{
  "index": "1939-B-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "9. Evaluate the definite integrals\n(i) \\( \\int_{1}^{3} \\frac{d x}{\\sqrt{(x-1)(3-x)}} \\),\n(ii) \\( \\int_{1}^{\\infty} \\frac{d x}{e^{x+1}+e^{3-x}} \\).",
  "solution": "Solution. Part (i). Since the integrand is not defined at either bound of integration, one should write\n\\[\n\\begin{aligned}\n& \\int_{1}^{3} \\frac{d x}{\\sqrt{(x-1)(3-x)}}=\\lim _{\\substack{\\epsilon-0+\\\\\n\\delta-0+}} \\int_{1+\\epsilon}^{3-\\delta} \\frac{d x}{\\sqrt{(x-1)(3-x)}} \\\\\n= & \\left.\\lim _{\\substack{\\epsilon-0+\\\\\n\\delta-0+}} \\int_{1+\\epsilon}^{3-\\delta} \\frac{d x}{\\sqrt{1-(x-2)^{2}}}=\\lim _{\\substack{\\epsilon-0+\\\\\n\\delta-0+}} \\arcsin (x-2)\\right]_{1+\\epsilon}^{3-\\delta} \\\\\n= & \\lim _{\\substack{\\epsilon \\rightarrow 0+\\\\\n\\delta \\rightarrow 0+}}[\\arcsin (1-\\delta)-\\arcsin (\\epsilon-1)]=\\frac{\\pi}{2}+\\frac{\\pi}{2}=\\pi .\n\\end{aligned}\n\\]\n\nPart (ii). The difficulty here is with the infinite interval of integration. Let \\( y=x-1 \\); then\n\\[\n\\begin{array}{c}\n\\int \\frac{d x}{e^{x+1}+e^{3-x}}=\\frac{1}{e^{2}} \\int \\frac{d x}{e^{x-1}+e^{1-x}}=\\frac{1}{e^{2}} \\int \\frac{d y}{e^{y}+e^{-y}} \\\\\n=\\frac{1}{e^{2}} \\int \\frac{e^{y} d y}{e^{2 y}+1}=\\frac{1}{e^{2}} \\arctan e^{y}+c\n\\end{array}\n\\]\n\nHence\n\\[\n\\begin{aligned}\n\\int_{1}^{\\infty} \\frac{d x}{e^{x+1}+e^{3-x}} & =\\lim _{N \\rightarrow \\infty} \\int_{1}^{N} \\frac{d x}{e^{x+1}+e^{3-x}} \\\\\n& =\\lim _{N \\rightarrow \\infty} \\frac{1}{e^{2}}\\left[\\arctan e^{N-1}-\\arctan e^{0}\\right] \\\\\n& =\\frac{1}{e^{2}}\\left[\\frac{\\pi}{2}-\\frac{\\pi}{4}\\right]=\\frac{\\pi}{4 e^{2}} .\n\\end{aligned}\n\\]",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "N",
    "c",
    "\\\\epsilon",
    "\\\\delta"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "y": "ordinate",
        "N": "endpoint",
        "c": "constant",
        "\\epsilon": "tolerance",
        "\\delta": "deviation"
      },
      "question": "9. Evaluate the definite integrals\n(i) \\( \\int_{1}^{3} \\frac{d variable}{\\sqrt{(variable-1)(3-variable)}} \\),\n(ii) \\( \\int_{1}^{\\infty} \\frac{d variable}{e^{variable+1}+e^{3-variable}} \\).",
      "solution": "Solution. Part (i). Since the integrand is not defined at either bound of integration, one should write\n\\[\n\\begin{aligned}\n& \\int_{1}^{3} \\frac{d variable}{\\sqrt{(variable-1)(3-variable)}}=\\lim _{\\substack{tolerance-0+\\\\\ndeviation-0+}} \\int_{1+tolerance}^{3-deviation} \\frac{d variable}{\\sqrt{(variable-1)(3-variable)}} \\\\\n= & \\left.