{
  "index": "1985-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Evaluate $\\int_0^\\infty t^{-1/2}e^{-1985(t+t^{-1})}\\,dt$. You may\nassume that $\\int_{-\\infty}^\\infty e^{-x^2}\\,dx = \\sqrt{\\pi}$.",
  "solution": "Solution (adapted from [Bernau]). For \\( a>0 \\), let\n\\[\nI(a)=\\int_{0}^{\\infty} t^{-1 / 2} e^{-a\\left(t+t^{-1}\\right)} d t .\n\\]\n\nThe integral converges, since the integrand is bounded by \\( t^{-1 / 2} \\) on \\( (0,1] \\) and by \\( e^{-a t} \\) on \\( [1, \\infty) \\). Hence\n\\[\nI(a)=\\lim _{B \\rightarrow \\infty}\\left[\\int_{1 / B}^{1} t^{-1 / 2} e^{-a\\left(t+t^{-1}\\right)} d t+\\int_{1}^{B} t^{-1 / 2} e^{-a\\left(t+t^{-1}\\right)} d t\\right] .\n\\]\n\nSubstitute \\( 1 / t \\) for \\( t \\) in the first integral to conclude\n\\[\nI(a)=\\lim _{B \\rightarrow \\infty} \\int_{1}^{B}\\left(t^{-1 / 2}+t^{-3 / 2}\\right) e^{-a\\left(t+t^{-1}\\right)} d t\n\\]\n\nNow use the substitution \\( u=a^{1 / 2}\\left(t^{1 / 2}-t^{-1 / 2}\\right) \\) to obtain\n\\[\n\\begin{aligned}\nI(a) & =2 a^{-1 / 2} \\lim _{B \\rightarrow \\infty} \\int_{0}^{a^{1 / 2}\\left(B^{1 / 2}-B^{-1 / 2}\\right)} e^{-u^{2}-2 a} d u \\\\\n& =2 a^{-1 / 2} e^{-2 a} \\int_{0}^{\\infty} e^{-u^{2}} d u \\\\\n& =\\sqrt{\\frac{\\pi}{a}} e^{-2 a}\n\\end{aligned}\n\\]\nso \\( I(1985)=\\sqrt{\\frac{\\pi}{1985}} e^{-3970} \\).\nRemark. The modified Bessel function of the second kind (also known as Macdonald's function) has the integral representation\n\\[\nK_{\\nu}(z)=\\int_{0}^{\\infty} e^{-z \\cosh t} \\cosh (\\nu t) d t\n\\]\nfor \\( \\operatorname{Re}(z)>0 \\) [O, p. 250]. When \\( \\nu=1 / 2 \\), the substitution \\( u=e^{t} \\) relates this to expressions occurring in the solution above; to be precise, \\( I(a)=2 K_{1 / 2}(2 a) \\) for all \\( a>0 \\). Thus \\( K_{1 / 2}(z)=\\sqrt{\\frac{\\pi}{2 z}} e^{-z} \\) for \\( z>0 \\). Similar formulas exist for \\( K_{n+1 / 2}(z) \\) for each integer \\( n \\). For arbitrary \\( \\nu \\), the function \\( w=K_{\\nu}(z) \\) is a solution of the differential equation\n\\[\nz^{2} w^{\\prime \\prime}+z w^{\\prime}-\\left(z^{2}+\\nu^{2}\\right) w=0 .\n\\]",
  "vars": [
    "t",
    "x",
    "u",
    "z",
    "w",
    "I"
  ],
  "params": [
    "a",
    "B",
    "n",
    "\\\\nu"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "t": "varitime",
        "x": "varixval",
        "u": "varuauxi",
        "z": "varzenit",
        "w": "varomega",
        "I": "intfunca",
        "a": "paramcoef",
        "B": "parambigh",
        "n": "paramindx",
        "\\nu": "paramnucl"
      },
      "question": "Evaluate $\\int_0^\\infty varitime^{-1/2}e^{-1985(varitime+varitime^{-1})}\\,dvaritime$. You may\nassume that $\\int_{-\\infty}^\\infty e^{-varixval^2}\\,dvarixval = \\sqrt{\\pi}$.",
