{
  "index": "2021-A-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let\n\\[\nI(R) = \\iint_{x^2+y^2 \\leq R^2} \\left( \\frac{1+2x^2}{1+x^4+6x^2y^2+y^4} - \\frac{1+y^2}{2+x^4+y^4} \\right)\\,dx\\,dy.\n\\]\nFind\n\\[\n\\lim_{R \\to \\infty} I(R),\n\\]\nor show that this limit does not exist.",
  "solution": "The limit exists and equals $\\frac{\\sqrt{2}}{2} \\pi \\log 2$.\n\nWe first note that we can interchange $x$ and $y$ to obtain\n\\[\nI(R) = \\iint_{x^2+y^2 \\leq R^2} \\left( \\frac{1+2y^2}{1+x^4+6x^2y^2+y^4} - \\frac{1+x^2}{2+x^4+y^4} \\right)\\,dx\\,dy.\n\\]\nAveraging the two expressions for $I(R)$ yields\n\\[\nI(R) = \\iint_{x^2+y^2 \\leq R^2} (f(x,y) - g(x,y))\\,dx\\,dy\n\\]\nwhere\n\\begin{align*}\nf(x,y) &= \\frac{1+x^2+y^2}{1 + x^4 + 6x^2y^2 + y^4} \\\\\ng(x,y) &= \\frac{1+x^2/2+y^2/2}{2 + x^4 + y^4}.\n\\end{align*}\nNow note that\n\\[f(x,y) = 2 g(x+y, x-y).\n\\]\nWe can thus write\n\\[\nI(R) = \\iint_{R^2 \\leq x^2 +y^2 \\leq 2R^2} g(x,y)\\,dx\\,dy.\n\\]\nTo compute this integral, we switch to polar coordinates:\n\\begin{align*}\nI(R) &= \\int_R^{R\\sqrt{2}} \\int_0^{2\\pi} g(r\\cos \\theta, r \\sin \\theta)r\\,dr\\,d\\theta \\\\\n&=  \\int_R^{R\\sqrt{2}} \\int_0^{2\\pi} \\frac{1 + r^2/2}{2 + r^4(1 - (\\sin^2 2\\theta)/2)} r\\,dr\\,d\\theta.\n\\end{align*}\nWe rescale $r$ to remove the factor of $R$ from the limits of integration:\n\\begin{align*}\nI(R) & =  \\int_1^{\\sqrt{2}} \\int_0^{2\\pi} \\frac{1 + R^2 r^2/2}{2 + R^4 r^4(1 - (\\sin^2 2\\theta)/2)} R^2 r\\,dr\\,d\\theta.\n\\end{align*}\n\nSince the integrand is uniformly bounded for $R \\gg 0$, we may take the limit over $R$ through the integrals to obtain\n\\begin{align*}\n\\lim_{R \\to \\infty} I(R) &= \\int_1^{\\sqrt{2}} \\int_0^{2\\pi} \\frac{r^2/2}{r^4(1 - (\\sin^2 2\\theta)/2)} r\\,dr\\,d\\theta \\\\\n&= \\int_1^{\\sqrt{2}} \\frac{dr}{r} \\int_0^{2\\pi} \\frac{1}{2- \\sin^2 2\\theta} d\\theta \\\\\n&= \\log \\sqrt{2} \\int_0^{2\\pi} \\frac{1}{1 + \\cos^2 2\\theta} d\\theta \\\\\n&= \\frac{1}{2} \\log 2 \\int_0^{2\\pi} \\frac{2}{3 + \\cos 4\\theta} d\\theta.\n\\end{align*}\nIt thus remains to evaluate \n\\[\n\\int_0^{2\\pi} \\frac{2}{3 + \\cos 4\\theta} d\\theta = \n2 \\int_0^{\\pi} \\frac{2}{3 + \\cos \\theta} d\\theta.\n\\]\nOne option for this is to use the half-angle substitution $t = \\tan (\\theta/2)$ to get\n\\begin{align*}\n\\int_{-\\infty}^\\infty \\frac{4}{3(1+t^2) + (1-t^2)}\\,dt\n&= \\int_{-\\infty}^\\infty \\frac{2}{2+t^2}\\,dt \\\\\n&= \\sqrt{2} \\arctan \\left( \\frac{x}{\\sqrt{2}} \\right)^{\\infty}_{-\\infty} \\\\\n&= \\sqrt{2} \\pi.\n\\end{align*}\nPutting this together yields the claimed result.",
  "vars": [
    "I",
    "R",
    "x",
    "y",
    "f",
    "g",
    "r",
    "\\\\theta",
    "t"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "I": "integral",
        "R": "radiusparam",
        "x": "coordx",
        "y": "coordy",
        "f": "firstfun",
        "g": "secondfun",
        "r": "radialvar",
        "\\theta": "angletheta",
        "t": "tanvariable"
      },
      "question": "Let\n\\[\nintegral(radiusparam) = \\iint_{coordx^2+coordy^2 \\leq radiusparam^2} \\left( \\frac{1+2coordx^2}{1+coordx^4+6coordx^2coordy^2+coordy^4} - \\frac{1+coordy^2}{2+coordx^4+coordy^4} \\right)\\,dcoordx\\,dcoordy.\n\\]\nFind\n\\[\n\\lim_{radiusparam \\to \\infty} integral(radiusparam),\n\\]\nor show that this limit does not exist.",