\\lim _{\\substack{tolerance-0+\\\\\ndeviation-0+}} \\int_{1+tolerance}^{3-deviation} \\frac{d variable}{\\sqrt{1-(variable-2)^{2}}}=\\lim _{\\substack{tolerance-0+\\\\\ndeviation-0+}} \\arcsin (variable-2)\\right]_{1+tolerance}^{3-deviation} \\\\\n= & \\lim _{\\substack{tolerance \\rightarrow 0+\\\\\ndeviation \\rightarrow 0+}}[\\arcsin (1-deviation)-\\arcsin (tolerance-1)]=\\frac{\\pi}{2}+\\frac{\\pi}{2}=\\pi .\n\\end{aligned}\n\\]\n\nPart (ii). The difficulty here is with the infinite interval of integration. Let \\( ordinate=variable-1 \\); then\n\\[\n\\begin{array}{c}\n\\int \\frac{d variable}{e^{variable+1}+e^{3-variable}}=\\frac{1}{e^{2}} \\int \\frac{d variable}{e^{variable-1}+e^{1-variable}}=\\frac{1}{e^{2}} \\int \\frac{d ordinate}{e^{ordinate}+e^{-ordinate}} \\\\\n=\\frac{1}{e^{2}} \\int \\frac{e^{ordinate} d ordinate}{e^{2 ordinate}+1}=\\frac{1}{e^{2}} \\arctan e^{ordinate}+constant\n\\end{array}\n\\]\n\nHence\n\\[\n\\begin{aligned}\n\\int_{1}^{\\infty} \\frac{d variable}{e^{variable+1}+e^{3-variable}} & =\\lim _{endpoint \\rightarrow \\infty} \\int_{1}^{endpoint} \\frac{d variable}{e^{variable+1}+e^{3-variable}} \\\\\n& =\\lim _{endpoint \\rightarrow \\infty} \\frac{1}{e^{2}}\\left[\\arctan e^{endpoint-1}-\\arctan e^{0}\\right] \\\\\n& =\\frac{1}{e^{2}}\\left[\\frac{\\pi}{2}-\\frac{\\pi}{4}\\right]=\\frac{\\pi}{4 e^{2}} .\n\\end{aligned}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marblestone",
        "y": "lanternpost",
        "N": "rainclouds",
        "c": "driftwood",
        "\\epsilon": "hazelnut",
        "\\delta": "buttercup"
      },
      "question": "9. Evaluate the definite integrals\n(i) \\( \\int_{1}^{3} \\frac{d marblestone}{\\sqrt{(marblestone-1)(3-marblestone)}} \\),\n(ii) \\( \\int_{1}^{\\infty} \\frac{d marblestone}{e^{marblestone+1}+e^{3-marblestone}} \\).",
      "solution": "Solution. Part (i). Since the integrand is not defined at either bound of integration, one should write\n\\[\n\\begin{aligned}\n& \\int_{1}^{3} \\frac{d marblestone}{\\sqrt{(marblestone-1)(3-marblestone)}}=\\lim _{\\substack{hazelnut-0+\\\\ buttercup-0+}} \\int_{1+hazelnut}^{3-buttercup} \\frac{d marblestone}{\\sqrt{(marblestone-1)(3-marblestone)}} \\\\\n= & \\left.\\lim _{\\substack{hazelnut-0+\\\\ buttercup-0+}} \\int_{1+hazelnut}^{3-buttercup} \\frac{d marblestone}{\\sqrt{1-(marblestone-2)^{2}}}=\\lim _{\\substack{hazelnut-0+\\\\ buttercup-0+}} \\arcsin (marblestone-2)\\right]_{1+hazelnut}^{3-buttercup} \\\\\n= & \\lim _{\\substack{hazelnut \\rightarrow 0+\\\\ buttercup \\rightarrow 0+}}[\\arcsin (1-buttercup)-\\arcsin (hazelnut-1)]=\\frac{\\pi}{2}+\\frac{\\pi}{2}=\\pi .