      "solution": "Solution (adapted from [Bernau]). For \\( paramcoef>0 \\), let\n\\[\nintfunca(paramcoef)=\\int_{0}^{\\infty} varitime^{-1 / 2} e^{-paramcoef\\left(varitime+varitime^{-1}\\right)} d varitime .\n\\]\n\nThe integral converges, since the integrand is bounded by \\( varitime^{-1 / 2} \\) on \\( (0,1] \\) and by \\( e^{-paramcoef varitime} \\) on \\( [1, \\infty) \\). Hence\n\\[\nintfunca(paramcoef)=\\lim _{parambigh \\rightarrow \\infty}\\left[\\int_{1 / parambigh}^{1} varitime^{-1 / 2} e^{-paramcoef\\left(varitime+varitime^{-1}\\right)} d varitime+\\int_{1}^{parambigh} varitime^{-1 / 2} e^{-paramcoef\\left(varitime+varitime^{-1}\\right)} d varitime\\right] .\n\\]\n\nSubstitute \\( 1 / varitime \\) for \\( varitime \\) in the first integral to conclude\n\\[\nintfunca(paramcoef)=\\lim _{parambigh \\rightarrow \\infty} \\int_{1}^{parambigh}\\left(varitime^{-1 / 2}+varitime^{-3 / 2}\\right) e^{-paramcoef\\left(varitime+varitime^{-1}\\right)} d varitime .\n\\]\n\nNow use the substitution \\( varuauxi=paramcoef^{1 / 2}\\left(varitime^{1 / 2}-varitime^{-1 / 2}\\right) \\) to obtain\n\\[\n\\begin{aligned}\nintfunca(paramcoef) & =2 \\, paramcoef^{-1 / 2} \\lim _{parambigh \\rightarrow \\infty} \\int_{0}^{paramcoef^{1 / 2}\\left(parambigh^{1 / 2}-parambigh^{-1 / 2}\\right)} e^{-varuauxi^{2}-2 \\, paramcoef} d varuauxi \\\\\n& =2 \\, paramcoef^{-1 / 2} e^{-2 \\, paramcoef} \\int_{0}^{\\infty} e^{-varuauxi^{2}} d varuauxi \\\\\n& =\\sqrt{\\frac{\\pi}{paramcoef}} e^{-2 \\, paramcoef}\n\\end{aligned}\n\\]\nso \\( intfunca(1985)=\\sqrt{\\frac{\\pi}{1985}} e^{-3970} \\).\n\nRemark. The modified Bessel function of the second kind (also known as Macdonald's function) has the integral representation\n\\[\nK_{paramnucl}(varzenit)=\\int_{0}^{\\infty} e^{-varzenit \\cosh varitime} \\cosh (paramnucl\\, varitime) d varitime\n\\]\nfor \\( \\operatorname{Re}(varzenit)>0 \\) [O, p. 250]. When \\( paramnucl=1 / 2 \\), the substitution \\( varuauxi=e^{varitime} \\) relates this to expressions occurring in the solution above; to be precise, \\( intfunca(paramcoef)=2 K_{1 / 2}(2 \\, paramcoef) \\) for all \\( paramcoef>0 \\). Thus \\( K_{1 / 2}(varzenit)=\\sqrt{\\frac{\\pi}{2 \\, varzenit}} e^{-varzenit} \\) for \\( varzenit>0 \\). Similar formulas exist for \\( K_{paramindx+1 / 2}(varzenit) \\) for each integer \\( paramindx \\). For arbitrary \\( paramnucl \\), the function \\( varomega=K_{paramnucl}(varzenit) \\) is a solution of the differential equation\n\\[\nvarzenit^{2} \\, varomega^{\\prime \\prime}+varzenit \\, varomega^{\\prime}-\\left(varzenit^{2}+paramnucl^{2}\\right) varomega=0 .\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "t": "marigolds",
        "x": "snowflake",
        "u": "lighthouse",
        "z": "beanstalk",
        "w": "hairbrush",
        "I": "tortoise",
        "a": "cinnamon",
        "B": "driftwood",
        "n": "scarecrow",
        "\\nu": "raindrops"
      },
      "question": "Evaluate $\\int_0^\\infty marigolds^{-1/2}e^{-1985(marigolds+marigolds^{-1})}\\,dmarigolds$. You may\nassume that $\\int_{-\\infty}^\\infty e^{-snowflake^2}\\,dsnowflake = \\sqrt{\\pi}$.",
      "solution": "Solution (adapted from [Bernau]). For \\( cinnamon>0 \\), let\n\\[\ntortoise(cinnamon)=\\int_{0}^{\\infty} marigolds^{-1 / 2} e^{-cinnamon\\left(marigolds+marigolds^{-1}\\right)} d marigolds .\n\\]\n\nThe integral converges, since the integrand is bounded by \\( marigolds^{-1 / 2} \\) on \\( (0,1] \\) and by \\( e^{-cinnamon marigolds} \\) on \\( [1, \\infty) \\). Hence\n\\[\ntortoise(cinnamon)=\\lim _{driftwood \\rightarrow \\infty}\\left[\\int_{1 / driftwood}^{1} marigolds^{-1 / 2} e^{-cinnamon\\left(marigolds+marigolds^{-1}\\right)} d marigolds+\\int_{1}^{driftwood} marigolds^{-1 / 2} e^{-cinnamon\\left(marigolds+marigolds^{-1}\\right)} d marigolds\\right] .