice of the parameter angleth, the quadratic form parama coordx^{2}+2 paramb coordx coordy+paramc coordy^{2} becomes\n\\[\nrotcoefu rotuvar^{2}+rotcoefv rotvvar^{2}.\n\\]\n\nThe discriminant parama paramc-paramb^{2} is an invariant under (1), so\n\\[\nrotcoefu\\,rotcoefv=parama paramc-paramb^{2}.\n\\]\n\nSince the original form is positive definite, we also have rotcoefu>0 and rotcoefv>0. The transformation is area-preserving, so the required integral in the problem becomes\n\\[\n\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d rotuvar d rotvvar}{\\left(paramp+rotcoefu rotuvar^{2}+rotcoefv rotvvar^{2}\\right)^{2}}.\n\\]\n\nUnder the substitutions\n\\[\nrotuvar=\\sqrt{\\frac{paramp}{rotcoefu}}\\,varscale, \\qquad rotvvar=\\sqrt{\\frac{paramp}{rotcoefv}}\\,varscalt\n\\]\nthis becomes\n\\[\n\\frac{1}{paramp\\,\\sqrt{rotcoefu\\,rotcoefv}}\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d varscale d varscalt}{\\left(1+varscale^{2}+varscalt^{2}\\right)^{2}}.\n\\]\n\nWe need only show that the value of the integral in (3) is \\pi. Passing to polar coordinates\n\\[\n\\begin{array}{l}\nvarscale=radiusr \\cos angleth \\\\\nvarscalt=radiusr \\sin angleth,\n\\end{array}\n\\]\nwe get\n\\[\n\\begin{aligned}\n\\int_{0}^{2 \\pi} \\int_{0}^{\\infty} \\frac{radiusr\\,d radiusr d angleth}{\\left(1+radiusr^{2}\\right)^{2}} &= 2 \\pi \\int_{0}^{\\infty} \\frac{radiusr d radiusr}{\\left(1+radiusr^{2}\\right)^{2}} \\\\\n&=2 \\pi\\left[\\frac{-1}{2\\left(1+radiusr^{2}\\right)}\\right]_{0}^{\\infty}=\\pi.\n\\end{aligned}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "banister",
        "k": "pinecone",
        "x": "starlight",
        "y": "lighthouse",
        "u": "waterfall",
        "v": "paperclip",
        "s": "toadstool",
        "t": "afterglow",
        "r": "chocolate",
        "\\theta": "\\navigation",
        "N_k": "cornflake",
        "T_k": "broomstick",
        "x_k": "partridge",
        "p": "moonstone",
        "a": "raincloud",
        "b": "peppermint",
        "c": "thunderous",
        "A": "driftwood",
        "C": "sugarplum",
        "\\alpha": "\\buttercup"
      },
      "question": "2. Answer either (i) or (ii):\n(i) Prove that\n\\[\n\\sum_{banister=2}^{\\infty} \\frac{\\cos (\\log \\log banister)}{\\log banister}\n\\]\ndiverges.\n(page 277)\n(ii) Assume that \\( moonstone>0, raincloud>0 \\), and \\( raincloud\\,thunderous-peppermint^{2}>0 \\), and show that\n\\[\n\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d starlight d lighthouse}{\\left(moonstone+raincloud starlight^{2}+2 peppermint starlight lighthouse+thunderous lighthouse^{2}\\right)^{2}}=\\pi moonstone^{-1}\\left(raincloud\\,thunderous-peppermint^{2}\\right)^{-12}\n\\]",
      "solution": "Solution. A convergent series cannot have blocks of terms whose sum is arbitrarily large. We shall show that the given series has such blocks.\n\nFor a positive integer \\( pinecone \\) consider the set \\( cornflake \\) of integers \\( banister \\) such that\n\\[\n2 \\pi pinecone-\\frac{1}{3} \\pi \\leq \\log \\log banister \\leq 2 \\pi pinecone\n\\]\nand let\n\\[\nbroomstick=\\sum_{cornflake} \\frac{\\cos (\\log \\log banister)}{\\log banister}\n\\]\n\nWe shall prove that \\( broomstick \\rightarrow \\infty \\) as \\( pinecone \\rightarrow \\infty \\). Now \\( cornflake=\\left\\{banister: \\exp \\exp \\left(2 \\pi pinecone-\\frac{1}{3} \\pi\\right)\\right. \\leq banister \\leq \\exp \\exp (2 \\pi pinecone)\\} \\) and therefore \\( \\left|cornflake\\right| \\), the number of elements in \\( cornflake \\), satisfies\n\\[\n\\left|cornflake\\right| \\geq \\exp \\exp 2 \\pi pinecone-\\exp (\\buttercup \\exp 2 \\pi pinecone)-1\n\\]\nwhere \\( \\buttercup=\\exp \\left(-\\frac{1}{3} \\pi\\right) \\).\nEach term in the sum (1) is at least\n\\[\n\\frac{\\cos \\left(-\\frac{1}{3} \\pi\\right)}{\\exp 2 \\pi pinecone}=\\frac{1}{2 \\exp 2 \\pi pinecone}\n\\]\nso\n\\[\nbroomstick \\geq \\frac{1}{2 partridge}\\left(\\exp \\left(partridge\\right)-\\exp \\left(\\buttercup partridge\\right)-1\\right)\n\\]\nwhere \\( partridge=\\exp 2 \\pi pinecone \\).\nNow\n\\[\n\\lim _{starlight \\rightarrow \\infty} \\frac{1}{starlight}(\\exp starlight-\\exp \\buttercup starlight)=\\lim _{starlight \\rightarrow \\infty}(\\exp starlight-\\buttercup \\exp \\buttercup starlight)=\\infty\n\\]\nby L'Hospital's rule, using the fact that \\( \\buttercup<1 \\). Since \\( partridge \\rightarrow \\infty \\) as \\( pinecone \\rightarrow \\infty \\), it follows that \\( broomstick \\rightarrow \\infty \\) as \\( pinecone \\rightarrow \\infty \\), and this proves that the given series diverges.