{
  "index": "1982-A-3",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Problem A-3\nEvaluate\n\\[\n\\int_{0}^{\\infty} \\frac{\\operatorname{Arctan}(\\pi x)-\\operatorname{Arctan} x}{x} d x\n\\]",
  "solution": "\\begin{array}{l}\n\\text { A-3. }\\\\\n\\begin{aligned}\n\\int_{0}^{x} \\frac{\\operatorname{Arctan}(\\pi x)}{x} \\operatorname{Arctan} x d x & =\\left.\\int_{01}^{x} \\frac{1}{x} \\operatorname{Arctan}(u x)\\right|_{u=1} ^{u=\\pi} d x \\\\\n& =\\int_{0}^{x} \\int_{1}^{\\pi} \\frac{1}{1+(x u)^{2}} d u d x=\\int_{1}^{\\pi} \\int_{0}^{x} \\frac{1}{1+(x u)^{2}} d x d u \\\\\n& =\\int_{1}^{\\pi} \\frac{1}{u} \\cdot \\frac{\\pi}{2} d u=\\frac{\\pi}{2} \\ln \\pi\n\\end{aligned}\n\\end{array}",
  "vars": [
    "x",
    "u"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "positivereal",
        "u": "auxiliary"
      },
      "question": "\\[\n\\int_{0}^{\\infty} \\frac{\\operatorname{Arctan}(\\pi \\mathrm{positivereal})-\\operatorname{Arctan} \n\\mathrm{positivereal}}{\\mathrm{positivereal}} \\, d\\mathrm{positivereal}\n\\]",
      "solution": "\\begin{array}{l}\n\\text { A-3. }\\\\\n\\begin{aligned}\n\\int_{0}^{\\mathrm{positivereal}} \\frac{\\operatorname{Arctan}(\\pi \n\\mathrm{positivereal})}{\\mathrm{positivereal}} \\, \\operatorname{Arctan} \\mathrm{positivereal}\n\\, d\\mathrm{positivereal} & =\\left.\\int_{01}^{\\mathrm{positivereal}} \\frac{1}{\\mathrm{positivereal}} \\, \\operatorname{Arctan}(\\mathrm{auxiliary}\\,\\mathrm{positivereal})\\right|_{\\mathrm{auxiliary}=1} ^{\\mathrm{auxiliary}=\\pi} d\\mathrm{positivereal} \\\\\n& =\\int_{0}^{\\mathrm{positivereal}} \\int_{1}^{\\pi} \\frac{1}{1+(\\mathrm{positivereal}\\,\\mathrm{auxiliary})^{2}} \\, d\\mathrm{auxiliary} \\, d\\mathrm{positivereal}=\\int_{1}^{\\pi} \\int_{0}^{\\mathrm{positivereal}} \\frac{1}{1+(\\mathrm{positivereal}\\,\\mathrm{auxiliary})^{2}} \\, d\\mathrm{positivereal} \\, d\\mathrm{auxiliary} \\\\\n& =\\int_{1}^{\\pi} \\frac{1}{\\mathrm{auxiliary}} \\cdot \\frac{\\pi}{2} \\, d\\mathrm{auxiliary}=\\frac{\\pi}{2} \\ln \\pi\n\\end{aligned}\n\\end{array}"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "longitude",
        "u": "capacity"
      },
      "question": "Problem A-3\nEvaluate\n\\[\n\\int_{0}^{\\infty} \\frac{\\operatorname{Arctan}(\\pi longitude)-\\operatorname{Arctan} longitude}{longitude} d longitude\n\\]",
