{
  "index": "1980-A-3",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Problem A-3\nEvaluate\n\\[\n\\int_{0}^{\\pi / 2} \\frac{d x}{1+(\\tan x)^{\\sqrt{2}}}\n\\]",
  "solution": "A-3.\nLet \\( I \\) be the given definite integral and \\( \\sqrt{2}=r \\). We show that \\( I=\\pi / 4 \\). Using \\( x=(\\pi / 2)-u \\), one has\n\\[\nI=\\int_{\\pi / 2}^{0} \\frac{-d u}{1+\\cot ^{r} u}=\\int_{0}^{\\pi / 2} \\frac{\\tan ^{r} u d u}{\\tan ^{r} u+1} .\n\\]\n\nHence\n\\[\n2 I=\\int_{0}^{\\pi / 2} \\frac{1+\\tan ^{r} x}{1+\\tan ^{r} x} d x=\\int_{0}^{\\pi / 2} d x=\\pi / 2 \\quad \\text { and } \\quad I=\\pi / 4\n\\]",
  "vars": [
    "x",
    "u"
  ],
  "params": [
    "I",
    "r"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "anglevar",
        "u": "altangle",
        "I": "integralvalue",
        "r": "sqrtconst"
      },
      "question": "Problem A-3\nEvaluate\n\\[\n\\int_{0}^{\\pi / 2} \\frac{d anglevar}{1+(\\tan anglevar)^{\\sqrt{2}}}\n\\]",
      "solution": "A-3.\nLet \\( integralvalue \\) be the given definite integral and \\( \\sqrt{2}=sqrtconst \\). We show that \\( integralvalue=\\pi / 4 \\). Using \\( anglevar=(\\pi / 2)-altangle \\), one has\n\\[\nintegralvalue=\\int_{\\pi / 2}^{0} \\frac{-d altangle}{1+\\cot ^{sqrtconst} altangle}=\\int_{0}^{\\pi / 2} \\frac{\\tan ^{sqrtconst} altangle d altangle}{\\tan ^{sqrtconst} altangle+1} .\n\\]\n\nHence\n\\[\n2 integralvalue=\\int_{0}^{\\pi / 2} \\frac{1+\\tan ^{sqrtconst} anglevar}{1+\\tan ^{sqrtconst} anglevar} d anglevar=\\int_{0}^{\\pi / 2} d anglevar=\\pi / 2 \\quad \\text { and } \\quad integralvalue=\\pi / 4\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "buttercup",
        "u": "hummingbird",
        "I": "waterspout",
        "r": "thunderbolt"
      },
      "question": "Problem A-3\nEvaluate\n\\[\n\\int_{0}^{\\pi / 2} \\frac{d buttercup}{1+(\\tan buttercup)^{\\sqrt{2}}}\n\\]\n",
      "solution": "A-3.\nLet \\( waterspout \\) be the given definite integral and \\( \\sqrt{2}=thunderbolt \\). We show that \\( waterspout=\\pi / 4 \\). Using \\( buttercup=(\\pi / 2)-hummingbird \\), one has\n\\[\nwaterspout=\\int_{\\pi / 2}^{0} \\frac{-d hummingbird}{1+\\cot ^{thunderbolt} hummingbird}=\\int_{0}^{\\pi / 2} \\frac{\\tan ^{thunderbolt} hummingbird d hummingbird}{\\tan ^{thunderbolt} hummingbird+1} .\n\\]\n\nHence\n\\[\n2 waterspout=\\int_{0}^{\\pi / 2} \\frac{1+\\tan ^{thunderbolt} buttercup}{1+\\tan ^{thunderbolt} buttercup} d buttercup=\\int_{0}^{\\pi / 2} d buttercup=\\pi / 2 \\quad \\text { and } \\quad waterspout=\\pi / 4\n\\]\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "u": "outsideangle",
        "I": "derivativevalue",
        "r": "squarednumber"
      },
      "question": "Problem A-3\nEvaluate\n\\[\n\\int_{0}^{\\pi / 2} \\frac{d verticalaxis}{1+(\\tan verticalaxis)^{\\sqrt{2}}}\n\\]",