\n\\]\n\nSubstitute \\( 1 / marigolds \\) for \\( marigolds \\) in the first integral to conclude\n\\[\ntortoise(cinnamon)=\\lim _{driftwood \\rightarrow \\infty} \\int_{1}^{driftwood}\\left(marigolds^{-1 / 2}+marigolds^{-3 / 2}\\right) e^{-cinnamon\\left(marigolds+marigolds^{-1}\\right)} d marigolds\n\\]\n\nNow use the substitution \\( lighthouse=cinnamon^{1 / 2}\\left(marigolds^{1 / 2}-marigolds^{-1 / 2}\\right) \\) to obtain\n\\[\n\\begin{aligned}\ntortoise(cinnamon) & =2 cinnamon^{-1 / 2} \\lim _{driftwood \\rightarrow \\infty} \\int_{0}^{cinnamon^{1 / 2}\\left(driftwood^{1 / 2}-driftwood^{-1 / 2}\\right)} e^{-lighthouse^{2}-2 cinnamon} d lighthouse \\\\\n& =2 cinnamon^{-1 / 2} e^{-2 cinnamon} \\int_{0}^{\\infty} e^{-lighthouse^{2}} d lighthouse \\\\\n& =\\sqrt{\\frac{\\pi}{cinnamon}} e^{-2 cinnamon}\n\\end{aligned}\n\\]\nso \\( tortoise(1985)=\\sqrt{\\frac{\\pi}{1985}} e^{-3970} \\).\n\nRemark. The modified Bessel function of the second kind (also known as Macdonald's function) has the integral representation\n\\[\nK_{raindrops}(beanstalk)=\\int_{0}^{\\infty} e^{-beanstalk \\cosh marigolds} \\cosh (raindrops marigolds) d marigolds\n\\]\nfor \\( \\operatorname{Re}(beanstalk)>0 \\) [O, p. 250]. When \\( raindrops=1 / 2 \\), the substitution \\( lighthouse=e^{marigolds} \\) relates this to expressions occurring in the solution above; to be precise, \\( tortoise(cinnamon)=2 K_{1 / 2}(2 cinnamon) \\) for all \\( cinnamon>0 \\). Thus \\( K_{1 / 2}(beanstalk)=\\sqrt{\\frac{\\pi}{2 beanstalk}} e^{-beanstalk} \\) for \\( beanstalk>0 \\). Similar formulas exist for \\( K_{scarecrow+1 / 2}(beanstalk) \\) for each integer \\( scarecrow \\). For arbitrary \\( raindrops \\), the function \\( hairbrush=K_{raindrops}(beanstalk) \\) is a solution of the differential equation\n\\[\nbeanstalk^{2} hairbrush^{\\prime \\prime}+beanstalk hairbrush^{\\prime}-\\left(beanstalk^{2}+raindrops^{2}\\right) hairbrush=0 .\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "t": "spacelocus",
        "x": "momentum",
        "u": "original",
        "z": "grounded",
        "w": "problematic",
        "I": "derivative",
        "a": "variable",
        "B": "unbounded",
        "n": "fraction",
        "\\nu": "chaosity"
      },
      "question": "Evaluate $\\int_0^\\infty spacelocus^{-1/2}e^{-1985(spacelocus+spacelocus^{-1})}\\,d spacelocus$. You may\nassume that $\\int_{-\\infty}^\\infty e^{-momentum^2}\\,d momentum = \\sqrt{\\pi}$.",
      "solution": "Solution (adapted from [Bernau]). For \\( variable>0 \\), let\n\\[\nderivative(variable)=\\int_{0}^{\\infty} spacelocus^{-1 / 2} e^{-variable\\left(spacelocus+spacelocus^{-1}\\right)} d spacelocus .\n\\]\n\nThe integral converges, since the integrand is bounded by \\( spacelocus^{-1 / 2} \\) on \\( (0,1] \\) and by \\( e^{-variable spacelocus} \\) on \\( [1, \\infty) \\). Hence\n\\[\nderivative(variable)=\\lim _{unbounded \\rightarrow \\infty}\\left[\\int_{1 / unbounded}^{1} spacelocus^{-1 / 2} e^{-variable\\left(spacelocus+spacelocus^{-1}\\right)} d spacelocus+\\int_{1}^{unbounded} spacelocus^{-1 / 2} e^{-variable\\left(spacelocus+spacelocus^{-1}\\right)} d spacelocus\\right] .