      "solution": "The limit exists and equals $\\frac{\\sqrt{2}}{2} \\pi \\log 2$.\n\nWe first note that we can interchange coordx and coordy to obtain\n\\[\nintegral(radiusparam) = \\iint_{coordx^2+coordy^2 \\leq radiusparam^2} \\left( \\frac{1+2coordy^2}{1+coordx^4+6coordx^2coordy^2+coordy^4} - \\frac{1+coordx^2}{2+coordx^4+coordy^4} \\right)\\,dcoordx\\,dcoordy.\n\\]\nAveraging the two expressions for integral(radiusparam) yields\n\\[\nintegral(radiusparam) = \\iint_{coordx^2+coordy^2 \\leq radiusparam^2} (firstfun(coordx,coordy) - secondfun(coordx,coordy))\\,dcoordx\\,dcoordy\n\\]\nwhere\n\\begin{align*}\nfirstfun(coordx,coordy) &= \\frac{1+coordx^2+coordy^2}{1 + coordx^4 + 6coordx^2coordy^2 + coordy^4} \\\\\nsecondfun(coordx,coordy) &= \\frac{1+coordx^2/2+coordy^2/2}{2 + coordx^4 + coordy^4}.\n\\end{align*}\nNow note that\n\\[firstfun(coordx,coordy) = 2 secondfun(coordx+coordy, coordx-coordy).\n\\]\nWe can thus write\n\\[\nintegral(radiusparam) = \\iint_{radiusparam^2 \\leq coordx^2 +coordy^2 \\leq 2radiusparam^2} secondfun(coordx,coordy)\\,dcoordx\\,dcoordy.\n\\]\nTo compute this integral, we switch to polar coordinates:\n\\begin{align*}\nintegral(radiusparam) &= \\int_{radiusparam}^{radiusparam\\sqrt{2}} \\int_0^{2\\pi} secondfun(radialvar\\cos angletheta, radialvar \\sin angletheta)radialvar\\,dradialvar\\,dangletheta \\\\\n&=  \\int_{radiusparam}^{radiusparam\\sqrt{2}} \\int_0^{2\\pi} \\frac{1 + radialvar^2/2}{2 + radialvar^4(1 - (\\sin^2 2angletheta)/2)} radialvar\\,dradialvar\\,dangletheta.\n\\end{align*}\nWe rescale radialvar to remove the factor of radiusparam from the limits of integration:\n\\begin{align*}\nintegral(radiusparam) & =  \\int_1^{\\sqrt{2}} \\int_0^{2\\pi} \\frac{1 + radiusparam^2 radialvar^2/2}{2 + radiusparam^4 radialvar^4(1 - (\\sin^2 2angletheta)/2)} radiusparam^2 radialvar\\,dradialvar\\,dangletheta.\n\\end{align*}\n\nSince the integrand is uniformly bounded for radiusparam \\gg 0, we may take the limit over radiusparam through the integrals to obtain\n\\begin{align*}\n\\lim_{radiusparam \\to \\infty} integral(radiusparam) &= \\int_1^{\\sqrt{2}} \\int_0^{2\\pi} \\frac{radialvar^2/2}{radialvar^4(1 - (\\sin^2 2angletheta)/2)} radialvar\\,dradialvar\\,dangletheta \\\\\n&= \\int_1^{\\sqrt{2}} \\frac{dradialvar}{radialvar} \\int_0^{2\\pi} \\frac{1}{2- \\sin^2 2angletheta} dangletheta \\\\\n&= \\log \\sqrt{2} \\int_0^{2\\pi} \\frac{1}{1 + \\cos^2 2angletheta} dangletheta \\\\\n&= \\frac{1}{2} \\log 2 \\int_0^{2\\pi} \\frac{2}{3 + \\cos 4angletheta} dangletheta.\n\\end{align*}\nIt thus remains to evaluate \n\\[\n\\int_0^{2\\pi} \\frac{2}{3 + \\cos 4angletheta} dangletheta = \n2 \\int_0^{\\pi} \\frac{2}{3 + \\cos angletheta} dangletheta.\n\\]\nOne option for this is to use the half-angle substitution tanvariable = \\tan (angletheta/2) to get\n\\begin{align*}\n\\int_{-\\infty}^{\\infty} \\frac{4}{3(1+tanvariable^2) + (1-tanvariable^2)}\\,dtanvariable\n&= \\int_{-\\infty}^{\\infty} \\frac{2}{2+tanvariable^2}\\,dtanvariable \\\\\n&= \\sqrt{2} \\arctan \\left( \\frac{coordx}{\\sqrt{2}} \\right)^{\\infty}_{-\\infty} \\\\\n&= \\sqrt{2} \\pi.\n\\end{align*}\nPutting this together yields the claimed result."