      "solution": "A-3.\nLet \\( derivativevalue \\) be the given definite integral and \\( \\sqrt{2}=squarednumber \\). We show that \\( derivativevalue=\\pi / 4 \\). Using \\( verticalaxis=(\\pi / 2)-outsideangle \\), one has\n\\[\nderivativevalue=\\int_{\\pi / 2}^{0} \\frac{-d outsideangle}{1+\\cot ^{squarednumber} outsideangle}=\\int_{0}^{\\pi / 2} \\frac{\\tan ^{squarednumber} outsideangle d outsideangle}{\\tan ^{squarednumber} outsideangle+1} .\n\\]\nHence\n\\[\n2 derivativevalue=\\int_{0}^{\\pi / 2} \\frac{1+\\tan ^{squarednumber} verticalaxis}{1+\\tan ^{squarednumber} verticalaxis} d verticalaxis=\\int_{0}^{\\pi / 2} d verticalaxis=\\pi / 2 \\quad \\text { and } \\quad derivativevalue=\\pi / 4\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "u": "hjgrksla",
        "I": "mnvfrtle",
        "r": "pxqzsndk"
      },
      "question": "Problem A-3\nEvaluate\n\\[\n\\int_{0}^{\\pi / 2} \\frac{d qzxwvtnp}{1+(\\tan qzxwvtnp)^{\\sqrt{2}}}\n\\]",
      "solution": "A-3.\nLet \\( mnvfrtle \\) be the given definite integral and \\( \\sqrt{2}=pxqzsndk \\). We show that \\( mnvfrtle=\\pi / 4 \\). Using \\( qzxwvtnp=(\\pi / 2)-hjgrksla \\), one has\n\\[\nmnvfrtle=\\int_{\\pi / 2}^{0} \\frac{-d hjgrksla}{1+\\cot ^{pxqzsndk} hjgrksla}=\\int_{0}^{\\pi / 2} \\frac{\\tan ^{pxqzsndk} hjgrksla d hjgrksla}{\\tan ^{pxqzsndk} hjgrksla+1} .\n\\]\n\nHence\n\\[\n2 mnvfrtle=\\int_{0}^{\\pi / 2} \\frac{1+\\tan ^{pxqzsndk} qzxwvtnp}{1+\\tan ^{pxqzsndk} qzxwvtnp} d qzxwvtnp=\\int_{0}^{\\pi / 2} d qzxwvtnp=\\pi / 2 \\quad \\text { and } \\quad mnvfrtle=\\pi / 4\n\\]\n"
    },
    "kernel_variant": {
      "question": "Fix an integer n ≥ 1, a real parameter m > 0, and let r be an arbitrary positive real number (for instance r = √2, but the statement is to be proved for every r > 0).  Evaluate the n–fold integral  \n\n  Iₙ(m,r) = ∫_{0}^{π/2}⋯∫_{0}^{π/2}  \n                      ∏_{k=1}^{n} (sin^{m-1}x_k · cos^{m-1}x_k)  \n                      ────────────────────────────────────────────── dx₁…dxₙ,  \n                        1 + ( ∏_{k=1}^{n} tan^{r}x_k )\n\nand show that it can be expressed in closed Gamma–function form.  In particular,\n\n(a) prove that Iₙ(m,r) is completely independent of the chosen exponent r > 0, and  \n\n(b) show that the exact value is  \n\n  Iₙ(m,r) = 2^{-(n+1)} · [ Γ(m/2)² / Γ(m) ]^{\\,n}.",
      "solution": "Step 1.  Notation and a key functional identity.  \nPut  \n  P(x₁,…,xₙ) = ∏_{k=1}^{n} tan^{r}x_k,   f(t) = 1/(1+t).  \nObserve the elementary identity  \n\n  f(t) + f(1/t) = 1   for every t>0.  (1)\n\nStep 2.  A volume–preserving involution of the integration cube.  \nDefine the map  \n\n  Φ : [0,π/2]^{n} → [0,π/2]^{n}, Φ(x₁,…,xₙ) = (π/2–x₁,…,π/2–xₙ).\n\nΦ is an involution (Φ² = identity) whose Jacobian determinant equals 1, so it preserves Lebesgue measure.  For every point X = (x₁,…,xₙ),\n\n  P(Φ(X)) = ∏_{k=1}^{n} tan^{r}(π/2–x_k) = ∏_{k=1}^{n} cot^{r}x_k = 1/P(X). (2)\n\nStep 3.  Summing the integrand over the Φ–paired points.  \nLet  \n\n  μ(X) = ∏_{k=1}^{n} sin^{m-1}x_k · cos^{m-1}x_k  (the weight in the numerator).\n\nBecause sin(π/2–x) = cos x and cos(π/2–x) = sin x, we have\n\n  μ(Φ(X)) = μ(X);                  (3)\n\nhence the weight is Φ–invariant.\n\nNow add the integrand at X and at Φ(X): by (1)–(3)\n\n  μ(X)·f(P(X)) + μ(Φ(X))·f(P(Φ(X)))\n     = μ(X)·[ f(P(X)) + f(1/P(X)) ]\n     = μ(X).                (4)\n\nStep 4.  Integrating relation (4).  \nIntegrate both sides of (4) over the whole n–cube:\n\n  Iₙ(m,r) + Iₙ(m,r) = ∫_{[0,π/2]^n} μ(X) dX.  (5)\n\nTherefore  \n\n  2·Iₙ(m,r) = [ ∫_{0}^{π/2} sin^{m-1}x · cos^{m-1}x dx ]^{n}. (6)\n\n(The right–hand side factorises because μ is a product.)\n\nStep 5.  The one–dimensional Beta integral.  \nCompute once and for all  \n\n  J(m) = ∫_{0}^{π/2} sin^{m-1}x · cos^{m-1}x dx.\n\nPut u = sin²x ⇒ du = 2 sin x cos x dx.  Then\n\n  J(m) = ½ ∫_{0}^{1} u^{(m-2)/2}(1-u)^{(m-2)/2} du\n         = ½ B( m/2, m/2 )\n         = ½ Γ(m/2)² / Γ(m).  (7)\n\nStep 6.  Finishing the evaluation.  \nInsert (7) into (6):\n\n  2·Iₙ(m,r) = [ ½ Γ(m/2)² / Γ(m) ]^{n}.\n\nHence  \n\n  Iₙ(m,r) = 2^{-(n+1)} · [ Γ(m/2)² / Γ(m) ]^{\\,n}.  (8)\n\nEquation (8) is manifestly free of r; this proves both requested items (a) and (b).  ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.645658",
        "was_fixed": false,
        "difficulty_analysis": "• Dimensional escalation: the original single integral has been replaced by an n–fold integral over the n-cube [0, π/2]^n.  \n\n• Extra parameters: two continuous parameters (m and r) and one discrete parameter (n) must be handled simultaneously.  \n\n• Invariance arguments in higher dimensions: the solution hinges on constructing a measure-preserving involution Φ and pairing points X with Φ(X), a non-trivial generalisation of the simple x ↔ π/2–x trick in one dimension.\n\n• Beta/Gamma technology: evaluation of the remaining factor demands familiarity with Beta–function substitutions and Gamma–function identities, not needed in the original problem.\n\n• Independence proof: besides computing the integral, one must prove that the result does not depend on the exponent r, an additional conceptual layer absent from the original.\n\nThese cumulative complexities require deeper insight, multiple advanced techniques (symmetry in high dimension, special functions, measure-preserving maps), and a longer chain of logical steps, making the enhanced variant significantly more challenging than both the original and the existing kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer n ≥ 1, a real parameter m > 0, and let r be an arbitrary positive real number (for instance r = √2, but the statement is to be proved for every r > 0).  Evaluate the n–fold integral  \n\n  Iₙ(m,r) = ∫_{0}^{π/2}⋯∫_{0}^{π/2}  \n                      ∏_{k=1}^{n} (sin^{m-1}x_k · cos^{m-1}x_k)  \n                      ────────────────────────────────────────────── dx₁…dxₙ,  \n                        1 + ( ∏_{k=1}^{n} tan^{r}x_k )\n\nand show that it can be expressed in closed Gamma–function form.  In particular,\n\n(a) prove that Iₙ(m,r) is completely independent of the chosen exponent r > 0, and  \n\n(b) show that the exact value is  \n\n  Iₙ(m,r) = 2^{-(n+1)} · [ Γ(m/2)² / Γ(m) ]^{\\,n}.",