\n\\]\n\nSubstitute \\( 1 / spacelocus \\) for \\( spacelocus \\) in the first integral to conclude\n\\[\nderivative(variable)=\\lim _{unbounded \\rightarrow \\infty} \\int_{1}^{unbounded}\\left(spacelocus^{-1 / 2}+spacelocus^{-3 / 2}\\right) e^{-variable\\left(spacelocus+spacelocus^{-1}\\right)} d spacelocus\n\\]\n\nNow use the substitution \\( original=variable^{1 / 2}\\left(spacelocus^{1 / 2}-spacelocus^{-1 / 2}\\right) \\) to obtain\n\\[\n\\begin{aligned}\nderivative(variable) & =2 variable^{-1 / 2} \\lim _{unbounded \\rightarrow \\infty} \\int_{0}^{variable^{1 / 2}\\left(unbounded^{1 / 2}-unbounded^{-1 / 2}\\right)} e^{-original^{2}-2 variable} d original \\\\\n& =2 variable^{-1 / 2} e^{-2 variable} \\int_{0}^{\\infty} e^{-original^{2}} d original \\\\\n& =\\sqrt{\\frac{\\pi}{variable}} e^{-2 variable}\n\\end{aligned}\n\\]\nso \\( derivative(1985)=\\sqrt{\\frac{\\pi}{1985}} e^{-3970} \\).\nRemark. The modified Bessel function of the second kind (also known as Macdonald's function) has the integral representation\n\\[\nK_{chaosity}(grounded)=\\int_{0}^{\\infty} e^{-grounded \\cosh spacelocus} \\cosh (chaosity spacelocus) d spacelocus\n\\]\nfor \\( \\operatorname{Re}(grounded)>0 \\) [O, p. 250]. When \\( chaosity=1 / 2 \\), the substitution \\( original=e^{spacelocus} \\) relates this to expressions occurring in the solution above; to be precise, \\( derivative(variable)=2 K_{1 / 2}(2 variable) \\) for all \\( variable>0 \\). Thus \\( K_{1 / 2}(grounded)=\\sqrt{\\frac{\\pi}{2 grounded}} e^{-grounded} \\) for \\( grounded>0 \\). Similar formulas exist for \\( K_{fraction+1 / 2}(grounded) \\) for each integer \\( fraction \\). For arbitrary \\( chaosity \\), the function \\( problematic=K_{chaosity}(grounded) \\) is a solution of the differential equation\n\\[\ngrounded^{2} problematic^{\\prime \\prime}+grounded problematic^{\\prime}-\\left(grounded^{2}+chaosity^{2}\\right) problematic=0 .\n\\]"
    },
    "garbled_string": {
      "map": {
        "t": "zynoqpgr",
        "x": "kufyxele",
        "u": "pagostir",
        "z": "mavriond",
        "w": "salphuxe",
        "I": "mahxplor",
        "a": "fenotroc",
        "B": "cukvlzpq",
        "n": "werglasm",
        "\\nu": "quilmony"
      },
      "question": "Evaluate $\\int_0^\\infty zynoqpgr^{-1/2}e^{-1985(zynoqpgr+zynoqpgr^{-1})}\\,dzynoqpgr$. You may assume that $\\int_{-\\infty}^\\infty e^{-kufyxele^2}\\,dkufyxele = \\sqrt{\\pi}$.",
      "solution": "Solution (adapted from [Bernau]). For $ fenotroc>0 $, let\n\\[\nmahxplor(fenotroc)=\\int_{0}^{\\infty} zynoqpgr^{-1 / 2} e^{-fenotroc\\left(zynoqpgr+zynoqpgr^{-1}\\right)} d zynoqpgr .\n\\]\n\nThe integral converges, since the integrand is bounded by $ zynoqpgr^{-1 / 2} $ on $ (0,1] $ and by $ e^{-fenotroc zynoqpgr} $ on $ [1, \\infty) $. Hence\n\\[\nmahxplor(fenotroc)=\\lim _{cukvlzpq \\rightarrow \\infty}\\left[\\int_{1 / cukvlzpq}^{1} zynoqpgr^{-1 / 2} e^{-fenotroc\\left(zynoqpgr+zynoqpgr^{-1}\\right)} d zynoqpgr+\\int_{1}^{cukvlzpq} zynoqpgr^{-1 / 2} e^{-fenotroc\\left(zynoqpgr+zynoqpgr^{-1}\\right)} d zynoqpgr\\right] .