      "solution": "\\begin{array}{l}\n\\text { A-3. }\\\\\n\\begin{aligned}\n\\int_{0}^{longitude} \\frac{\\operatorname{Arctan}(\\pi longitude)}{longitude} \\operatorname{Arctan} longitude d longitude & =\\left.\\int_{01}^{longitude} \\frac{1}{longitude} \\operatorname{Arctan}(capacity longitude)\\right|_{capacity=1} ^{capacity=\\pi} d longitude \\\\\n& =\\int_{0}^{longitude} \\int_{1}^{\\pi} \\frac{1}{1+(longitude capacity)^{2}} d capacity d longitude=\\int_{1}^{\\pi} \\int_{0}^{longitude} \\frac{1}{1+(longitude capacity)^{2}} d longitude d capacity \\\\\n& =\\int_{1}^{\\pi} \\frac{1}{capacity} \\cdot \\frac{\\pi}{2} d capacity=\\frac{\\pi}{2} \\ln \\pi\n\\end{aligned}\n\\end{array}"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "u": "steadfast"
      },
      "question": "Problem A-3\nEvaluate\n\\[\n\\int_{0}^{\\infty} \\frac{\\operatorname{Arctan}(\\pi constantval)-\\operatorname{Arctan} constantval}{constantval} d constantval\n\\]",
      "solution": "\\begin{array}{l}\n\\text { A-3. }\\\\\n\\begin{aligned}\n\\int_{0}^{constantval} \\frac{\\operatorname{Arctan}(\\pi constantval)}{constantval} \\operatorname{Arctan} constantval d constantval & =\\left.\\int_{01}^{constantval} \\frac{1}{constantval} \\operatorname{Arctan}(steadfast constantval)\\right|_{steadfast=1} ^{steadfast=\\pi} d constantval \\\\\n& =\\int_{0}^{constantval} \\int_{1}^{\\pi} \\frac{1}{1+(constantval steadfast)^{2}} d steadfast d constantval=\\int_{1}^{\\pi} \\int_{0}^{constantval} \\frac{1}{1+(constantval steadfast)^{2}} d constantval d steadfast \\\\\n& =\\int_{1}^{\\pi} \\frac{1}{steadfast} \\cdot \\frac{\\pi}{2} d steadfast=\\frac{\\pi}{2} \\ln \\pi\n\\end{aligned}\n\\end{array}"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "u": "hjgrksla"
      },
      "question": "Problem A-3\nEvaluate\n\\[\n\\int_{0}^{\\infty} \\frac{\\operatorname{Arctan}(\\pi qzxwvtnp)-\\operatorname{Arctan} qzxwvtnp}{qzxwvtnp} d qzxwvtnp\n\\]",
      "solution": "\\begin{array}{l}\n\\text { A-3. }\\\\\n\\begin{aligned}\n\\int_{0}^{qzxwvtnp} \\frac{\\operatorname{Arctan}(\\pi qzxwvtnp)}{qzxwvtnp} \\operatorname{Arctan} qzxwvtnp d qzxwvtnp & =\\left.\\int_{01}^{qzxwvtnp} \\frac{1}{qzxwvtnp} \\operatorname{Arctan}(hjgrksla qzxwvtnp)\\right|_{hjgrksla=1} ^{hjgrksla=\\pi} d qzxwvtnp \\\\\n& =\\int_{0}^{qzxwvtnp} \\int_{1}^{\\pi} \\frac{1}{1+(qzxwvtnp hjgrksla)^{2}} d hjgrksla d qzxwvtnp=\\int_{1}^{\\pi} \\int_{0}^{qzxwvtnp} \\frac{1}{1+(qzxwvtnp hjgrksla)^{2}} d qzxwvtnp d hjgrksla \\\\\n& =\\int_{1}^{\\pi} \\frac{1}{hjgrksla} \\cdot \\frac{\\pi}{2} d hjgrksla=\\frac{\\pi}{2} \\ln \\pi\n\\end{aligned}\n\\end{array}"
    },
    "kernel_variant": {
      "question": "Evaluate the improper integral  \n\\[\nJ \\;=\\;\\int_{-\\infty}^{0}\\;\n        \\frac{\\displaystyle\\Arctan\\!\\bigl(8x\\bigr)\\;-\\;\\Arctan\\!\\bigl(2x\\bigr)}\n             {x\\,(1+x^{2})^{2}}\\,dx .\n\\]",