\n\\end{aligned}\n\\]\n\nPart (ii). The difficulty here is with the infinite interval of integration. Let \\( lanternpost=marblestone-1 \\); then\n\\[\n\\begin{array}{c}\n\\int \\frac{d marblestone}{e^{marblestone+1}+e^{3-marblestone}}=\\frac{1}{e^{2}} \\int \\frac{d marblestone}{e^{marblestone-1}+e^{1-marblestone}}=\\frac{1}{e^{2}} \\int \\frac{d lanternpost}{e^{lanternpost}+e^{-lanternpost}} \\\\\n=\\frac{1}{e^{2}} \\int \\frac{e^{lanternpost} d lanternpost}{e^{2 lanternpost}+1}=\\frac{1}{e^{2}} \\arctan e^{lanternpost}+driftwood\n\\end{array}\n\\]\n\nHence\n\\[\n\\begin{aligned}\n\\int_{1}^{\\infty} \\frac{d marblestone}{e^{marblestone+1}+e^{3-marblestone}} & =\\lim _{rainclouds \\rightarrow \\infty} \\int_{1}^{rainclouds} \\frac{d marblestone}{e^{marblestone+1}+e^{3-marblestone}} \\\\\n& =\\lim _{rainclouds \\rightarrow \\infty} \\frac{1}{e^{2}}\\left[\\arctan e^{rainclouds-1}-\\arctan e^{0}\\right] \\\\\n& =\\frac{1}{e^{2}}\\left[\\frac{\\pi}{2}-\\frac{\\pi}{4}\\right]=\\frac{\\pi}{4 e^{2}} .\n\\end{aligned}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "y": "staticnumber",
        "N": "smallcount",
        "c": "variable",
        "\\\\epsilon": "massivegap",
        "\\\\delta": "hugechange"
      },
      "question": "Problem:\n<<<\n9. Evaluate the definite integrals\n(i) \\( \\int_{1}^{3} \\frac{d constantvalue}{\\sqrt{(constantvalue-1)(3-constantvalue)}} \\),\n(ii) \\( \\int_{1}^{\\infty} \\frac{d constantvalue}{e^{constantvalue+1}+e^{3-constantvalue}} \\).\n>>>\n",
      "solution": "Solution:\n<<<\nSolution. Part (i). Since the integrand is not defined at either bound of integration, one should write\n\\[\n\\begin{aligned}\n& \\int_{1}^{3} \\frac{d constantvalue}{\\sqrt{(constantvalue-1)(3-constantvalue)}}=\\lim _{\\substack{massivegap-0+\\\\\nhugechange-0+}} \\int_{1+massivegap}^{3-hugechange} \\frac{d constantvalue}{\\sqrt{(constantvalue-1)(3-constantvalue)}} \\\\\n= & \\left.\\lim _{\\substack{massivegap-0+\\\\\nhugechange-0+}} \\int_{1+massivegap}^{3-hugechange} \\frac{d constantvalue}{\\sqrt{1-(constantvalue-2)^{2}}}=\\lim _{\\substack{massivegap-0+\\\\\nhugechange-0+}} \\arcsin (constantvalue-2)\\right]_{1+massivegap}^{3-hugechange} \\\\\n= & \\lim _{\\substack{massivegap \\rightarrow 0+\\\\\nhugechange \\rightarrow 0+}}[\\arcsin (1-hugechange)-\\arcsin (massivegap-1)]=\\frac{\\pi}{2}+\\frac{\\pi}{2}=\\pi .\n\\end{aligned}\n\\]\n\nPart (ii). The difficulty here is with the infinite interval of integration. Let \\( staticnumber=constantvalue-1 \\); then\n\\[\n\\begin{array}{c}\n\\int \\frac{d constantvalue}{e^{constantvalue+1}+e^{3-constantvalue}}=\\frac{1}{e^{2}} \\int \\frac{d constantvalue}{e^{constantvalue-1}+e^{1-constantvalue}}=\\frac{1}{e^{2}} \\int \\frac{d staticnumber}{e^{staticnumber}+e^{-staticnumber}} \\\\\n=\\frac{1}{e^{2}} \\int \\frac{e^{staticnumber} d staticnumber}{e^{2 staticnumber}+1}=\\frac{1}{e^{2}} \\arctan e^{staticnumber}+variable\n\\end{array}\n\\]\n\nHence\n\\[\n\\begin{aligned}\n\\int_{1}^{\\infty} \\frac{d