\n\nSolution. By a rotation of axes,\n\\[\n\\begin{array}{c}\nstarlight=waterfall \\cos \\navigation+paperclip \\sin \\navigation \\\\\nlighthouse=-waterfall \\sin \\navigation+paperclip \\cos \\navigation\n\\end{array}\n\\]\nwith proper choice of the parameter \\( \\navigation \\), the quadratic form \\( raincloud starlight^{2}+2 peppermint starlight lighthouse+thunderous lighthouse^{2} \\) becomes\n\\[\ndriftwood waterfall^{2}+sugarplum paperclip^{2} .\n\\]\n\nThe discriminant \\( raincloud thunderous-peppermint^{\\mathbf{2}} \\) is an invariant under (1), so\n\\[\ndriftwood sugarplum=raincloud thunderous-peppermint^{2} .\n\\]\n\nSince the original form is positive definite, we also have \\( driftwood>0 \\) and \\( sugarplum>0 \\). The transformation is area-preserving, so the required integral in the problem becomes\n\\[\n\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d waterfall d paperclip}{\\left(moonstone+driftwood waterfall^{2}+sugarplum paperclip^{2}\\right)^{2}} .\n\\]\n\nUnder the substitutions\n\\[\nwaterfall=\\sqrt{\\frac{moonstone}{driftwood}} toadstool, \\quad paperclip=\\sqrt{\\frac{moonstone}{sugarplum}} afterglow\n\\]\nthis becomes\n\\[\n\\frac{1}{moonstone \\sqrt{driftwood sugarplum}} \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d toadstool d afterglow}{\\left(1+toadstool^{2}+afterglow^{2}\\right)^{2}}\n\\]\n\nWe need only show that the value of the integral in (3) is \\( \\pi \\). Passing to polar coordinates\n\\[\n\\begin{array}{l}\ntoadstool=chocolate \\cos \\navigation \\\\\nafterglow=chocolate \\sin \\navigation,\n\\end{array}\n\\]\nwe get\n\\[\n\\begin{aligned}\n\\int_{0}^{2 \\pi} \\int_{0}^{\\infty} \\frac{chocolate d chocolate d \\navigation}{\\left(1+chocolate^{2}\\right)^{2}} & =2 \\pi \\int_{0}^{\\infty} \\frac{chocolate d chocolate}{\\left(1+chocolate^{2}\\right)^{2}} \\\\\n& =2 \\pi\\left[\\frac{-1}{2\\left(1+chocolate^{2}\\right)}\\right]_{0}^{\\infty}=\\pi\n\\end{aligned}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "staticvalue",
        "k": "continuous",
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "u": "stationary",
        "v": "motionless",
        "s": "fixedpoint",
        "t": "timeless",
        "r": "centerpoint",
        "\\theta": "linearity",
        "N_k": "nullcollection",
        "T_k": "zeroproduct",
        "x_k": "equilibrium",
        "p": "rigidity",
        "a": "indifference",
        "b": "decoupled",
        "c": "simplicity",
        "A": "weakness",
        "C": "fragility",
        "\\alpha": "ultimateend"
      },
      "question": "2. Answer either (i) or (ii):\n(i) Prove that\n\\[\n\\sum_{staticvalue=2}^{\\infty} \\frac{\\cos (\\log \\log staticvalue)}{\\log staticvalue}\n\\]\ndiverges.\n(page 277)\n(ii) Assume that \\( rigidity>0, indifference>0 \\), and \\( indifference simplicity-decoupled^{2}>0 \\), and show that\n\\[\n\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d verticalaxis d horizontalaxis}{\\left(rigidity+indifference verticalaxis^{2}+2 decoupled verticalaxis horizontalaxis+simplicity horizontalaxis^{2}\\right)^{2}}=\\pi rigidity^{-1}\\left(indifference simplicity-decoupled^{2}\\right)^{-12}\n\\]",
      "solution": "Solution. A convergent series cannot have blocks of terms whose sum is arbitrarily large. We shall show that the given series has such blocks.\n\nFor a positive integer \\( continuous \\) consider the set \\( nullcollection \\) of integers \\( staticvalue \\) such that\n\\[\n2 \\pi continuous-\\frac{1}{3} \\pi \\leq \\log \\log staticvalue \\leq 2 \\pi continuous\n\\]\nand let\n\\[\nzeroproduct=\\sum_{nullcollection} \\frac{\\cos (\\log \\log staticvalue)}{\\log staticvalue}\n\\]\n\nWe shall prove that \\( zeroproduct \\rightarrow \\infty \\) as \\( continuous \\rightarrow \\infty \\). Now \\( nullcollection=\\left\\{staticvalue: \\exp \\exp \\left(2 \\pi continuous-\\frac{1}{3} \\pi\\right)\\right. \\leq staticvalue \\leq \\exp \\exp (2 \\pi continuous)\\} \\) and therefore \\( \\left|nullcollection\\right| \\), the number of elements in \\( nullcollection \\), satisfies\n\\[\n\\left|nullcollection\\right| \\geq \\exp \\exp 2 \\pi continuous-\\exp (ultimateend \\exp 2 \\pi continuous)-1\n\\]\nwhere \\( ultimateend=\\exp \\left(-\\frac{1}{3} \\pi\\right) \\).