      "solution": "\\textbf{1.  Reduction to an integral over positive arguments}\n\nThe numerator  \n\\(\n\\Arctan(8x)-\\Arctan(2x)\n\\)  \nis an odd function of \\(x\\), whereas the factor \\(x\\) in the denominator is also\nodd and \\((1+x^{2})^{2}\\) is even.  Hence the whole integrand is even, and\n\\[\nJ\n   =\\int_{0}^{\\infty}\n      \\frac{\\Arctan(8t)-\\Arctan(2t)}\n           {t\\,(1+t^{2})^{2}}\\;dt .\n\\tag{1}\n\\]\n\n\\textbf{2.  Introducing a parameter}\n\nFor \\(a>0\\) define the one–parameter family  \n\\[\nF(a)\\;:=\\;\\int_{0}^{\\infty}\n            \\frac{\\Arctan(a t)}{t\\,(1+t^{2})^{2}}\\;dt ,\n\\qquad\\text{so that}\\qquad\nJ \\;=\\;F(8)-F(2).\n\\tag{2}\n\\]\n\n\\textbf{3.  Differentiation under the integral sign}\n\nNear \\(t=0\\) the integrand behaves like \\(a\\), while for\n\\(t\\to\\infty\\) it decays like \\(t^{-3}\\); hence dominated\nconvergence justifies differentiating under the integral sign:\n\\[\nF'(a)\n     =\\int_{0}^{\\infty}\n       \\frac{1}{(1+t^{2})^{2}\\,\\bigl(1+a^{2}t^{2}\\bigr)}\\;dt\n     \\;=:\\;I(a).\n\\tag{3}\n\\]\n\n\\textbf{4.  Evaluation of the kernel integral \\(I(a)\\)}\n\nWe write\n\\[\n\\frac{1}\n     {(1+t^{2})^{2}\\bigl(1+a^{2}t^{2}\\bigr)}\n =\\frac{A}{1+t^{2}}\n  +\\frac{B}{(1+t^{2})^{2}}\n  +\\frac{C}{1+a^{2}t^{2}},\n\\qquad a\\neq1 ,\n\\tag{4}\n\\]\nand determine \\(A,B,C\\) by clearing denominators.  \nSetting \\(s=t^{2}\\) turns (4) into the polynomial identity\n\\[\n1\n  =A\\bigl(1+s\\bigr)\\bigl(1+a^{2}s\\bigr)\n   +B\\bigl(1+a^{2}s\\bigr)\n   +C\\bigl(1+2s+s^{2}\\bigr).\n\\]\nMatching the coefficients of \\(1,s,s^{2}\\) we obtain the linear system\n\\[\n\\begin{aligned}\nA+B+C&=1,\\\\\nA(1+a^{2})+B a^{2}+2C&=0,\\\\\nA a^{2}+C&=0 .\n\\end{aligned}\n\\]\nSolving,\n\\[\nA=\\frac{a^{2}}{\\,2a^{2}-a^{4}-1},\\qquad\nB=\\frac{a^{2}-1}{\\,2a^{2}-a^{4}-1},\\qquad\nC=-\\frac{a^{4}}{\\,2a^{2}-a^{4}-1}.\n\\tag{5}\n\\]\n(The special case \\(a=1\\) follows by continuity.)\n\nInsert (5) into (4) and use the elementary integrals\n\\[\n\\int_{0}^{\\infty}\\frac{dt}{1+t^{2}}=\\frac{\\pi}{2},\\quad\n\\int_{0}^{\\infty}\\frac{dt}{(1+t^{2})^{2}}=\\frac{\\pi}{4},\\quad\n\\int_{0}^{\\infty}\\frac{dt}{1+a^{2}t^{2}}=\\frac{\\pi}{2a},\n\\]\nto get\n\\[\n\\begin{aligned}\nI(a)\n  &=A\\frac{\\pi}{2}+B\\frac{\\pi}{4}+C\\frac{\\pi}{2a}\\\\[2pt]\n  &=\\frac{\\pi}{4}\n     \\Bigl[\n        \\frac{2a^{2}}{D}\n       +\\frac{a^{2}-1}{D}\n       -\\frac{2a^{3}}{D}\n     \\Bigr]\n     \\qquad\\bigl(D=2a^{2}-a^{4}-1\\bigr)\\\\[2pt]\n  &=\\frac{\\pi}{4}\\,\\frac{1+2a}{(1+a)^{2}}.\n\\end{aligned}\n\\tag{6}\n\\]\n(One may verify the last simplification algebraically or by noticing\nthat both sides coincide for three distinct values of \\(a\\) and obey\nthe same first–order differential equation.)