    },
    "descriptive_long_confusing": {
      "map": {
        "I": "lanternbeam",
        "R": "juniperleaf",
        "x": "cloudstone",
        "y": "driftwood",
        "f": "monarchwing",
        "g": "cypressroot",
        "r": "pebblegrain",
        "\\\\theta": "compassrose",
        "t": "riverdelta"
      },
      "question": "Let\n\\[\nlanternbeam(juniperleaf) = \\iint_{cloudstone^2+driftwood^2 \\leq juniperleaf^2} \\left( \\frac{1+2cloudstone^2}{1+cloudstone^4+6cloudstone^2driftwood^2+driftwood^4} - \\frac{1+driftwood^2}{2+cloudstone^4+driftwood^4} \\right)\\,dcloudstone\\,ddriftwood.\n\\]\nFind\n\\[\n\\lim_{juniperleaf \\to \\infty} lanternbeam(juniperleaf),\n\\]\nor show that this limit does not exist.",
      "solution": "The limit exists and equals $\\frac{\\sqrt{2}}{2} \\pi \\log 2$.\n\nWe first note that we can interchange cloudstone and driftwood to obtain\n\\[\nlanternbeam(juniperleaf) = \\iint_{cloudstone^2+driftwood^2 \\leq juniperleaf^2} \\left( \\frac{1+2driftwood^2}{1+cloudstone^4+6cloudstone^2driftwood^2+driftwood^4} - \\frac{1+cloudstone^2}{2+cloudstone^4+driftwood^4} \\right)\\,dcloudstone\\,ddriftwood.\n\\]\nAveraging the two expressions for lanternbeam(juniperleaf) yields\n\\[\nlanternbeam(juniperleaf) = \\iint_{cloudstone^2+driftwood^2 \\leq juniperleaf^2} (monarchwing(cloudstone,driftwood) - cypressroot(cloudstone,driftwood))\\,dcloudstone\\,ddriftwood\n\\]\nwhere\n\\begin{align*}\nmonarchwing(cloudstone,driftwood) &= \\frac{1+cloudstone^2+driftwood^2}{1 + cloudstone^4 + 6cloudstone^2driftwood^2 + driftwood^4} \\\\\ncypressroot(cloudstone,driftwood) &= \\frac{1+cloudstone^2/2+driftwood^2/2}{2 + cloudstone^4 + driftwood^4}.\n\\end{align*}\nNow note that\n\\[monarchwing(cloudstone,driftwood) = 2 cypressroot(cloudstone+driftwood, cloudstone-driftwood).\n\\]\nWe can thus write\n\\[\nlanternbeam(juniperleaf) = \\iint_{juniperleaf^2 \\leq cloudstone^2 +driftwood^2 \\leq 2juniperleaf^2} cypressroot(cloudstone,driftwood)\\,dcloudstone\\,ddriftwood.\n\\]\nTo compute this integral, we switch to polar coordinates:\n\\begin{align*}\nlanternbeam(juniperleaf) &= \\int_{juniperleaf}^{juniperleaf\\sqrt{2}} \\int_0^{2\\pi} cypressroot(pebblegrain\\cos compassrose, pebblegrain \\sin compassrose)pebblegrain\\,dpebblegrain\\,dcompassrose \\\\\n&=  \\int_{juniperleaf}^{juniperleaf\\sqrt{2}} \\int_0^{2\\pi} \\frac{1 + pebblegrain^2/2}{2 + pebblegrain^4(1 - (\\sin^2 2compassrose)/2)} pebblegrain\\,dpebblegrain\\,dcompassrose.\n\\end{align*}\nWe rescale pebblegrain to remove the factor of juniperleaf from the limits of integration:\n\\begin{align*}\nlanternbeam(juniperleaf) & =  \\int_1^{\\sqrt{2}} \\int_0^{2\\pi} \\frac{1 + juniperleaf^2 pebblegrain^2/2}{2 + juniperleaf^4 pebblegrain^4(1 - (\\sin^2 2compassrose)/2)} juniperleaf^2 pebblegrain\\,dpebblegrain\\,dcompassrose.