constantvalue}{e^{constantvalue+1}+e^{3-constantvalue}} & =\\lim _{smallcount \\rightarrow \\infty} \\int_{1}^{smallcount} \\frac{d constantvalue}{e^{constantvalue+1}+e^{3-constantvalue}} \\\\\n& =\\lim _{smallcount \\rightarrow \\infty} \\frac{1}{e^{2}}\\left[\\arctan e^{smallcount-1}-\\arctan e^{0}\\right] \\\\\n& =\\frac{1}{e^{2}}\\left[\\frac{\\pi}{2}-\\frac{\\pi}{4}\\right]=\\frac{\\pi}{4 e^{2}} .\n\\end{aligned}\n\\]\n>>>\n"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "N": "vmbtczay",
        "c": "plsnrqiw",
        "\\epsilon": "lkrdmvse",
        "\\delta": "wqfjzupk"
      },
      "question": "9. Evaluate the definite integrals\n(i) \\( \\int_{1}^{3} \\frac{d qzxwvtnp}{\\sqrt{(qzxwvtnp-1)(3-qzxwvtnp)}} \\),\n(ii) \\( \\int_{1}^{\\infty} \\frac{d qzxwvtnp}{e^{qzxwvtnp+1}+e^{3-qzxwvtnp}} \\).",
      "solution": "Solution. Part (i). Since the integrand is not defined at either bound of integration, one should write\n\\[\n\\begin{aligned}\n& \\int_{1}^{3} \\frac{d qzxwvtnp}{\\sqrt{(qzxwvtnp-1)(3-qzxwvtnp)}}=\\lim _{\\substack{lkrdmvse-0+\\\\\nwqfjzupk-0+}} \\int_{1+lkrdmvse}^{3-wqfjzupk} \\frac{d qzxwvtnp}{\\sqrt{(qzxwvtnp-1)(3-qzxwvtnp)}} \\\\\n= & \\left.\\lim _{\\substack{lkrdmvse-0+\\\\\nwqfjzupk-0+}} \\int_{1+lkrdmvse}^{3-wqfjzupk} \\frac{d qzxwvtnp}{\\sqrt{1-(qzxwvtnp-2)^{2}}}=\\lim _{\\substack{lkrdmvse-0+\\\\\nwqfjzupk-0+}} \\arcsin (qzxwvtnp-2)\\right]_{1+lkrdmvse}^{3-wqfjzupk} \\\\\n= & \\lim _{\\substack{lkrdmvse \\rightarrow 0+\\\\\nwqfjzupk \\rightarrow 0+}}[\\arcsin (1-wqfjzupk)-\\arcsin (lkrdmvse-1)]=\\frac{\\pi}{2}+\\frac{\\pi}{2}=\\pi .\n\\end{aligned}\n\\]\n\nPart (ii). The difficulty here is with the infinite interval of integration. Let \\( hjgrksla=qzxwvtnp-1 \\); then\n\\[\n\\begin{array}{c}\n\\int \\frac{d qzxwvtnp}{e^{qzxwvtnp+1}+e^{3-qzxwvtnp}}=\\frac{1}{e^{2}} \\int \\frac{d qzxwvtnp}{e^{qzxwvtnp-1}+e^{1-qzxwvtnp}}=\\frac{1}{e^{2}} \\int \\frac{d hjgrksla}{e^{hjgrksla}+e^{-hjgrksla}} \\\\\n=\\frac{1}{e^{2}} \\int \\frac{e^{hjgrksla} d hjgrksla}{e^{2 hjgrksla}+1}=\\frac{1}{e^{2}} \\arctan e^{hjgrksla}+plsnrqiw\n\\end{array}\n\\]\n\nHence\n\\[\n\\begin{aligned}\n\\int_{1}^{\\infty} \\frac{d qzxwvtnp}{e^{qzxwvtnp+1}+e^{3-qzxwvtnp}} & =\\lim _{vmbtczay \\rightarrow \\infty} \\int_{1}^{vmbtczay} \\frac{d qzxwvtnp}{e^{qzxwvtnp+1}+e^{3-qzxwvtnp}} \\\\\n& =\\lim _{vmbtczay \\rightarrow \\infty} \\frac{1}{e^{2}}\\left[\\arctan e^{vmbtczay-1}-\\arctan e^{0}\\right] \\\\\n& =\\frac{1}{e^{2}}\\left[\\frac{\\pi}{2}-\\frac{\\pi}{4}\\right]=\\frac{\\pi}{4 e^{2}} .\n\\end{aligned}\n\\]"
    },
    "kernel_variant": {