\n\n\\textbf{5.  Integration of \\(F'(a)\\)}\n\nBecause \\(F(0)=0\\), integrate (6) from \\(0\\) to \\(a\\):\n\\[\n\\begin{aligned}\nF(a)\n &=\\frac{\\pi}{4}\\int_{0}^{a}\\frac{1+2u}{(1+u)^{2}}\\;du\\\\[4pt]\n &=\\frac{\\pi}{4}\\int_{1}^{1+a}\n     \\Bigl(\\frac{2}{w}-\\frac{1}{w^{2}}\\Bigr)\\,dw\n     \\quad(w=1+u)\\\\[4pt]\n &=\\frac{\\pi}{4}\\Bigl(2\\ln w+\\frac{1}{w}\\Bigr)_{1}^{1+a}\\\\[4pt]\n &=\\frac{\\pi}{2}\\,\\ln(1+a)\n   +\\frac{\\pi}{4}\\Bigl(\\frac{1}{1+a}-1\\Bigr).\n\\end{aligned}\n\\tag{7}\n\\]\n\n\\textbf{6.  Computing \\(J\\)}\n\n\\[\n\\begin{aligned}\nJ\n  &=F(8)-F(2)\\\\[2pt]\n  &=\\frac{\\pi}{2}\\bigl[\\ln(9)-\\ln(3)\\bigr]\n    +\\frac{\\pi}{4}\\Bigl(\\frac{1}{9}-\\frac{1}{3}\\Bigr)\\\\[4pt]\n  &=\\frac{\\pi}{2}\\,\\ln 3\\;-\\;\\frac{\\pi}{18}.\n\\end{aligned}\n\\]\n\n\\[\n\\boxed{\\,J=\\displaystyle\\frac{\\pi}{2}\\,\\ln 3-\\frac{\\pi}{18}\\,}.\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.663340",
        "was_fixed": false,
        "difficulty_analysis": "•  Extra denominator.  Unlike the original kernel (which had only the factor \\(x\\) in the denominator), the new integral contains the additional factor \\(1+x^{2}\\).  This destroys the simple double–integral trick and forces a more elaborate approach.\n\n•  Parameter calculus.  The solution introduces a whole family \\(F(a)\\) and differentiates under the integral sign.  Mastering such a technique and justifying each step is beyond the scope of the routine manipulations that solve the original problem.\n\n•  Partial–fraction decomposition and classical integrals.  Computing \\(F'(a)\\) requires splitting the rational kernel into two pieces and knowing (or deriving) the classical integrals\n  \\(\\int_{0}^{\\infty}\\!\\!dt/(1+t^{2})=\\pi/2\\) and \n  \\(\\int_{0}^{\\infty}\\!\\!dt/(1+a^{2}t^{2})=\\pi/(2a)\\).\n\n•  Logarithmic antiderivative.  Integrating \\(F'(a)\\) with respect to \\(a\\) introduces an additional layer of analysis (and a boundary-value argument) before the final logarithmic expression appears.\n\n•  Overall structure.  Compared with the original single–line evaluation leading straight to \\((\\pi/2)\\ln\\pi\\), the enhanced variant demands:   \n  1) a symmetry reduction,  \n  2) creation of a parameter family,  \n  3) differentiation under the integral sign,  \n  4) decomposition of rational functions,  \n  5) two classical definite integrals, and  \n  6) an integration in the parameter space.  \n\nThese extra steps substantially deepen both the technical and conceptual workload, making the enhanced kernel variant markedly harder than the original."