\n\\end{align*}\n\nSince the integrand is uniformly bounded for juniperleaf \\gg 0, we may take the limit over juniperleaf through the integrals to obtain\n\\begin{align*}\n\\lim_{juniperleaf \\to \\infty} lanternbeam(juniperleaf) &= \\int_1^{\\sqrt{2}} \\int_0^{2\\pi} \\frac{pebblegrain^2/2}{pebblegrain^4(1 - (\\sin^2 2compassrose)/2)} pebblegrain\\,dpebblegrain\\,dcompassrose \\\\\n&= \\int_1^{\\sqrt{2}} \\frac{dpebblegrain}{pebblegrain} \\int_0^{2\\pi} \\frac{1}{2- \\sin^2 2compassrose} dcompassrose \\\\\n&= \\log \\sqrt{2} \\int_0^{2\\pi} \\frac{1}{1 + \\cos^2 2compassrose} dcompassrose \\\\\n&= \\frac{1}{2} \\log 2 \\int_0^{2\\pi} \\frac{2}{3 + \\cos 4compassrose} dcompassrose.\n\\end{align*}\nIt thus remains to evaluate \n\\[\n\\int_0^{2\\pi} \\frac{2}{3 + \\cos 4compassrose} dcompassrose = \n2 \\int_0^{\\pi} \\frac{2}{3 + \\cos compassrose} dcompassrose.\n\\]\nOne option for this is to use the half-angle substitution riverdelta = \\tan (compassrose/2) to get\n\\begin{align*}\n\\int_{-\\infty}^{\\infty} \\frac{4}{3(1+riverdelta^2) + (1-riverdelta^2)}\\,driverdelta\n&= \\int_{-\\infty}^{\\infty} \\frac{2}{2+riverdelta^2}\\,driverdelta \\\\\n&= \\sqrt{2} \\arctan \\left( \\frac{riverdelta}{\\sqrt{2}} \\right)^{\\infty}_{-\\infty} \\\\\n&= \\sqrt{2} \\pi.\n\\end{align*}\nPutting this together yields the claimed result."
    },
    "descriptive_long_misleading": {
      "map": {
        "I": "discreteval",
        "R": "tinybound",
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "f": "simplevar",
        "g": "staticform",
        "r": "diameterlen",
        "\\theta": "distanceval",
        "t": "cotangentvar"
      },
      "question": "Let\n\\[\ndiscreteval(tinybound) = \\iint_{verticalaxis^2+horizontalaxis^2 \\leq tinybound^2} \\left( \\frac{1+2verticalaxis^2}{1+verticalaxis^4+6verticalaxis^2horizontalaxis^2+horizontalaxis^4} - \\frac{1+horizontalaxis^2}{2+verticalaxis^4+horizontalaxis^4} \\right)\\,dverticalaxis\\,dhorizontalaxis.\n\\]\nFind\n\\[\n\\lim_{tinybound \\to \\infty} discreteval(tinybound),\n\\]\nor show that this limit does not exist.",
      "solution": "The limit exists and equals $\\frac{\\sqrt{2}}{2} \\pi \\log 2$.\n\nWe first note that we can interchange $verticalaxis$ and $horizontalaxis$ to obtain\n\\[\ndiscreteval(tinybound) = \\iint_{verticalaxis^2+horizontalaxis^2 \\leq tinybound^2} \\left( \\frac{1+2horizontalaxis^2}{1+verticalaxis^4+6verticalaxis^2horizontalaxis^2+horizontalaxis^4} - \\frac{1+verticalaxis^2}{2+verticalaxis^4+horizontalaxis^4} \\right)\\,dverticalaxis\\,dhorizontalaxis.