      "solution": "Step 1.  Notation and a key functional identity.  \nPut  \n  P(x₁,…,xₙ) = ∏_{k=1}^{n} tan^{r}x_k,   f(t) = 1/(1+t).  \nObserve the elementary identity  \n\n  f(t) + f(1/t) = 1   for every t>0.  (1)\n\nStep 2.  A volume–preserving involution of the integration cube.  \nDefine the map  \n\n  Φ : [0,π/2]^{n} → [0,π/2]^{n}, Φ(x₁,…,xₙ) = (π/2–x₁,…,π/2–xₙ).\n\nΦ is an involution (Φ² = identity) whose Jacobian determinant equals 1, so it preserves Lebesgue measure.  For every point X = (x₁,…,xₙ),\n\n  P(Φ(X)) = ∏_{k=1}^{n} tan^{r}(π/2–x_k) = ∏_{k=1}^{n} cot^{r}x_k = 1/P(X). (2)\n\nStep 3.  Summing the integrand over the Φ–paired points.  \nLet  \n\n  μ(X) = ∏_{k=1}^{n} sin^{m-1}x_k · cos^{m-1}x_k  (the weight in the numerator).\n\nBecause sin(π/2–x) = cos x and cos(π/2–x) = sin x, we have\n\n  μ(Φ(X)) = μ(X);                  (3)\n\nhence the weight is Φ–invariant.\n\nNow add the integrand at X and at Φ(X): by (1)–(3)\n\n  μ(X)·f(P(X)) + μ(Φ(X))·f(P(Φ(X)))\n     = μ(X)·[ f(P(X)) + f(1/P(X)) ]\n     = μ(X).                (4)\n\nStep 4.  Integrating relation (4).  \nIntegrate both sides of (4) over the whole n–cube:\n\n  Iₙ(m,r) + Iₙ(m,r) = ∫_{[0,π/2]^n} μ(X) dX.  (5)\n\nTherefore  \n\n  2·Iₙ(m,r) = [ ∫_{0}^{π/2} sin^{m-1}x · cos^{m-1}x dx ]^{n}. (6)\n\n(The right–hand side factorises because μ is a product.)\n\nStep 5.  The one–dimensional Beta integral.  \nCompute once and for all  \n\n  J(m) = ∫_{0}^{π/2} sin^{m-1}x · cos^{m-1}x dx.\n\nPut u = sin²x ⇒ du = 2 sin x cos x dx.  Then\n\n  J(m) = ½ ∫_{0}^{1} u^{(m-2)/2}(1-u)^{(m-2)/2} du\n         = ½ B( m/2, m/2 )\n         = ½ Γ(m/2)² / Γ(m).  (7)\n\nStep 6.  Finishing the evaluation.  \nInsert (7) into (6):\n\n  2·Iₙ(m,r) = [ ½ Γ(m/2)² / Γ(m) ]^{n}.\n\nHence  \n\n  Iₙ(m,r) = 2^{-(n+1)} · [ Γ(m/2)² / Γ(m) ]^{\\,n}.  (8)\n\nEquation (8) is manifestly free of r; this proves both requested items (a) and (b).  ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.513261",
        "was_fixed": false,
        "difficulty_analysis": "• Dimensional escalation: the original single integral has been replaced by an n–fold integral over the n-cube [0, π/2]^n.  \n\n• Extra parameters: two continuous parameters (m and r) and one discrete parameter (n) must be handled simultaneously.  \n\n• Invariance arguments in higher dimensions: the solution hinges on constructing a measure-preserving involution Φ and pairing points X with Φ(X), a non-trivial generalisation of the simple x ↔ π/2–x trick in one dimension.\n\n• Beta/Gamma technology: evaluation of the remaining factor demands familiarity with Beta–function substitutions and Gamma–function identities, not needed in the original problem.\n\n• Independence proof: besides computing the integral, one must prove that the result does not depend on the exponent r, an additional conceptual layer absent from the original.\n\nThese cumulative complexities require deeper insight, multiple advanced techniques (symmetry in high dimension, special functions, measure-preserving maps), and a longer chain of logical steps, making the enhanced variant significantly more challenging than both the original and the existing kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}