\nEach term in the sum (1) is at least\n\\[\n\\frac{\\cos \\left(-\\frac{1}{3} \\pi\\right)}{\\exp 2 \\pi continuous}=\\frac{1}{2 \\exp 2 \\pi continuous}\n\\]\nso\n\\[\nzeroproduct \\geq \\frac{1}{2 equilibrium}\\left(\\exp \\left(equilibrium\\right)-\\exp \\left(ultimateend equilibrium\\right)-1\\right)\n\\]\nwhere \\( equilibrium=\\exp 2 \\pi continuous \\).\nNow\n\\[\n\\lim _{verticalaxis \\rightarrow \\infty} \\frac{1}{verticalaxis}(\\exp verticalaxis-\\exp ultimateend verticalaxis)=\\lim _{verticalaxis \\rightarrow \\infty}(\\exp verticalaxis-ultimateend \\exp ultimateend verticalaxis)=\\infty\n\\]\nby L'Hospital's rule, using the fact that \\( ultimateend<1 \\). Since \\( equilibrium \\rightarrow \\infty \\) as \\( continuous \\rightarrow \\infty \\), it follows that \\( zeroproduct \\rightarrow \\infty \\) as \\( continuous \\rightarrow \\infty \\), and this proves that the given series diverges.\n\nSolution. By a rotation of axes,\n\\[\n\\begin{array}{c}\nverticalaxis=stationary \\cos linearity+motionless \\sin linearity \\\\\nhorizontalaxis=-stationary \\sin linearity+motionless \\cos linearity\n\\end{array}\n\\]\nwith proper choice of the parameter \\( linearity \\), the quadratic form \\( indifference verticalaxis^{2}+2 decoupled verticalaxis horizontalaxis+simplicity horizontalaxis^{2} \\) becomes\n\\[\nweakness stationary^{2}+fragility motionless^{2} .\n\\]\n\nThe discriminant \\( indifference simplicity-decoupled^{\\mathbf{2}} \\) is an invariant under (1), so\n\\[\nweakness fragility=indifference simplicity-decoupled^{2} .\n\\]\n\nSince the original form is positive definite, we also have \\( weakness>0 \\) and \\( fragility>0 \\). The transformation is area-preserving, so the required integral in the problem becomes\n\\[\n\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d stationary d motionless}{\\left(rigidity+weakness stationary^{2}+fragility motionless^{2}\\right)^{2}} .\n\\]\n\nUnder the substitutions\n\\[\nstationary=\\sqrt{\\frac{rigidity}{weakness}} fixedpoint, \\quad motionless=\\sqrt{\\frac{rigidity}{fragility}} timeless\n\\]\nthis becomes\n\\[\n\\frac{1}{rigidity \\sqrt{weakness fragility}} \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d fixedpoint d timeless}{\\left(1+fixedpoint^{2}+timeless^{2}\\right)^{2}}\n\\]\n\nWe need only show that the value of the integral in (3) is \\( \\pi \\). Passing to polar coordinates\n\\[\n\\begin{array}{l}\nfixedpoint=centerpoint \\cos linearity \\\\\ntimeless=centerpoint \\sin linearity,\n\\end{array}\n\\]\nwe get\n\\[\n\\begin{aligned}\n\\int_{0}^{2 \\pi} \\int_{0}^{\\infty} \\frac{centerpoint d centerpoint d linearity}{\\left(1+centerpoint^{2}\\right)^{2}} & =2 \\pi \\int_{0}^{\\infty} \\frac{centerpoint d centerpoint}{\\left(1+centerpoint^{2}\\right)^{2}} \\\\\n& =2 \\pi\\left[\\frac{-1}{2\\left(1+centerpoint^{2}\\right)}\\right]_{0}^{\\infty}=\\pi\n\\end{aligned}\n\\]"
    },
    "garbled_string": {
      "map": {
        "n": "jxqvmndt",
        "k": "zploqsru",
        "x": "frblwteu",
        "y": "akdzmnya",
        "u": "gkvsheqc",
        "v": "ncrdpyam",
        "s": "fdytsovl",
        "t": "wqnclzra",
        "r": "jkagmntp",
        "\\theta": "hzwyrbev",
        "N_k": "zmvhwqrt",
        "T_k": "ytrvqpsd",
        "x_k": "bfmglzrc",
        "p": "orlkapgi",
        "a": "egrvotum",
        "b": "clnyaszi",
        "c": "whsojbex",
        "A": "xtafycnb",
        "C": "vzpxqlir",
        "\\alpha": "pmurzcha"
      },
      "question": "2. Answer either (i) or (ii):\n(i) Prove that\n\\[\n\\sum_{jxqvmndt=2}^{\\infty} \\frac{\\cos (\\log \\log jxqvmndt)}{\\log jxqvmndt}\n\\]\ndiverges.\n(page 277)\n(ii) Assume that \\( orlkapgi>0, egrvotum>0 \\), and \\( egrvotum whsojbex-clnyaszi^{2}>0 \\), and show that\n\\[\n\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d frblwteu d akdzmnya}{\\left(orlkapgi+egrvotum frblwteu^{2}+2 clnyaszi frblwteu akdzmnya+whsojbex akdzmnya^{2}\\right)^{2}}=\\pi orlkapgi^{-1}\\left(egrvotum whsojbex-clnyaszi^{2}\\right)^{-12}\n\\]",