      }
    },
    "original_kernel_variant": {
      "question": "Evaluate the improper integral  \n\\[\nJ \\;=\\;\\int_{-\\infty}^{0}\\;\n        \\frac{\\displaystyle\\Arctan\\!\\bigl(8x\\bigr)\\;-\\;\\Arctan\\!\\bigl(2x\\bigr)}\n             {x\\,(1+x^{2})^{2}}\\,dx .\n\\]",
      "solution": "\\textbf{1.  Reduction to an integral over positive arguments}\n\nThe numerator  \n\\(\n\\Arctan(8x)-\\Arctan(2x)\n\\)  \nis an odd function of \\(x\\), whereas the factor \\(x\\) in the denominator is also\nodd and \\((1+x^{2})^{2}\\) is even.  Hence the whole integrand is even, and\n\\[\nJ\n   =\\int_{0}^{\\infty}\n      \\frac{\\Arctan(8t)-\\Arctan(2t)}\n           {t\\,(1+t^{2})^{2}}\\;dt .\n\\tag{1}\n\\]\n\n\\textbf{2.  Introducing a parameter}\n\nFor \\(a>0\\) define the one–parameter family  \n\\[\nF(a)\\;:=\\;\\int_{0}^{\\infty}\n            \\frac{\\Arctan(a t)}{t\\,(1+t^{2})^{2}}\\;dt ,\n\\qquad\\text{so that}\\qquad\nJ \\;=\\;F(8)-F(2).\n\\tag{2}\n\\]\n\n\\textbf{3.  Differentiation under the integral sign}\n\nNear \\(t=0\\) the integrand behaves like \\(a\\), while for\n\\(t\\to\\infty\\) it decays like \\(t^{-3}\\); hence dominated\nconvergence justifies differentiating under the integral sign:\n\\[\nF'(a)\n     =\\int_{0}^{\\infty}\n       \\frac{1}{(1+t^{2})^{2}\\,\\bigl(1+a^{2}t^{2}\\bigr)}\\;dt\n     \\;=:\\;I(a).\n\\tag{3}\n\\]\n\n\\textbf{4.  Evaluation of the kernel integral \\(I(a)\\)}\n\nWe write\n\\[\n\\frac{1}\n     {(1+t^{2})^{2}\\bigl(1+a^{2}t^{2}\\bigr)}\n =\\frac{A}{1+t^{2}}\n  +\\frac{B}{(1+t^{2})^{2}}\n  +\\frac{C}{1+a^{2}t^{2}},\n\\qquad a\\neq1 ,\n\\tag{4}\n\\]\nand determine \\(A,B,C\\) by clearing denominators.  \nSetting \\(s=t^{2}\\) turns (4) into the polynomial identity\n\\[\n1\n  =A\\bigl(1+s\\bigr)\\bigl(1+a^{2}s\\bigr)\n   +B\\bigl(1+a^{2}s\\bigr)\n   +C\\bigl(1+2s+s^{2}\\bigr).\n\\]\nMatching the coefficients of \\(1,s,s^{2}\\) we obtain the linear system\n\\[\n\\begin{aligned}\nA+B+C&=1,\\\\\nA(1+a^{2})+B a^{2}+2C&=0,\\\\\nA a^{2}+C&=0 .\n\\end{aligned}\n\\]\nSolving,\n\\[\nA=\\frac{a^{2}}{\\,2a^{2}-a^{4}-1},\\qquad\nB=\\frac{a^{2}-1}{\\,2a^{2}-a^{4}-1},\\qquad\nC=-\\frac{a^{4}}{\\,2a^{2}-a^{4}-1}.\n\\tag{5}\n\\]\n(The special case \\(a=1\\) follows by continuity.)\n\nInsert (5) into (4) and use the elementary integrals\n\\[\n\\int_{0}^{\\infty}\\frac{dt}{1+t^{2}}=\\frac{\\pi}{2},\\quad\n\\int_{0}^{\\infty}\\frac{dt}{(1+t^{2})^{2}}=\\frac{\\pi}{4},\\quad\n\\int_{0}^{\\infty}\\frac{dt}{1+a^{2}t^{2}}=\\frac{\\pi}{2a},\n\\]\nto get\n\\[\n\\begin{aligned}\nI(a)\n  &=A\\frac{\\pi}{2}+B\\frac{\\pi}{4}+C\\frac{\\pi}{2a}\\\\[2pt]\n  &=\\frac{\\pi}{4}\n     \\Bigl[\n        \\frac{2a^{2}}{D}\n       +\\frac{a^{2}-1}{D}\n       -\\frac{2a^{3}}{D}\n     \\Bigr]\n     \\qquad\\bigl(D=2a^{2}-a^{4}-1\\bigr)\\\\[2pt]\n  &=\\frac{\\pi}{4}\\,\\frac{1+2a}{(1+a)^{2}}.