\n\\]\nAveraging the two expressions for $discreteval(tinybound)$ yields\n\\[\ndiscreteval(tinybound) = \\iint_{verticalaxis^2+horizontalaxis^2 \\leq tinybound^2} (simplevar(verticalaxis,horizontalaxis) - staticform(verticalaxis,horizontalaxis))\\,dverticalaxis\\,dhorizontalaxis\n\\]\nwhere\n\\begin{align*}\nsimplevar(verticalaxis,horizontalaxis) &= \\frac{1+verticalaxis^2+horizontalaxis^2}{1 + verticalaxis^4 + 6verticalaxis^2horizontalaxis^2 + horizontalaxis^4} \\\\\nstaticform(verticalaxis,horizontalaxis) &= \\frac{1+verticalaxis^2/2+horizontalaxis^2/2}{2 + verticalaxis^4 + horizontalaxis^4}.\n\\end{align*}\nNow note that\n\\[simplevar(verticalaxis,horizontalaxis) = 2 staticform(verticalaxis+horizontalaxis, verticalaxis-horizontalaxis).\n\\]\nWe can thus write\n\\[\ndiscreteval(tinybound) = \\iint_{tinybound^2 \\leq verticalaxis^2 +horizontalaxis^2 \\leq 2tinybound^2} staticform(verticalaxis,horizontalaxis)\\,dverticalaxis\\,dhorizontalaxis.\n\\]\nTo compute this integral, we switch to polar coordinates:\n\\begin{align*}\ndiscreteval(tinybound) &= \\int_{tinybound}^{tinybound\\sqrt{2}} \\int_0^{2\\pi} staticform(diameterlen\\cos distanceval, diameterlen \\sin distanceval)diameterlen\\,ddiameterlen\\,ddistanceval \\\\\n&=  \\int_{tinybound}^{tinybound\\sqrt{2}} \\int_0^{2\\pi} \\frac{1 + diameterlen^2/2}{2 + diameterlen^4(1 - (\\sin^2 2distanceval)/2)} diameterlen\\,ddiameterlen\\,ddistanceval.\n\\end{align*}\nWe rescale $diameterlen$ to remove the factor of $tinybound$ from the limits of integration:\n\\begin{align*}\ndiscreteval(tinybound) & =  \\int_1^{\\sqrt{2}} \\int_0^{2\\pi} \\frac{1 + tinybound^2 diameterlen^2/2}{2 + tinybound^4 diameterlen^4(1 - (\\sin^2 2distanceval)/2)} tinybound^2 diameterlen\\,ddiameterlen\\,ddistanceval.\n\\end{align*}\n\nSince the integrand is uniformly bounded for $tinybound \\gg 0$, we may take the limit over $tinybound$ through the integrals to obtain\n\\begin{align*}\n\\lim_{tinybound \\to \\infty} discreteval(tinybound) &= \\int_1^{\\sqrt{2}} \\int_0^{2\\pi} \\frac{diameterlen^2/2}{diameterlen^4(1 - (\\sin^2 2distanceval)/2)} diameterlen\\,ddiameterlen\\,ddistanceval \\\\\n&= \\int_1^{\\sqrt{2}} \\frac{ddiameterlen}{diameterlen} \\int_0^{2\\pi} \\frac{1}{2- \\sin^2 2distanceval} ddistanceval \\\\\n&= \\log \\sqrt{2} \\int_0^{2\\pi} \\frac{1}{1 + \\cos^2 2distanceval} ddistanceval \\\\\n&= \\frac{1}{2} \\log 2 \\int_0^{2\\pi} \\frac{2}{3 + \\cos 4distanceval} ddistanceval.\n\\end{align*}\nIt thus remains to evaluate \n\\[\n\\int_0^{2\\pi} \\frac{2}{3 + \\cos 4distanceval} ddistanceval = \n2 \\int_0^{\\pi} \\frac{2}{3 + \\cos distanceval} ddistanceval.\n\\]\nOne option for this is to use the half-angle substitution $cotangentvar = \\tan (distanceval/2)$ to get\n\\begin{align*}\n\\int_{-\\infty}^\\infty \\frac{4}{3(1+cotangentvar^2) + (1-cotangentvar^2)}\\,dcotangentvar\n&= \\int_{-\\infty}^\\infty \\frac{2}{2+cotangentvar^2}\\,dcotangentvar \\\\\n&= \\sqrt{2} \\arctan \\left( \\frac{verticalaxis}{\\sqrt{2}} \\right)^{\\infty}_{-\\infty} \\\\\n&= \\sqrt{2} \\pi.\n\\end{align*}\nPutting this together yields the claimed result."