\n\\]\n\nSubstitute $ 1 / zynoqpgr $ for $ zynoqpgr $ in the first integral to conclude\n\\[\nmahxplor(fenotroc)=\\lim _{cukvlzpq \\rightarrow \\infty} \\int_{1}^{cukvlzpq}\\left(zynoqpgr^{-1 / 2}+zynoqpgr^{-3 / 2}\\right) e^{-fenotroc\\left(zynoqpgr+zynoqpgr^{-1}\\right)} d zynoqpgr\n\\]\n\nNow use the substitution $ pagostir=fenotroc^{1 / 2}\\left(zynoqpgr^{1 / 2}-zynoqpgr^{-1 / 2}\\right) $ to obtain\n\\[\n\\begin{aligned}\nmahxplor(fenotroc) & =2 fenotroc^{-1 / 2} \\lim _{cukvlzpq \\rightarrow \\infty} \\int_{0}^{fenotroc^{1 / 2}\\left(cukvlzpq^{1 / 2}-cukvlzpq^{-1 / 2}\\right)} e^{-pagostir^{2}-2 fenotroc} d pagostir \\\\\n& =2 fenotroc^{-1 / 2} e^{-2 fenotroc} \\int_{0}^{\\infty} e^{-pagostir^{2}} d pagostir \\\\\n& =\\sqrt{\\frac{\\pi}{fenotroc}} e^{-2 fenotroc}\n\\end{aligned}\n\\]\nso $ mahxplor(1985)=\\sqrt{\\frac{\\pi}{1985}} e^{-3970} $.\n\nRemark. The modified Bessel function of the second kind (also known as Macdonald's function) has the integral representation\n\\[\nK_{quilmony}(mavriond)=\\int_{0}^{\\infty} e^{-mavriond \\cosh zynoqpgr} \\cosh (quilmony zynoqpgr) d zynoqpgr\n\\]\nfor $ \\operatorname{Re}(mavriond)>0 $ [O, p. 250]. When $ quilmony=1 / 2 $, the substitution $ pagostir=e^{zynoqpgr} $ relates this to expressions occurring in the solution above; to be precise, $ mahxplor(fenotroc)=2 K_{1 / 2}(2 fenotroc) $ for all $ fenotroc>0 $. Thus $ K_{1 / 2}(mavriond)=\\sqrt{\\frac{\\pi}{2 mavriond}} e^{-mavriond} $ for $ mavriond>0 $. Similar formulas exist for $ K_{werglasm+1 / 2}(mavriond) $ for each integer $ werglasm $. For arbitrary $ quilmony $, the function $ salphuxe=K_{quilmony}(mavriond) $ is a solution of the differential equation\n\\[\nmavriond^{2} salphuxe^{\\prime \\prime}+mavriond salphuxe^{\\prime}-\\left(mavriond^{2}+quilmony^{2}\\right) salphuxe=0 .\n\\]"
    },
    "kernel_variant": {
      "question": "Evaluate\n\\[\n\\mathcal{I}\\;=\\;\n\\int_{0}^{\\infty}\nt^{-\\tfrac12}\\,\n\\bigl(t^{\\tfrac12}+t^{-\\tfrac12}\\bigr)^{3}\\;\n\\exp\\!\\Bigl[-\\,2024\\,\\bigl(t+t^{-1}\\bigr)+2023\\,\\bigl(t-t^{-1}\\bigr)\\Bigr]\\,\ndt.\n\\]\nExpress your answer in closed form, reducing it as far as possible and (if necessary) writing the result in terms of the modified Bessel functions \\(K_{\\nu}\\).\n\n--------------------------------------------------------------------",
      "solution": "Step 1.  Get rid of the fractional powers  \nExpand the algebraic factor and absorb the leading \\(t^{-1/2}\\):\n\\[\n(t^{1/2}+t^{-1/2})^{3}=t^{3/2}+3t^{1/2}+3t^{-1/2}+t^{-3/2},\n\\]\nhence\n\\[\nt^{-1/2}(t^{1/2}+t^{-1/2})^{3}\n      =t+3+3t^{-1}+t^{-2}.\n\\]\nTherefore\n\\[\n\\mathcal{I}\n   =\\int_{0}^{\\infty}\\!\n      \\Bigl(t+3+3t^{-1}+t^{-2}\\Bigr)\\,\n      e^{-2024(t+t^{-1})+2023(t-t^{-1})}\\,dt.\n\\]\n\nStep 2.  Re-write the exponential so that each summand fits the same template  \n\\[\n-2024(t+t^{-1})+2023(t-t^{-1})\n      =-\\bigl(2024-2023\\bigr)t-\\bigl(2024+2023\\bigr)t^{-1}\n      =-t-4047\\,t^{-1}.\n\\]\nThus\n\\[\n\\mathcal{I}=\\int_{0}^{\\infty}\n             \\bigl(t+3+3t^{-1}+t^{-2}\\bigr)\\,\n             e^{-\\bigl(t+4047\\,t^{-1}\\bigr)}\\,dt .\n\\]\n\nStep 3.  