      "solution": "Solution. A convergent series cannot have blocks of terms whose sum is arbitrarily large. We shall show that the given series has such blocks.\n\nFor a positive integer \\( zploqsru \\) consider the set \\( zmvhwqrt \\) of integers \\( jxqvmndt \\) such that\n\\[\n2 \\pi zploqsru-\\frac{1}{3} \\pi \\leq \\log \\log jxqvmndt \\leq 2 \\pi zploqsru\n\\]\nand let\n\\[\nytrvqpsd=\\sum_{zmvhwqrt} \\frac{\\cos (\\log \\log jxqvmndt)}{\\log jxqvmndt}\n\\]\n\nWe shall prove that \\( ytrvqpsd \\rightarrow \\infty \\) as \\( zploqsru \\rightarrow \\infty \\). Now \\( zmvhwqrt=\\{jxqvmndt: \\exp \\exp (2 \\pi zploqsru-\\frac{1}{3} \\pi) \\leq jxqvmndt \\leq \\exp \\exp (2 \\pi zploqsru)\\} \\) and therefore \\( |zmvhwqrt| \\), the number of elements in \\( zmvhwqrt \\), satisfies\n\\[\n|zmvhwqrt| \\geq \\exp \\exp 2 \\pi zploqsru-\\exp (pmurzcha \\exp 2 \\pi zploqsru)-1\n\\]\nwhere \\( pmurzcha=\\exp \\left(-\\frac{1}{3} \\pi\\right) \\).\nEach term in the sum (1) is at least\n\\[\n\\frac{\\cos \\left(-\\frac{1}{3} \\pi\\right)}{\\exp 2 \\pi zploqsru}=\\frac{1}{2 \\exp 2 \\pi zploqsru}\n\\]\nso\n\\[\nytrvqpsd \\geq \\frac{1}{2 bfmglzrc}\\left(\\exp (bfmglzrc)-\\exp (pmurzcha bfmglzrc)-1\\right)\n\\]\nwhere \\( bfmglzrc=\\exp 2 \\pi zploqsru \\).\nNow\n\\[\n\\lim_{frblwteu \\to \\infty} \\frac{1}{frblwteu}\\bigl(\\exp frblwteu-\\exp pmurzcha frblwteu\\bigr)=\\lim_{frblwteu \\to \\infty}\\bigl(\\exp frblwteu-pmurzcha \\exp pmurzcha frblwteu\\bigr)=\\infty\n\\]\nby L'Hospital's rule, using the fact that \\( pmurzcha<1 \\). Since \\( bfmglzrc \\rightarrow \\infty \\) as \\( zploqsru \\rightarrow \\infty \\), it follows that \\( ytrvqpsd \\rightarrow \\infty \\) as \\( zploqsru \\rightarrow \\infty \\), and this proves that the given series diverges.\n\nSolution. By a rotation of axes,\n\\[\n\\begin{array}{c}\nfrblwteu=gkvsheqc \\cos hzwyrbev+ncrdpyam \\sin hzwyrbev \\\\\nakdzmnya=-gkvsheqc \\sin hzwyrbev+ncrdpyam \\cos hzwyrbev\n\\end{array}\n\\]\nwith proper choice of the parameter \\( hzwyrbev \\), the quadratic form \\( egrvotum frblwteu^{2}+2 clnyaszi frblwteu akdzmnya+whsojbex akdzmnya^{2} \\) becomes\n\\[\nxtafycnb gkvsheqc^{2}+vzpxqlir ncrdpyam^{2} .\n\\]\n\nThe discriminant \\( egrvotum whsojbex-clnyaszi^{2} \\) is an invariant under (1), so\n\\[\nxtafycnb vzpxqlir=egrvotum whsojbex-clnyaszi^{2} .\n\\]\n\nSince the original form is positive definite, we also have \\( xtafycnb>0 \\) and \\( vzpxqlir>0 \\). The transformation is area-preserving, so the required integral in the problem becomes\n\\[\n\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d gkvsheqc d ncrdpyam}{\\left(orlkapgi+xtafycnb gkvsheqc^{2}+vzpxqlir ncrdpyam^{2}\\right)^{2}} .\n\\]\n\nUnder the substitutions\n\\[\ngkvsheqc=\\sqrt{\\frac{orlkapgi}{xtafycnb}} fdytsovl, \\quad ncrdpyam=\\sqrt{\\frac{orlkapgi}{vzpxqlir}} wqnclzra\\]\nthis becomes\n\\[\n\\frac{1}{orlkapgi \\sqrt{xtafycnb vzpxqlir}} \\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} \\frac{d fdytsovl d wqnclzra}{\\left(1+fdytsovl^{2}+wqnclzra^{2}\\right)^{2}}\n\\]\n\nWe need only show that the value of the integral in (3) is \\( \\pi \\). Passing to polar coordinates\n\\[\n\\begin{array}{l}\nfdytsovl=jkagmntp \\cos hzwyrbev \\\\\nwqnclzra=jkagmntp \\sin hzwyrbev,\n\\end{array}\n\\]\nwe get\n\\[\n\\begin{aligned}\n\\int_{0}^{2 \\pi} \\int_{0}^{\\infty} \\frac{jkagmntp d jkagmntp d hzwyrbev}{\\left(1+jkagmntp^{2}\\right)^{2}} &=2 \\pi \\int_{0}^{\\infty} \\frac{jkagmntp d jkagmntp}{\\left(1+jkagmntp^{2}\\right)^{2}} \\\\\n&=2 \\pi\\left[\\frac{-1}{2\\left(1+jkagmntp^{2}\\right)}\\right]_{0}^{\\infty}=\\pi\n\\end{aligned}\n\\]"
    },
    "kernel_variant": {
      "question": "Answer either part $\\text{(i)}$ or part $\\text{(ii)}$.\n\n(i) (Primes and almost–periodic phases)  \nLet $\\beta\\in\\mathbb R$.  Show that the series  \n\\[\n\\tag{$\\star$}\\label{star}\n\\sum_{\\substack{p\\ \\text{prime}\\\\ p\\ge 3}}\n\\frac{\\cos\\!\\bigl(\\beta+\\log\\log p\\bigr)}{\\log p}\n\\]\nis divergent.  Moreover prove that its sequence of partial sums is\nunbounded both above and below.\n\n(ii) (Higher–dimensional quadratic–form integral)  \nLet  \n\\[\nn\\ge 2,\\qquad m>\\frac n 2,\\qquad q>0,\n\\]\nand let $M$ be an $n\\times n$ real symmetric positive–definite matrix\nwith $\\det M>0$.  Prove the evaluation\n\\[\n\\int_{\\mathbb R^{n}}\n\\frac{d^{n}x}{\\bigl(q+x^{\\mathsf T} M x\\bigr)^{m}}\n=\\frac{\\pi^{\\,n/2}\\,\\Gamma\\!\\bigl(m-\\tfrac n 2\\bigr)}\n      {\\Gamma(m)\\,q^{\\,m-n/2}\\sqrt{\\det M}}.\n\\]\n\n────────────────────────────────────────────────────────",