\n\\end{aligned}\n\\tag{6}\n\\]\n(One may verify the last simplification algebraically or by noticing\nthat both sides coincide for three distinct values of \\(a\\) and obey\nthe same first–order differential equation.)\n\n\\textbf{5.  Integration of \\(F'(a)\\)}\n\nBecause \\(F(0)=0\\), integrate (6) from \\(0\\) to \\(a\\):\n\\[\n\\begin{aligned}\nF(a)\n &=\\frac{\\pi}{4}\\int_{0}^{a}\\frac{1+2u}{(1+u)^{2}}\\;du\\\\[4pt]\n &=\\frac{\\pi}{4}\\int_{1}^{1+a}\n     \\Bigl(\\frac{2}{w}-\\frac{1}{w^{2}}\\Bigr)\\,dw\n     \\quad(w=1+u)\\\\[4pt]\n &=\\frac{\\pi}{4}\\Bigl(2\\ln w+\\frac{1}{w}\\Bigr)_{1}^{1+a}\\\\[4pt]\n &=\\frac{\\pi}{2}\\,\\ln(1+a)\n   +\\frac{\\pi}{4}\\Bigl(\\frac{1}{1+a}-1\\Bigr).\n\\end{aligned}\n\\tag{7}\n\\]\n\n\\textbf{6.  Computing \\(J\\)}\n\n\\[\n\\begin{aligned}\nJ\n  &=F(8)-F(2)\\\\[2pt]\n  &=\\frac{\\pi}{2}\\bigl[\\ln(9)-\\ln(3)\\bigr]\n    +\\frac{\\pi}{4}\\Bigl(\\frac{1}{9}-\\frac{1}{3}\\Bigr)\\\\[4pt]\n  &=\\frac{\\pi}{2}\\,\\ln 3\\;-\\;\\frac{\\pi}{18}.\n\\end{aligned}\n\\]\n\n\\[\n\\boxed{\\,J=\\displaystyle\\frac{\\pi}{2}\\,\\ln 3-\\frac{\\pi}{18}\\,}.\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.520880",
        "was_fixed": false,
        "difficulty_analysis": "•  Extra denominator.  Unlike the original kernel (which had only the factor \\(x\\) in the denominator), the new integral contains the additional factor \\(1+x^{2}\\).  This destroys the simple double–integral trick and forces a more elaborate approach.\n\n•  Parameter calculus.  The solution introduces a whole family \\(F(a)\\) and differentiates under the integral sign.  Mastering such a technique and justifying each step is beyond the scope of the routine manipulations that solve the original problem.\n\n•  Partial–fraction decomposition and classical integrals.  Computing \\(F'(a)\\) requires splitting the rational kernel into two pieces and knowing (or deriving) the classical integrals\n  \\(\\int_{0}^{\\infty}\\!\\!dt/(1+t^{2})=\\pi/2\\) and \n  \\(\\int_{0}^{\\infty}\\!\\!dt/(1+a^{2}t^{2})=\\pi/(2a)\\).\n\n•  Logarithmic antiderivative.  Integrating \\(F'(a)\\) with respect to \\(a\\) introduces an additional layer of analysis (and a boundary-value argument) before the final logarithmic expression appears.\n\n•  Overall structure.  Compared with the original single–line evaluation leading straight to \\((\\pi/2)\\ln\\pi\\), the enhanced variant demands:   \n  1) a symmetry reduction,  \n  2) creation of a parameter family,  \n  3) differentiation under the integral sign,  \n  4) decomposition of rational functions,  \n  5) two classical definite integrals, and  \n  6) an integration in the parameter space.  \n\nThese extra steps substantially deepen both the technical and conceptual workload, making the enhanced kernel variant markedly harder than the original."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}