Identify a standard integral  \nFor \\(\\alpha,\\beta>0\\) and any real \\(\\nu\\)\n\\[\n\\int_{0}^{\\infty}x^{\\nu-1}e^{-\\alpha x-\\beta/x}\\,dx\n      =2\\Bigl(\\frac{\\beta}{\\alpha}\\Bigr)^{\\nu/2}K_{\\nu}\\!\\bigl(2\\sqrt{\\alpha\\beta}\\bigr)\n      \\quad\\bigl(\\text{modified Bessel }K_{\\nu}\\bigr).\n\\]\nIn our case \\(\\alpha=1,\\;\\beta=4047\\) and we need the four exponents  \n\\(x^{k}\\) with \\(k=1,0,-1,-2\\).  Writing \\(\\nu=k+1\\) and letting\n\\[\n\\lambda:=2\\sqrt{4047},\n\\qquad\\beta:=4047,\n\\]\nwe obtain\n\\[\n\\begin{aligned}\nI_{1}&:=\\int_{0}^{\\infty} t^{ 1}\\,e^{-(t+\\beta t^{-1})}\\,dt\n      =2\\beta K_{2}(\\lambda),\\\\[2mm]\nI_{0}&:=\\int_{0}^{\\infty} t^{ 0}\\,e^{-(t+\\beta t^{-1})}\\,dt\n      =2\\beta^{1/2}K_{1}(\\lambda),\\\\[2mm]\nI_{-1}&:=\\int_{0}^{\\infty} t^{-1}\\,e^{-(t+\\beta t^{-1})}\\,dt\n      =2 K_{0}(\\lambda),\\\\[2mm]\nI_{-2}&:=\\int_{0}^{\\infty} t^{-2}\\,e^{-(t+\\beta t^{-1})}\\,dt\n      =2\\beta^{-1/2}K_{1}(\\lambda),\n\\end{aligned}\n\\]\nbecause \\(K_{-\\nu}=K_{\\nu}\\).\n\nStep 4.  Assemble the pieces  \n\\[\n\\mathcal{I}=I_{1}+3I_{0}+3I_{-1}+I_{-2}\n           =2\\Bigl[\\beta K_{2}(\\lambda)\n                    +(3\\beta^{1/2}+\\beta^{-1/2})K_{1}(\\lambda)\n                    +3K_{0}(\\lambda)\\Bigr].\n\\]\n\nStep 5.  Remove \\(K_{2}\\) so that only the two lowest orders remain  \nThe recurrence \\(K_{\\nu+1}(z)=K_{\\nu-1}(z)+\\tfrac{2\\nu}{z}K_{\\nu}(z)\\) with \\(\\nu=1\\) gives  \n\\(K_{2}(\\lambda)=K_{0}(\\lambda)+\\frac{2}{\\lambda}K_{1}(\\lambda)\\).\nHence\n\\[\n\\begin{aligned}\n\\mathcal{I}\n&=2\\Bigl[\\beta\\bigl(K_{0}(\\lambda)+\\tfrac{2}{\\lambda}K_{1}(\\lambda)\\bigr)\n          +(3\\beta^{1/2}+\\beta^{-1/2})K_{1}(\\lambda)\n          +3K_{0}(\\lambda)\\Bigr]\\\\[2mm]\n&=2(\\beta+3)K_{0}(\\lambda)\n  +2\\Bigl(\\tfrac{2\\beta}{\\lambda}+3\\beta^{1/2}+\\beta^{-1/2}\\Bigr)K_{1}(\\lambda).\n\\end{aligned}\n\\]\n\nStep 6.  Insert the numeric value \\(\\beta=4047,\\;\\lambda=2\\sqrt{4047}\\)  \n\\[\n\\boxed{%\n\\displaystyle\n\\mathcal{I}\n   =8100\\,K_{0}\\!\\bigl(2\\sqrt{4047}\\bigr)\n    +\\Bigl(8\\sqrt{4047}+ \\frac{2}{\\sqrt{4047}}\\Bigr)\\,\n      K_{1}\\!\\bigl(2\\sqrt{4047}\\bigr)\n}.\n\\]\n(Any equivalent reduction—e.g.\\ leaving the answer in the form of Step 4—is also acceptable.)\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.688003",
        "was_fixed": false,
        "difficulty_analysis": "1.  Extra algebraic factor  \n   •  The cubic \\((t^{1/2}+t^{-1/2})^{3}\\) produces four separate power terms, forcing the solver to decompose the integral into a linear combination of four different Mellin–type integrals.\n\n2.  Mixed linear combination in the exponent  \n   •  The simultaneous appearance of \\(t+t^{-1}\\) and \\(t-t^{-1}\\) means the exponent cannot be handled by the single substitution \\(u=t^{1/2}-t^{-1/2}\\) that solved the original problem.  One must instead recognise how to rewrite the exponential as \\(-\\alpha t-\\beta t^{-1}\\) with different \\(\\alpha,\\beta\\).\n\n3.  Higher–order Bessel functions  \n   •  The decomposition yields integrals involving \\(K_{0},K_{1},K_{2}\\) (orders 0, 1, 2) instead of the single half-integer \\(K_{1/2}\\) that collapses to an elementary exponential in the original problem.  