      "solution": "We give complete proofs of both parts.  Part $\\text{(ii)}$ is unchanged;\npart $\\text{(i)}$ now supplies the missing uniform prime–counting bound\nand a rigorous domination argument in the final block selection.\n\n--------------------------------------------------------\nSolution to part $\\text{(i)}$\n\nFix $\\beta\\in\\mathbb R$ and abbreviate  \n\\[\na_{p}:=\\frac{\\cos\\!\\bigl(\\beta+\\log\\log p\\bigr)}{\\log p},\\qquad\nS(N):=\\sum_{\\substack{3\\le p\\le N\\\\ p\\ \\text{prime}}}a_{p}\\qquad(N\\ge 3).\n\\]\n\nWe prove\n\n(A)  $\\bigl(S(N)\\bigr)_{N\\ge 3}$ is not Cauchy, hence \\eqref{star} diverges.\n\n(B)  $\\sup_{N\\ge 3}S(N)=+\\infty$ and $\\inf_{N\\ge 3}S(N)=-\\infty$.\n\nThe proof proceeds in four steps.\n\n--------------------------------------------------\nStep 1.  Windows where $\\cos(\\beta+\\theta)$ has fixed sign.\n\nChoose $\\delta\\in(0,\\pi/6)$ and set  \n\\[\nI_{k}^{+}:=\\bigl\\{\\theta\\in\\mathbb R:\\ |\\beta+\\theta-2\\pi k|<\\delta\\bigr\\},\n\\quad\nI_{k}^{-}:=\\bigl\\{\\theta\\in\\mathbb R:\\ |\\beta+\\theta-(\\pi+2\\pi k)|<\\delta\\bigr\\},\n\\qquad k\\in\\mathbb N.\n\\]\nBecause $|\\theta|<\\delta$ implies\n$\\cos\\theta\\ge\\cos\\delta=:c_{0}>1/2$, we have\n\\[\n\\cos(\\beta+\\theta)\\ge c_{0}\\quad(\\theta\\in I_{k}^{+}),\\qquad\n\\cos(\\beta+\\theta)\\le -c_{0}\\quad(\\theta\\in I_{k}^{-}).\n\\tag{1}\\label{cosBounds}\n\\]\n\n--------------------------------------------------\nStep 2.  Lifting the windows to intervals of primes.\n\nDefine  \n\\[\n\\begin{aligned}\nL_{k}^{+}&:=\\exp\\!\\bigl(\\exp(2\\pi k-\\beta-\\delta)\\bigr),&\nU_{k}^{+}&:=\\exp\\!\\bigl(\\exp(2\\pi k-\\beta+\\delta)\\bigr),\\\\[4pt]\nL_{k}^{-}&:=\\exp\\!\\bigl(\\exp(\\pi+2\\pi k-\\beta-\\delta)\\bigr),&\nU_{k}^{-}&:=\\exp\\!\\bigl(\\exp(\\pi+2\\pi k-\\beta+\\delta)\\bigr),\n\\end{aligned}\n\\qquad k\\in\\mathbb N.\n\\]\nIf $p\\in[L_{k}^{\\pm},U_{k}^{\\pm}]$ then\n$\\log\\log p\\in I_{k}^{\\pm}$, hence \\eqref{cosBounds} applies.\nPut\n\\[\n\\mathcal P_{k}^{\\pm}:=\\bigl\\{p\\ \\text{prime}:\\ L_{k}^{\\pm}\\le p\\le U_{k}^{\\pm}\\bigr\\}.\n\\]\n\n--------------------------------------------------\nStep 3.  Quantitative information on each block.\n\nFrom now on we restrict to $k\\ge k_{0}(\\beta,\\delta)$ so large that\n$L_{k}^{\\pm}>3$.  Discarding finitely many initial blocks has no effect\non divergence or unboundedness.\n\n3.1  Length of the interval.  \nSince $e^{\\delta}-e^{-\\delta}\\ge\\delta$ and\n$\\theta_{k}^{\\pm}:=2\\pi k-\\beta$ or $\\pi+2\\pi k-\\beta\\ge 1$ for\n$k\\ge k_{0}$, we have\n\\[\n\\frac{L_{k}^{\\pm}}{U_{k}^{\\pm}}\n=\\exp\\!\\Bigl(-\\exp(\\theta_{k}^{\\pm})\\bigl(e^{\\delta}-e^{-\\delta}\\bigr)\\Bigr)\n\\le\\exp\\!\\bigl(-\\exp(\\theta_{k}^{\\pm})\\delta\\bigr)\\le e^{-\\delta}<1.\n\\]\nHence\n\\[\nU_{k}^{\\pm}-L_{k}^{\\pm}=U_{k}^{\\pm}\\Bigl(1-\\frac{L_{k}^{\\pm}}{U_{k}^{\\pm}}\\Bigr)\n\\ge\\bigl(1-e^{-\\delta}\\bigr)U_{k}^{\\pm}=c_{1}\\,U_{k}^{\\pm}\n\\tag{2}\\label{length}\n\\]\nwith $c_{1}=1-e^{-\\delta}\\in(0,1)$.\n\n3.2  How many primes lie between $L_{k}^{\\pm}$ and $U_{k}^{\\pm}$?\n\nWe use the quantitative Prime Number Theorem: there exists $c>0$ such that  \n\\[\n\\pi(x)=\\operatorname{Li}(x)+O\\!\\bigl(x\\exp(-c\\sqrt{\\log x})\\bigr)\n\\qquad(x\\to\\infty).\n\\]\nLet $L<L^{\\ast}<U$ with $U\\le 2L$.  Writing the above formula for $L$\nand $U$ we obtain\n\\[\n\\pi(U)-\\pi(L)\n  =\\int_{L}^{U}\\frac{dt}{\\log t}\n   +O\\!\\bigl(U\\exp(-c\\sqrt{\\log U})\\bigr)\n   +O\\!\\bigl(L\\exp(-c\\sqrt{\\log L})\\bigr).\n\\]\nBecause $U-L\\ge c_{1}U$ by \\eqref{length} and $\\log t\\le\\log U$ in the\nrange of integration,\n\\[\n\\int_{L}^{U}\\frac{dt}{\\log t}\\ge\\frac{U-L}{\\log U}\n\\ge\\frac{c_{1}U}{\\log U}.\n\\]\nThe error term is $\\ll U/(\\log U)^{2}$, so for all sufficiently large\n$U$\n\\[\n\\pi(U)-\\pi(L)\\ge\\frac{c_{1}}{2}\\frac{U}{\\log U}.\n\\]\nApplying this with $(L,U)=(L_{k}^{\\pm},U_{k}^{\\pm})$ gives\n\\[\n|\\mathcal P_{k}^{\\pm}|\n=\\pi\\!\\bigl(U_{k}^{\\pm}\\bigr)-\\pi\\!\\bigl(L_{k}^{\\pm}\\bigr)\n\\ge c_{2}\\,\\frac{U_{k}^{\\pm}}{\\log U_{k}^{\\pm}}\n\\tag{3}\\label{primeCount}\n\\]\nfor some $c_{2}=c_{2}(\\delta)>0$ and all $k\\ge k_{0}$.\n\n3.3  A uniform bound on $1/\\log p$.  \nFor $p\\in\\mathcal P_{k}^{\\pm}$ we have\n$p\\le U_{k}^{\\pm}$, hence\n\\[\n\\frac1{\\log p}\\ge\\frac1{\\log U_{k}^{\\pm}}.