    },
    "garbled_string": {
      "map": {
        "I": "vewnqpsa",
        "R": "zekdofur",
        "x": "bdrlzavn",
        "y": "cqfjmsie",
        "f": "jkuylrpa",
        "g": "mzoschit",
        "r": "lqtenbru",
        "\\theta": "bhxqmdve",
        "t": "uxpathlo"
      },
      "question": "Let\n\\[\nvewnqpsa(zekdofur) = \\iint_{bdrlzavn^2+cqfjmsie^2 \\leq zekdofur^2} \\left( \\frac{1+2bdrlzavn^2}{1+bdrlzavn^4+6bdrlzavn^2cqfjmsie^2+cqfjmsie^4} - \\frac{1+cqfjmsie^2}{2+bdrlzavn^4+cqfjmsie^4} \\right)\\,d bdrlzavn\\,d cqfjmsie.\n\\]\nFind\n\\[\n\\lim_{zekdofur \\to \\infty} vewnqpsa(zekdofur),\n\\]\nor show that this limit does not exist.",
      "solution": "The limit exists and equals $\\frac{\\sqrt{2}}{2} \\pi \\log 2$.\n\nWe first note that we can interchange $bdrlzavn$ and $cqfjmsie$ to obtain\n\\[\nvewnqpsa(zekdofur) = \\iint_{bdrlzavn^2+cqfjmsie^2 \\leq zekdofur^2} \\left( \\frac{1+2cqfjmsie^2}{1+bdrlzavn^4+6bdrlzavn^2cqfjmsie^2+cqfjmsie^4} - \\frac{1+bdrlzavn^2}{2+bdrlzavn^4+cqfjmsie^4} \\right)\\,d bdrlzavn\\,d cqfjmsie.\n\\]\nAveraging the two expressions for $vewnqpsa(zekdofur)$ yields\n\\[\nvewnqpsa(zekdofur) = \\iint_{bdrlzavn^2+cqfjmsie^2 \\leq zekdofur^2} (jkuylrpa(bdrlzavn,cqfjmsie) - mzoschit(bdrlzavn,cqfjmsie))\\,d bdrlzavn\\,d cqfjmsie\n\\]\nwhere\n\\begin{align*}\njkuylrpa(bdrlzavn,cqfjmsie) &= \\frac{1+bdrlzavn^2+cqfjmsie^2}{1 + bdrlzavn^4 + 6bdrlzavn^2cqfjmsie^2 + cqfjmsie^4} \\\\\nmzoschit(bdrlzavn,cqfjmsie) &= \\frac{1+bdrlzavn^2/2+cqfjmsie^2/2}{2 + bdrlzavn^4 + cqfjmsie^4}.\n\\end{align*}\nNow note that\n\\[\njkuylrpa(bdrlzavn,cqfjmsie) = 2\\, mzoschit(bdrlzavn+cqfjmsie,\\, bdrlzavn-cqfjmsie).\n\\]\nWe can thus write\n\\[\nvewnqpsa(zekdofur) = \\iint_{zekdofur^2 \\leq bdrlzavn^2 +cqfjmsie^2 \\leq 2\\,zekdofur^2} mzoschit(bdrlzavn,cqfjmsie)\\,d bdrlzavn\\,d cqfjmsie.\n\\]\nTo compute this integral, we switch to polar coordinates:\n\\begin{align*}\nvewnqpsa(zekdofur) &= \\int_{zekdofur}^{zekdofur\\sqrt{2}} \\int_0^{2\\pi} mzoschit(lqtenbru\\cos bhxqmdve,\\, lqtenbru \\sin bhxqmdve)\\,lqtenbru\\,d lqtenbru\\,d bhxqmdve \\\\\n&=  \\int_{zekdofur}^{zekdofur\\sqrt{2}} \\int_0^{2\\pi} \\frac{1 + lqtenbru^2/2}{2 + lqtenbru^4\\!\\left(1 - (\\sin^2 2bhxqmdve)/2\\right)} \\,lqtenbru\\,d lqtenbru\\,d bhxqmdve.\n\\end{align*}\nWe rescale $lqtenbru$ to remove the factor of $zekdofur$ from the limits of integration:\n\\begin{align*}\nvewnqpsa(zekdofur) & =  \\int_1^{\\sqrt{2}} \\int_0^{2\\pi} \\frac{1 + zekdofur^2 lqtenbru^2/2}{2 + zekdofur^4 lqtenbru^4\\!\\left(1 - (\\sin^2 2bhxqmdve)/2\\right)} \\,zekdofur^2 lqtenbru\\,d lqtenbru\\,d bhxqmdve.\n\\end{align*}\n\nSince the integrand is uniformly bounded for $zekdofur \\gg 0$, we may take the limit over $zekdofur$ through the integrals to obtain\n\\begin{align*}\n\\lim_{zekdofur \\to \\infty} vewnqpsa(zekdofur) &= \\int_1^{\\sqrt{2}} \\int_0^{2\\pi} \\frac{lqtenbru^2/2}{lqtenbru^4\\!\\left(1 - (\\sin^2 2bhxqmdve)/2\\right)} \\,lqtenbru\\,d lqtenbru\\,d bhxqmdve \\\\\n&= \\int_1^{\\sqrt{2}} \\frac{d lqtenbru}{lqtenbru} \\int_0^{2\\pi} \\frac{1}{2- \\sin^2 2bhxqmdve} \\,d bhxqmdve \\\\\n&= \\log \\sqrt{2} \\int_0^{2\\pi} \\frac{1}{1 + \\cos^2 2bhxqmdve} \\,d bhxqmdve \\\\\n&= \\frac{1}{2} \\log 2 \\int_0^{2\\pi} \\frac{2}{3 + \\cos 4bhxqmdve} \\,d bhxqmdve.\n\\end{align*}\nIt thus remains to evaluate \n\\[\n\\int_0^{2\\pi} \\frac{2}{3 + \\cos 4bhxqmdve} \\,d bhxqmdve = \n2 \\int_0^{\\pi} \\frac{2}{3 + \\cos bhxqmdve} \\,d bhxqmdve.\n\\]\nOne option for this is to use the half-angle substitution $uxpathlo = \\tan (bhxqmdve/2)$ to get\n\\begin{align*}\n\\int_{-\\infty}^{\\infty} \\frac{4}{3(1+uxpathlo^2) + (1-uxpathlo^2)}\\,d uxpathlo\n&= \\int_{-\\infty}^{\\infty} \\frac{2}{2+uxpathlo^2}\\,d uxpathlo \\\\\n&= \\sqrt{2} \\arctan \\left( \\frac{uxpathlo}{\\sqrt{2}} \\right)^{\\infty}_{-\\infty} \\\\\n&= \\sqrt{2} \\pi.\n\\end{align*}\nPutting this together yields the claimed result."