The solver must know both the general integral representation of \\(K_{\\nu}\\) and the three-term recurrence to simplify the final combination.\n\n4.  Multiple interacting concepts  \n   •  Mastery of binomial expansions of fractional powers, asymptotic convergence arguments, the Mellin transform, special-function identities and recurrences are all required.  No single “lucky’’ substitution will dispatch the integral; several layers of technique must be coordinated.\n\nOverall, the enhanced variant introduces a non-trivial algebraic prefactor and a skew exponent, pushes the calculation up to higher (non-half-integer) orders of the modified Bessel function, and obliges the solver to manipulate those functions with recurrence relations—rendering the task substantially more intricate than either the original problem or the simpler current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Evaluate\n\\[\n\\mathcal{I}\\;=\\;\n\\int_{0}^{\\infty}\nt^{-\\tfrac12}\\,\n\\bigl(t^{\\tfrac12}+t^{-\\tfrac12}\\bigr)^{3}\\;\n\\exp\\!\\Bigl[-\\,2024\\,\\bigl(t+t^{-1}\\bigr)+2023\\,\\bigl(t-t^{-1}\\bigr)\\Bigr]\\,\ndt.\n\\]\nExpress your answer in closed form, reducing it as far as possible and (if necessary) writing the result in terms of the modified Bessel functions \\(K_{\\nu}\\).\n\n--------------------------------------------------------------------",
      "solution": "Step 1.  Get rid of the fractional powers  \nExpand the algebraic factor and absorb the leading \\(t^{-1/2}\\):\n\\[\n(t^{1/2}+t^{-1/2})^{3}=t^{3/2}+3t^{1/2}+3t^{-1/2}+t^{-3/2},\n\\]\nhence\n\\[\nt^{-1/2}(t^{1/2}+t^{-1/2})^{3}\n      =t+3+3t^{-1}+t^{-2}.\n\\]\nTherefore\n\\[\n\\mathcal{I}\n   =\\int_{0}^{\\infty}\\!\n      \\Bigl(t+3+3t^{-1}+t^{-2}\\Bigr)\\,\n      e^{-2024(t+t^{-1})+2023(t-t^{-1})}\\,dt.\n\\]\n\nStep 2.  Re-write the exponential so that each summand fits the same template  \n\\[\n-2024(t+t^{-1})+2023(t-t^{-1})\n      =-\\bigl(2024-2023\\bigr)t-\\bigl(2024+2023\\bigr)t^{-1}\n      =-t-4047\\,t^{-1}.\n\\]\nThus\n\\[\n\\mathcal{I}=\\int_{0}^{\\infty}\n             \\bigl(t+3+3t^{-1}+t^{-2}\\bigr)\\,\n             e^{-\\bigl(t+4047\\,t^{-1}\\bigr)}\\,dt .\n\\]\n\nStep 3.  Identify a standard integral  \nFor \\(\\alpha,\\beta>0\\) and any real \\(\\nu\\)\n\\[\n\\int_{0}^{\\infty}x^{\\nu-1}e^{-\\alpha x-\\beta/x}\\,dx\n      =2\\Bigl(\\frac{\\beta}{\\alpha}\\Bigr)^{\\nu/2}K_{\\nu}\\!\\bigl(2\\sqrt{\\alpha\\beta}\\bigr)\n      \\quad\\bigl(\\text{modified Bessel }K_{\\nu}\\bigr).\n\\]\nIn our case \\(\\alpha=1,\\;\\beta=4047\\) and we need the four exponents  \n\\(x^{k}\\) with \\(k=1,0,-1,-2\\).  Writing \\(\\nu=k+1\\) and letting\n\\[\n\\lambda:=2\\sqrt{4047},\n\\qquad\\beta:=4047,\n\\]\nwe obtain\n\\[\n\\begin{aligned}\nI_{1}&:=\\int_{0}^{\\infty} t^{ 1}\\,e^{-(t+\\beta t^{-1})}\\,dt\n      =2\\beta K_{2}(\\lambda),\\\\[2mm]\nI_{0}&:=\\int_{0}^{\\infty} t^{ 0}\\,e^{-(t+\\beta t^{-1})}\\,dt\n      =2\\beta^{1/2}K_{1}(\\lambda),\\\\[2mm]\nI_{-1}&:=\\int_{0}^{\\infty} t^{-1}\\,e^{-(t+\\beta t^{-1})}\\,dt\n      =2 K_{0}(\\lambda),\\\\[2mm]\nI_{-2}&:=\\int_{0}^{\\infty} t^{-2}\\,e^{-(t+\\beta t^{-1})}\\,dt\n      =2\\beta^{-1/2}K_{1}(\\lambda),\n\\end{aligned}\n\\]\nbecause \\(K_{-\\nu}=K_{\\nu}\\).