\n\\tag{4}\\label{logBound}\n\\]\n\n3.4  Total contribution of one block.  \nCombining \\eqref{cosBounds}, \\eqref{primeCount} and \\eqref{logBound},\n\\[\n\\begin{aligned}\nS_{k}^{+}:=\\sum_{p\\in\\mathcal P_{k}^{+}}a_{p}\n&\\ge c_{0}c_{2}\\,\\frac{U_{k}^{+}}{(\\log U_{k}^{+})^{2}}\n   =:C_{k}^{+},\\\\[4pt]\nS_{k}^{-}:=\\sum_{p\\in\\mathcal P_{k}^{-}}a_{p}\n&\\le -c_{0}c_{2}\\,\\frac{U_{k}^{-}}{(\\log U_{k}^{-})^{2}}\n   =:C_{k}^{-}.\n\\end{aligned}\n\\tag{5}\\label{blockContribution}\n\\]\nBecause $U_{k}^{\\pm}\n=\\exp\\!\\bigl(\\exp(\\theta_{k}^{\\pm}+\\delta)\\bigr)$ grows\ndouble–exponentially with $k$,  \n\\[\n|C_{k}^{\\pm}|\\xrightarrow[k\\to\\infty]{}\\infty.\n\\tag{6}\\label{blockGrowth}\n\\]\n\n--------------------------------------------------\nStep 4.  Choosing blocks that dominate \\emph{all} previous contributions.\n\nLet  \n\\[\nT_{k}:=\\sup_{N<L_{k}^{+}}|S(N)|\n      =\\sup_{3\\le N<L_{k}^{+}}\\Bigl|\\sum_{\\substack{3\\le p\\le N}}a_{p}\\Bigr|.\n\\]\nWe first claim that\n\\[\nT_{k}\\ll\\frac{U_{k-1}^{+}}{\\bigl(\\log U_{k-1}^{+}\\bigr)^{2}}\n\\qquad(k\\ge k_{0}+1).\n\\tag{7}\\label{gapBound}\n\\]\nIndeed, break the range $3\\le p<L_{k}^{+}$ into \\emph{completed blocks}\n$\\mathcal P_{i}^{\\pm}$ with $i<k$ and the intermediate gaps\n$\\bigl(U_{i}^{\\pm},L_{i+1}^{\\pm}\\bigr)$.  The sum inside each completed\nblock is exactly $C_{i}^{\\pm}$, while inside a gap the absolute value of\neach term is at most $1/\\log p\\le1/\\log(U_{i+1}^{+})$ and there are\n$O\\!\\bigl(U_{i+1}^{+}/\\log U_{i+1}^{+}\\bigr)$ primes in the gap, so the\ntotal contribution of that gap is $O\\!\\bigl(U_{i+1}^{+}/(\\log\nU_{i+1}^{+})^{2}\\bigr)$.  Because $U_{i+1}^{+}$ increases with $i$, the\nlargest contribution arises when $i=k-2$; this yields \\eqref{gapBound}.\n\nNow construct inductively an increasing sequence\n$k_{1}<k_{2}<k_{3}<\\dots$ with $k_{1}\\ge k_{0}+1$ such that\n\\[\n|C_{k_{j}}^{+}|\\ >\\ T_{k_{j}}+|C_{k_{j}}^{-}|\n                   +\\sum_{i<j}\\bigl(|C_{k_{i}}^{+}|+|C_{k_{i}}^{-}|\\bigr),\n\\tag{8}\\label{dominate}\n\\]\nand similarly (for later use)\n\\[\n|C_{k_{j}}^{-}|\\ >\\ T_{k_{j}}+|C_{k_{j}}^{+}|\n                   +\\sum_{i<j}\\bigl(|C_{k_{i}}^{+}|+|C_{k_{i}}^{-}|\\bigr).\n\\tag{9}\n\\]\nThis is possible because of the double–exponential growth\n\\eqref{blockGrowth}: $|C_{k}^{\\pm}|$ eventually dominates both\n$T_{k}$ and the finite sum of earlier $|C_{i}^{\\pm}|$.\n\nConsider the partial sum at the end of a positive block,\n$N:=U_{k_{j}}^{+}$.  Using \\eqref{dominate},\n\\[\nS(N)=S\\!\\bigl(L_{k_{j}}^{+}\\bigr)+S_{k_{j}}^{+}\n     \\ge -T_{k_{j}}+|C_{k_{j}}^{+}|\n       -|C_{k_{j}}^{-}|\n       -\\sum_{i<j}\\bigl(|C_{k_{i}}^{+}|+|C_{k_{i}}^{-}|\\bigr)\n     >0.\n\\]\nFurthermore the last expression tends to $+\\infty$ as $j\\to\\infty$,\nhence $\\sup_{N}S(N)=+\\infty$.\nThe same reasoning with \\eqref{9} and $N:=U_{k_{j}}^{-}$ gives\n$\\inf_{N}S(N)=-\\infty$.\n\nUnboundedness on both sides precludes Cauchy convergence, completing the\nproof of (A)–(B).  \\hfill$\\square$\n\n--------------------------------------------------------\nSolution to part $\\text{(ii)}$  (unchanged)\n\nLet\n\\[\nI_{n,m}(q,M):=\\int_{\\mathbb R^{n}}\n                 \\frac{d^{n}x}{\\bigl(q+x^{\\mathsf T} M x\\bigr)^{m}}.\n\\]\n\nStep 1.  Orthogonal diagonalisation.  \nWrite $M=U^{\\mathsf T}DU$ with $U$ orthogonal and\n$D=\\operatorname{diag}(\\lambda_{1},\\dots,\\lambda_{n})$, $\\lambda_{j}>0$.\nSet $y=Ux$; then $d^{n}x=d^{n}y$ and\n$x^{\\mathsf T}Mx=y^{\\mathsf T}Dy$.\n\nStep 2.  Isotropic scaling.  \nDefine $z_{j}:=\\sqrt{\\lambda_{j}}\\,y_{j}$.\nThen $d^{n}y=(\\det M)^{-1/2}d^{n}z$ and\n$y^{\\mathsf T}Dy=\\lVert z\\rVert^{2}$.  Consequently\n\\[\nI_{n,m}(q,M)=\\frac1{\\sqrt{\\det M}}\n             \\int_{\\mathbb R^{n}}\\frac{d^{n}z}{\\bigl(q+\\lVert z\\rVert^{2}\\bigr)^{m}}\n             =\\frac{J_{n,m}(q)}{\\sqrt{\\det M}},\n\\]\nwhere $J_{n,m}(q)$ denotes the isotropic integral.\n\nStep 3.  Polar coordinates in $\\mathbb R^{n}$.  \nWith $\\omega_{n}:=2\\pi^{n/2}/\\Gamma(n/2)$ we obtain\n\\[\nJ_{n,m}(q)=\n\\omega_{n}\\int_{0}^{\\infty}\\frac{r^{\\,n-1}}{(q+r^{2})^{m}}\\;dr.\n\\]\n\nStep 4.  Beta–Gamma evaluation.  \nSubstitute $r=\\sqrt{q}\\,t$ and then $u=t^{2}$:\n\\[\n\\int_{0}^{\\infty}\\frac{r^{\\,n-1}}{(q+r^{2})^{m}}\\;dr\n=\\tfrac12\\,q^{\\,n/2-m}\\int_{0}^{\\infty}\\frac{u^{\\,n/2-1}}{(1+u)^{m}}\\,du\n=\\tfrac12\\,q^{\\,n/2-m}\\,B\\!\\Bigl(\\tfrac n 2,\\;m-\\tfrac n 2\\Bigr),\n\\]\nwhich converges because $m>n/2$ and equals\n\\[\n\\tfrac12\\,q^{\\,n/2-m}\\,\n\\frac{\\Gamma(n/2)\\,\\Gamma(m-n/2)}{\\Gamma(m)}.\n\\]\n\nStep 5.  Collecting constants.  \nMultiplying by $\\omega_{n}$ yields\n\\[\nJ_{n,m}(q)=\n\\pi^{\\,n/2}\\,\n\\frac{\\Gamma(m-n/2)}{\\Gamma(m)\\,q^{\\,m-n/2}},\n\\]\nand therefore\n\\[\nI_{n,m}(q,M)=\n\\frac{\\pi^{\\,n/2}\\,\\Gamma(m-n/2)}\n     {\\Gamma(m)\\,q^{\\,m-n/2}\\sqrt{\\det M}},\n\\]\nas required. \\hfill$\\square$\n\n────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.426323",