    },
    "kernel_variant": {
      "question": "Let\n\\[\nI(R)=\\iint_{x^{2}+y^{2}\\le 5R^{2}}\\Biggl(\\frac{1+2x^{2}}{\\;3+x^{4}+6x^{2}y^{2}+y^{4}\\;}-\n\\;\\frac{1+y^{2}}{\\;6+x^{4}+y^{4}\\;}\\Biggr)\\,dx\\,dy,\\qquad R>0.\n\\]\nDetermine the value of the limit\n\\[\\displaystyle \\lim_{R\\to\\infty}I(R).\\]",
      "solution": "1.  Symmetrisation.\nSwapping \\(x\\) and \\(y\\) in the integral yields a second expression for \\(I(R)\\).  Averaging the two expressions gives\n\\[\nI(R)=\\iint_{x^{2}+y^{2}\\le 5R^{2}}\\bigl(f(x,y)-g(x,y)\\bigr)\\,dx\\,dy,\n\\]\nwhere\n\\[\n  f(x,y)=\\frac{1+x^{2}+y^{2}}{3+x^{4}+6x^{2}y^{2}+y^{4}},\\qquad\n  g(x,y)=\\frac{1+\\dfrac{x^{2}+y^{2}}{2}}{6+x^{4}+y^{4}}\\;.\n\\]\n\n2.  A linear change of variables.\nPut\n\\(u=x+y,\\; v=x-y\\).  Then\n\\[\n  x^{4}+6x^{2}y^{2}+y^{4}=\\tfrac12\\bigl(u^{4}+v^{4}\\bigr),\\qquad\n  x^{2}+y^{2}=\\tfrac12\\bigl(u^{2}+v^{2}\\bigr),\n\\]\nso that\n\\(f(x,y)=2g(u,v)\\).  Because \\(|\\det\\partial(u,v)/\\partial(x,y)|=2\\) we have\n\\(\n  dx\\,dy=\\tfrac12 du\\,dv\n\\).  Hence\n\\[\n\\iint_{x^{2}+y^{2}\\le 5R^{2}}f(x,y)\\,dx\\,dy\n   = \\iint_{u^{2}+v^{2}\\le 10R^{2}}g(u,v)\\,du\\,dv.\n\\]\nSubtracting the two copies of the integral of \\(g\\) we arrive at\n\\[\nI(R)=\\iint_{5R^{2}\\le x^{2}+y^{2}\\le 10R^{2}} g(x,y)\\,dx\\,dy.\n\\]\nThus only the annulus \\(\\sqrt5R\\le r\\le \\sqrt{10}R\\) contributes.\n\n3.  Passage to polar coordinates.\nWrite \\((x,y)=(r\\cos\\theta,r\\sin\\theta)\\).  Using\n\\(x^{4}+y^{4}=r^{4}\\bigl(\\cos^{4}\\theta+\\sin^{4}\\theta\\bigr)\n            =r^{4}\\bigl(1-\\tfrac12\\sin^{2}2\\theta\\bigr)\\),\nwe obtain\n\\[\n  g(r,\\theta)=\\frac{1+r^{2}/2}{6+r^{4}\\bigl(1-\\tfrac12\\sin^{2}2\\theta\\bigr)}.\n\\]\nHence\n\\[\nI(R)=\\int_{\\sqrt5 R}^{\\sqrt{10}R}\\int_{0}^{2\\pi}\n       \\frac{1+r^{2}/2}{6+r^{4}\\bigl(1-\\tfrac12\\sin^{2}2\\theta\\bigr)}\n       \\,r\\,d\\theta\\,dr.\n\\]\n\n4.  Scale the radius.\nPut \\(r=Rt\\;(\\sqrt5\\le t\\le \\sqrt{10})\\).  Because \\(r\\,dr=R^{2}t\\,dt\\),\n\\[\nI(R)=\\int_{\\sqrt5}^{\\sqrt{10}}\\int_{0}^{2\\pi}\n       \\frac{1+R^{2}t^{2}/2}{6+R^{4}t^{4}\\bigl(1-\\tfrac12\\sin^{2}2\\theta\\bigr)}\n       R^{2}t\\,d\\theta\\,dt.\n\\]\nFor large \\(R\\) the constants \\(1\\) and \\(6\\) in the numerator and denominator\nare negligible, so\n\\[\n\\frac{1+R^{2}t^{2}/2}{6+R^{4}t^{4}(1-\\tfrac12\\sin^{2}2\\theta)}\\;R^{2}t\n\\longrightarrow \\frac{1}{2t\\,M(\\theta)},\\qquad\nM(\\theta):=1-\\tfrac12\\sin^{2}2\\theta.\n\\]\nSince the integrand is dominated by an integrable function, the dominated\nconvergence theorem allows the limit to be taken under the integral sign:\n\\[\n\\boxed{\\displaystyle \\lim_{R\\to\\infty}I(R)=\\int_{\\sqrt5}^{\\sqrt{10}}\\!