\n\nStep 4.  Assemble the pieces  \n\\[\n\\mathcal{I}=I_{1}+3I_{0}+3I_{-1}+I_{-2}\n           =2\\Bigl[\\beta K_{2}(\\lambda)\n                    +(3\\beta^{1/2}+\\beta^{-1/2})K_{1}(\\lambda)\n                    +3K_{0}(\\lambda)\\Bigr].\n\\]\n\nStep 5.  Remove \\(K_{2}\\) so that only the two lowest orders remain  \nThe recurrence \\(K_{\\nu+1}(z)=K_{\\nu-1}(z)+\\tfrac{2\\nu}{z}K_{\\nu}(z)\\) with \\(\\nu=1\\) gives  \n\\(K_{2}(\\lambda)=K_{0}(\\lambda)+\\frac{2}{\\lambda}K_{1}(\\lambda)\\).\nHence\n\\[\n\\begin{aligned}\n\\mathcal{I}\n&=2\\Bigl[\\beta\\bigl(K_{0}(\\lambda)+\\tfrac{2}{\\lambda}K_{1}(\\lambda)\\bigr)\n          +(3\\beta^{1/2}+\\beta^{-1/2})K_{1}(\\lambda)\n          +3K_{0}(\\lambda)\\Bigr]\\\\[2mm]\n&=2(\\beta+3)K_{0}(\\lambda)\n  +2\\Bigl(\\tfrac{2\\beta}{\\lambda}+3\\beta^{1/2}+\\beta^{-1/2}\\Bigr)K_{1}(\\lambda).\n\\end{aligned}\n\\]\n\nStep 6.  Insert the numeric value \\(\\beta=4047,\\;\\lambda=2\\sqrt{4047}\\)  \n\\[\n\\boxed{%\n\\displaystyle\n\\mathcal{I}\n   =8100\\,K_{0}\\!\\bigl(2\\sqrt{4047}\\bigr)\n    +\\Bigl(8\\sqrt{4047}+ \\frac{2}{\\sqrt{4047}}\\Bigr)\\,\n      K_{1}\\!\\bigl(2\\sqrt{4047}\\bigr)\n}.\n\\]\n(Any equivalent reduction—e.g.\\ leaving the answer in the form of Step 4—is also acceptable.)\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.539089",
        "was_fixed": false,
        "difficulty_analysis": "1.  Extra algebraic factor  \n   •  The cubic \\((t^{1/2}+t^{-1/2})^{3}\\) produces four separate power terms, forcing the solver to decompose the integral into a linear combination of four different Mellin–type integrals.\n\n2.  Mixed linear combination in the exponent  \n   •  The simultaneous appearance of \\(t+t^{-1}\\) and \\(t-t^{-1}\\) means the exponent cannot be handled by the single substitution \\(u=t^{1/2}-t^{-1/2}\\) that solved the original problem.  One must instead recognise how to rewrite the exponential as \\(-\\alpha t-\\beta t^{-1}\\) with different \\(\\alpha,\\beta\\).\n\n3.  Higher–order Bessel functions  \n   •  The decomposition yields integrals involving \\(K_{0},K_{1},K_{2}\\) (orders 0, 1, 2) instead of the single half-integer \\(K_{1/2}\\) that collapses to an elementary exponential in the original problem.  The solver must know both the general integral representation of \\(K_{\\nu}\\) and the three-term recurrence to simplify the final combination.\n\n4.  Multiple interacting concepts  \n   •  Mastery of binomial expansions of fractional powers, asymptotic convergence arguments, the Mellin transform, special-function identities and recurrences are all required.  No single “lucky’’ substitution will dispatch the integral; several layers of technique must be coordinated.\n\nOverall, the enhanced variant introduces a non-trivial algebraic prefactor and a skew exponent, pushes the calculation up to higher (non-half-integer) orders of the modified Bessel function, and obliges the solver to manipulate those functions with recurrence relations—rendering the task substantially more intricate than either the original problem or the simpler current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}