        "was_fixed": false,
        "difficulty_analysis": "1.  Series over primes (part (i))  \n   •  Requires quantitative versions of the Prime Number Theorem to guarantee enough primes in exponentially narrow intervals.  \n   •  Needs uniform‐distribution reasoning combined with trigonometric lower bounds.  \n   •  Forces the solver to control both magnitude and sign, proving unbounded oscillation, not mere divergence.  \n   These ingredients are far beyond the elementary “block-sum’’ argument sufficient for the original series over all integers.\n\n2.  n–dimensional integral (part (ii))  \n   •  Generalises the 2-dimensional, fixed-power case to arbitrary dimension n and arbitrary exponent m.  \n   •  Necessitates linear-algebraic diagonalisation of a symmetric matrix, an orthogonal change of variables, careful Jacobian tracking, polar co-ordinates in n dimensions, and Beta/Gamma function identities.  \n   •  The constraint m>n/2 (to ensure convergence) must be recognised and justified.  \n   The original problem is a special case n=2, m=2; here every parameter is variable, and the solution interweaves linear algebra, multivariable calculus, and special-function theory.\n\nOverall the enhanced kernel introduces substantially deeper theory, more variables, higher-dimensional geometry, and analytic-number-theoretic arguments, making it significantly harder than both the original problem and the previous kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Answer either part (i) or part (ii).\n\n(i)  (Primes and uniform distribution)   \nLet β be an arbitrary real number.  Show that the series taken over the odd prime numbers  \n\n                              \n  ∑_{p prime ≥ 3}  cos(β + log log p) ⁄ log p        (★)\n\ndiverges.  Moreover prove that the sequence of its partial sums is unbounded both above and below.\n\n(ii)  (Higher–dimensional rational integral with general exponent)   \nLet  \n\n• n ≥ 2 be an integer,   \n• m > n⁄2 a real number,   \n• q > 0, and   \n• M an n × n real, symmetric, positive–definite matrix (det M > 0).   \n\nProve the evaluation  \n\n  ∫_{ℝⁿ} dⁿx ⁄ (q + xᵀMx)ᵐ  \n       = π^{n⁄2} Γ(m – n⁄2) ⁄ ( Γ(m) q^{\\,m – n⁄2}√{det M} ).\n\n————————————————————————————————————",
      "solution": "We give complete proofs of both parts.  The proof of part (ii) is unchanged; the proof of part (i) is repaired and strengthened.\n\n────────────────────────────────────────\nSOLUTION TO (i)\n\nFix the real number β.  Write  \n\n  S := ∑_{p odd prime} a_p,   a_p := cos(β+log log p) ⁄ log p .\n\nWe establish two statements:\n\n(A) S diverges (its partial sums are not Cauchy);  \n(B) its partial sums are unbounded above and below.\n\nThe argument proceeds in four steps.\n\nStep 1.  Building “cosine—friendly’’ windows that work for every β.  \nChoose a small number 0<δ<π⁄6 and put\n\n  I_k^+ := {θ∈ℝ : |β+θ – 2πk| < δ},   k = 1,2,3,…  \n  I_k^- := {θ∈ℝ : |β+θ – (π+2πk)| < δ}, k = 1,2,3,…\n\nBecause |cos θ| ≥ cos δ > 1⁄2 whenever |θ|<δ, we have the uniform bounds\n\n  cos(β+θ) ≥ c := cos δ > ½  for θ∈I_k^+,  (1)  \n  cos(β+θ) ≤ –c                for θ∈I_k^–. (2)\n\nStep 2.  Translating the windows to sets of primes.  \nDefine for k≥1\n\n  L_k^+ := exp exp(2πk – β – δ),  U_k^+ := exp exp(2πk – β + δ),  \n  L_k^- := exp exp(π+2πk – β – δ), U_k^- := exp exp(π+2πk – β + δ),\n\nand  \n\n  𝒫_k^+ := {p prime : L_k^+ ≤ p ≤ U_k^+},   \n  𝒫_k^- := {p prime : L_k^- ≤ p ≤ U_k^-}.\n\nBy construction, log log p lies in I_k^+ (resp. I_k^–) whenever p∈𝒫_k^+ (resp. 𝒫_k^–); hence (1)–(2) apply.\n\nStep 3.  Quantitative size of every block.  \nThroughout the proof “c, C, …’’ denote positive constants that depend only on δ.\n\nPrime number theorem with remainder.  \nThe quantitative form π(x)=Li(x)+O( x e^{–√{log x}} ) (or any weaker remainder) yields\n\n  |𝒫_k^{±}| = (1+o(1))·(U_k^{±}–L_k^{±})⁄log U_k^{±}  (k→∞).  (3)\n\nBecause exp is monotone and e^{δ}–e^{–δ}≈2δ, we have\n\n  U_k^{±} – L_k^{±} ≥ C·U_k^{±}.                                     (4)\n\nHence, inserting (4) into (3),\n\n  |𝒫_k^{±}| ≥ C·U_k^{±} ⁄ log U_k^{±}.                             (5)\n\nEvery p in 𝒫_k^{±} satisfies log p ∼ exp exp(const·k) = exp(c₂k), so for those p\n\n  1⁄log p ≥ C⁄log U_k^{±}.                                           (6)\n\nBlock contribution.  Combining (1)–(2) and (5)–(6),\n\n  S_k^{+} := ∑_{p∈𝒫_k^{+}} a_p  \n     ≥ c·|𝒫_k^{+}|⁄log U_k^{+}  \n     ≥ C·U_k^{+} ⁄ (log U_k^{+})²,                          (7)\n\n  S_k^{–} := ∑_{p∈𝒫_k^{–}} a_p  \n     ≤ –C·U_k^{–} ⁄ (log U_k^{–})².                        (8)\n\nBecause U_k^{±}=exp exp(θ_k^{±}) with θ_k^{±} linear in k, each ratio on the right–hand\nside grows double–exponentially in k.  In particular\n\n  |S_k^{±}| → ∞ as k→∞.                                            (9)\n\nStep 4.  Divergence and two–sided unboundedness.  \nWrite the overall partial sums as\n\n  P_N := ∑_{p≤N} a_p.\n\nChoose an increasing sequence k₁<k₂<k₃<⋯ so that the intervals [L_{k_j}^+,U_{k_j}^+] and\n[L_{k_j}^–,U_{k_j}^–] are pairwise disjoint and satisfy U_{k_j}^± < L_{k_{j+1}}^±.\nBecause the blocks are disjoint, we can extend N through these intervals so that\n\n  P_{U_{k_j}^+} ≥ P_{L_{k_j}^+}+S_{k_j}^{+}\n         ≥ P_{L_{k_j}^+}+|S_{k_j}^{+}| → +∞ as j→∞,    (10)\n\nand similarly\n\n  P_{U_{k_j}^-} ≤ P_{L_{k_j}^-}+S_{k_j}^{–}\n         ≤ P_{L_{k_j}^-}–|S_{k_j}^{–}| → –∞ as j→∞.    (11)\n\nHence the set {P_N : N≥2} is unbounded above and below, and the series (★) cannot converge.  Statements (A) and (B) are proved. ∎\n\n\n\n────────────────────────────────────────\nSOLUTION TO (ii) (unchanged)\n\nSet  \n\n I_{n,m}(q,M) := ∫_{ℝⁿ} dⁿx ⁄ (q+xᵀMx)ᵐ.\n\nStep 1.  Orthogonal diagonalisation.  \nSince M is real symmetric positive–definite, write M=UᵀDU with U orthogonal and D=diag(λ₁,…,λ_n), λ_j>0.  Substitute y=Ux; then dⁿx=dⁿy and xᵀMx=yᵀDy.\n\nStep 2.  Isotropic scaling.  \nPut z_j:=√{λ_j} y_j.  Then dⁿy=(det M)^{–½}dⁿz and yᵀDy=∑z_j²=‖z‖².  Hence\n\n I_{n,m}(q,M)=det(M)^{–½} J_{n,m}(q),  \n J_{n,m}(q):=∫_{ℝⁿ} dⁿz ⁄ (q+‖z‖²)ᵐ.\n\nStep 3.  Polar coordinates in ℝⁿ.  \nWith ω_n:=2π^{n⁄2}/Γ(n⁄2),\n\n J_{n,m}(q)=ω_n ∫₀^∞ r^{n–1}/(q+r²)ᵐ dr.\n\nStep 4.  Beta–Gamma evaluation.  \nPut r=√q t and then u=t²:\n\n ∫₀^∞ r^{n–1}/(q+r²)ᵐ dr  \n  =½ q^{n⁄2–m} ∫₀^∞ u^{n⁄2–1}/(1+u)ᵐ du  \n  =½ q^{n⁄2–m} B(n⁄2, m–n⁄2)  \n  =½ q^{n⁄2–m} Γ(n⁄2)Γ(m–n⁄2)/Γ(m).\n\nStep 5.  Collect constants.  \nMultiplying by ω_n gives\n\n J_{n,m}(q)=π^{n⁄2} Γ(m–n⁄2)/( Γ(m) q^{\\,m–n⁄2} ),\n\nso\n\n I_{n,m}(q,M)=π^{n⁄2} Γ(m–n⁄2)/( Γ(m) q^{\\,m–n⁄2}√{det M} ),\n\nas required. ∎\n\n————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.370254",
        "was_fixed": false,
        "difficulty_analysis": "1.  Series over primes (part (i))  \n   •  Requires quantitative versions of the Prime Number Theorem to guarantee enough primes in exponentially narrow intervals.  \n   •  Needs uniform‐distribution reasoning combined with trigonometric lower bounds.  \n   •  Forces the solver to control both magnitude and sign, proving unbounded oscillation, not mere divergence.  \n   These ingredients are far beyond the elementary “block-sum’’ argument sufficient for the original series over all integers.\n\n2.  n–dimensional integral (part (ii))  \n   •  Generalises the 2-dimensional, fixed-power case to arbitrary dimension n and arbitrary exponent m.  \n   •  Necessitates linear-algebraic diagonalisation of a symmetric matrix, an orthogonal change of variables, careful Jacobian tracking, polar co-ordinates in n dimensions, and Beta/Gamma function identities.  \n   •  The constraint m>n/2 (to ensure convergence) must be recognised and justified.  \n   The original problem is a special case n=2, m=2; here every parameter is variable, and the solution interweaves linear algebra, multivariable calculus, and special-function theory.\n\nOverall the enhanced kernel introduces substantially deeper theory, more variables, higher-dimensional geometry, and analytic-number-theoretic arguments, making it significantly harder than both the original problem and the previous kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}