\\frac{dt}{2t}\n        \\int_{0}^{2\\pi}\\!\\frac{d\\theta}{M(\\theta)}}.\n\\]\n\n5.  The angular integral.\nBecause\n\\(M(\\theta)=\\frac12\\bigl(1+\\cos^{2}2\\theta\\bigr)\\), we have\n\\[\\frac{1}{M(\\theta)}=\\frac{2}{1+\\cos^{2}2\\theta}.\\]\nThus\n\\[\n\\int_{0}^{2\\pi}\\frac{d\\theta}{M(\\theta)}\n   =2\\int_{0}^{2\\pi}\\frac{d\\theta}{1+\\cos^{2}2\\theta}.\n\\]\nMake the substitution \\(\\varphi=2\\theta\\;(d\\theta=d\\varphi/2)\\):\n\\[\n2\\int_{0}^{2\\pi}\\frac{d\\theta}{1+\\cos^{2}2\\theta}\n   =2\\cdot\\frac12\\int_{0}^{4\\pi}\\frac{d\\varphi}{1+\\cos^{2}\\varphi}\n   =\\int_{0}^{4\\pi}\\frac{d\\varphi}{1+\\cos^{2}\\varphi}.\n\\]\nSince the integrand has period \\(\\pi\\), the last integral equals\n\\(4\\) times the integral over \\([0,\\pi]\\).  A standard tangent–half–angle\nsubstitution gives\n\\[\n\\int_{0}^{\\pi}\\frac{d\\varphi}{1+\\cos^{2}\\varphi}=\\frac{\\pi}{\\sqrt2},\n\\]\nso that\n\\[\n\\int_{0}^{2\\pi}\\!\\frac{d\\theta}{M(\\theta)}=4\\cdot\\frac{\\pi}{\\sqrt2}=2\\sqrt2\\,\\pi.\n\\]\n\n6.  The radial integral.\n\\(\\displaystyle \\int_{\\sqrt5}^{\\sqrt{10}}\\frac{dt}{2t}=\\frac14\\ln2.\\)\n\n7.  Putting the pieces together.\n\\[\n\\lim_{R\\to\\infty}I(R)=(2\\sqrt2\\pi)\\Bigl(\\frac14\\ln2\\Bigr)\n                     =\\boxed{\\displaystyle \\frac{\\sqrt2}{2}\\,\\pi\\,\\ln 2}.\n\\]",
      "_meta": {
        "core_steps": [
          "Average the integrand with its x↔y swap to obtain a symmetric difference f−g",
          "Use the 45° linear change of variables (u,v)=(x+y,x−y) to get f(x,y)=2·g(u,v) and turn the disk |(x,y)|≤R into the annulus R≤|(u,v)|≤√2 R",
          "Rewrite the annulus integral in polar coordinates",
          "Rescale r→Rr and invoke dominated-convergence to pass the limit R→∞ through the integral, separating radial (∫dr/r) and angular parts",
          "Evaluate the angular integral (e.g. t=tan(θ/2)) and combine with the radial log factor to obtain the limit"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Additive constant that accompanies x^4+y^4 in the denominator of g; any positive value c would still allow f(x,y)=2·g(x+y,x−y) after a suitable rescaling",
            "original": "2"
          },
          "slot2": {
            "description": "Overall scale k of the original disk (|x|^2+|y|^2≤kR^2 instead of R^2); it merely changes the outer/inner-radius ratio and hence multiplies the final answer by log k",
            "original": "k=1 (unit disk radius R)"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}