      "question": "Let  \n\n(I) For every integer k ≥ 2 set  \n   Iₖ := ∬_{[0,1]ᵏ} \\frac{dx₁\\cdots dxₖ}{1-\\bigl(x₁x₂\\cdots xₖ\\bigr)^{\\,2}}\\!.\n\n(II) For every real number s>0 set  \n   J(s):=\\displaystyle\\int_{0}^{\\infty}\\frac{x^{\\,s-1}\\,dx}{e^{x}+e^{-x}}\\!.\n\nThroughout Γ denotes the Gamma–function, ζ the Riemann zeta–function and  \nβ(s)=\\displaystyle\\sum_{n=0}^{\\infty}\\frac{(-1)^{n}}{(2n+1)^{s}} the Dirichlet beta–function.\n\n1. (a) Prove that Iₖ converges for every k≥2.  \n (b) Show that  \n   Iₖ=\\displaystyle\\sum_{n=0}^{\\infty}\\frac1{(2n+1)^{k}}=(1-2^{-k})\\,\\zeta(k).  \n  Give explicit numerical values for k=3 and k=4.\n\n2. (a) Prove that J(s) converges for every s>0 and that  \n   J(s)=\\Gamma(s)\\,\\beta(s).  \n (b) Compute J(3) and J(5) in closed form.\n\n3. (Parametrised variant and higher-order polygamma)  \n Fix a real parameter α>0 and set  \n   Iₖ(α):=\\displaystyle\\iint_{[0,1]^{k}}\\frac{dx₁\\cdots dxₖ}{1-\\bigl(x₁x₂\\cdots xₖ\\bigr)^{\\alpha}}\\!.  \n\n (a) Show that Iₖ(α) converges for every k≥2.\n\n (b) Prove that  \n   Iₖ(α)=\\alpha^{-k}\\,\\zeta\\!\\bigl(k,\\tfrac1\\alpha\\bigr),  \n  where ζ(s,q)=\\displaystyle\\sum_{n=0}^{\\infty}\\frac1{(n+q)^{s}} is the Hurwitz zeta–function.\n\n (c) Deduce Part 1 (b) by putting α=2 and using the identity  \n   ζ(k,\\tfrac12)=(2^{k}-1)\\,\\zeta(k).\n\n (d)  Denote by ψ^{(m)}(z)=\\dfrac{d^{m+1}}{dz^{m+1}}\\log\\Gamma(z) the m-th polygamma function.  \n  Show that for every even k=2m≥2 one has the exact formula  \n   I_{2m}(α)=\\dfrac{\\alpha^{-2m}}{(2m-1)!}\\;\\psi^{(2m-1)}\\!\\Bigl(\\frac1\\alpha\\Bigr). (*)\n\n  Using the known values  \n\n    ψ^{(3)}\\!\\bigl(\\tfrac12\\bigr)=\\pi^{4},  ψ^{(5)}\\!\\bigl(\\tfrac12\\bigr)=8\\pi^{6},  \n\n  obtain the explicit closed forms  \n\n    I₄(α)=\\dfrac{\\alpha^{-4}}{6}\\;\\psi^{(3)}\\!\\Bigl(\\dfrac1\\alpha\\Bigr),  \n    I₆(α)=\\dfrac{\\alpha^{-6}}{120}\\;\\psi^{(5)}\\!\\Bigl(\\dfrac1\\alpha\\Bigr),  \n\n  and verify in particular that  \n   I₄(2)=\\dfrac{\\pi^{4}}{96}, I₆(2)=\\dfrac{\\pi^{6}}{960}.",
      "solution": "Standard facts about the Gamma, zeta, beta, Hurwitz-zeta and polygamma functions are freely used.\n\n──────────────────────────────────\n1.  The basic integral Iₖ\n──────────────────────────────────\n1 (a) Convergence.  \nPut P=(x₁x₂⋯xₖ)².  Over the unit cube one has 0≤P≤1 and the only potential\nsingularity of the integrand 1/(1-P) occurs at the point (1,1,…,1) where P=1.\nWrite xᵢ=1-tᵢ with tᵢ∈[0,1].  Then\n\n 1-P=1-\\bigl(1-\\sum_{i=1}^{k}tᵢ+O(|t|^{2})\\bigr)=\\sum_{i=1}^{k}tᵢ+O(|t|^{2}).\n\nHence near the singular point the integrand behaves like 1/∑ tᵢ, whose\nk-dimensional integral over a small cube is finite as soon as k≥2.\nNo other singularities exist, so Iₖ converges for every k≥2.\n\n1 (b) Evaluation.  \nExpand 1/(1-P)=\\sum_{n=0}^{\\infty}P^{n}, interchange\nsum and integral (Tonelli) and compute\n\n Iₖ=\\sum_{n=0}^{\\infty}\\prod_{j=1}^{k}\\int_{0}^{1}x^{2n}\\,dx\n     =\\sum_{n=0}^{\\infty}\\Bigl(\\frac1{2n+1}\\Bigr)^{k}\n     =(1-2^{-k})\\,\\zeta(k).\n\nNumerical cases:\n\n k=3: I₃=(1-\\tfrac18)\\zeta(3)=\\tfrac78\\,\\zeta(3).  \n k=4: I₄=(1-\\tfrac1{16})\\zeta(4)=\\tfrac{15}{16}\\cdot\\frac{\\pi^{4}}{90}\n         =\\dfrac{\\pi^{4}}{96}.\n\n──────────────────────────────────\n2.  The integral  J(s)\n──────────────────────────────────\n2 (a) Convergence.  \nNear x=0 we have e^{x}+e^{-x}=2+O(x^{2}), so the integrand behaves like\nx^{s-1} and is integrable whenever s>0.\nFor x→∞ the denominator grows like e^{x}, so the integrand behaves like\nx^{s-1}e^{-x}, again integrable for every s.\n\nSeries representation:\n \\frac1{e^{x}+e^{-x}}\n     =\\frac{e^{-x}}{1+e^{-2x}}\n     =\\sum_{n=0}^{\\infty}(-1)^{n}\\,e^{-(2n+1)x},\nuniformly on [ε,∞) for any ε>0; thus\n\n J(s)=\\sum_{n=0}^{\\infty}(-1)^{n}\\int_{0}^{\\infty}x^{s-1}e^{-(2n+1)x}\\,dx\n     =\\Gamma(s)\\sum_{n=0}^{\\infty}\\frac{(-1)^{n}}{(2n+1)^{s}}\n     =\\Gamma(s)\\,\\beta(s).\n\n2 (b) Special values.\n β(3)=\\pi^{3}/32, β(5)=5\\pi^{5}/1536,\n\n Γ(3)=2, Γ(5)=24,\n\nso J(3)=\\pi^{3}/16 and J(5)=5\\pi^{5}/64.\n\n──────────────────────────────────\n3.  The parametrised integrals Iₖ(α)\n──────────────────────────────────\n3 (a) Convergence.  \nThe only critical point is (1,…,1).  Put xᵢ=1-tᵢ (tᵢ≥0);\nthen\n\n 1-\\bigl(x₁x₂\\cdots xₖ\\bigr)^{\\alpha}\n     =1-\\exp\\!\\bigl(\\alpha\\sum_{i}\\log(1-tᵢ)\\bigr)\n     =\\alpha\\sum_{i}tᵢ+O(|t|^{2}).\n\nThus the integrand behaves like 1/∑ tᵢ, which is k-integrable for k≥2.\n\n3 (b) Evaluation through the Hurwitz zeta.  \nExpand\n\n \\frac1{1-P^{\\alpha}}=\\sum_{n=0}^{\\infty}P^{\\alpha n}, P=x₁x₂\\cdots xₖ,\n\ninterchange sum and integral and obtain\n\n Iₖ(α)=\\sum_{n=0}^{\\infty}\\Bigl(\\frac1{\\alpha n+1}\\Bigr)^{k}\n         =\\alpha^{-k}\\sum_{n=0}^{\\infty}\\frac1{(n+\\frac1\\alpha)^{k}}\n         =\\alpha^{-k}\\,\\zeta\\!\\bigl(k,\\tfrac1\\alpha\\bigr).\n\n3 (c) Specialisation α=2.  \nBecause ζ(k,\\tfrac12)=(2^{k}-1)\\,ζ(k) we recover\n\n Iₖ(2)=2^{-k}(2^{k}-1)\\,ζ(k)=(1-2^{-k})\\,ζ(k)\n\nin agreement with Part 1 (b).\n\n3 (d) Even orders and the polygamma function.  \nRecall the higher-order\npolygamma: ψ^{(m)}(z)=\\dfrac{d^{m+1}}{dz^{m+1}}\\log\\Gamma(z).\nFor n≥0 one has  \n\n ψ^{(n)}(z)=(-1)^{\\,n+1}n!\\sum_{r=0}^{\\infty}\\frac1{(r+z)^{\\,n+1}}.\n\nTaking n=2m-1 (m≥1) gives  \n\n ζ(2m,z)=\\sum_{r=0}^{\\infty}\\frac1{(r+z)^{2m}}\n         =\\frac{1}{(2m-1)!}\\,\\psi^{(2m-1)}(z).\n\nInserting z=1/α into part (b) yields the exact identity\n\n I_{2m}(α)=\\frac{\\alpha^{-2m}}{(2m-1)!}\\,\n             \\psi^{(2m-1)}\\!\\Bigl(\\frac1\\alpha\\Bigr)\\!,  m≥1. (*)\n\nExplicit cases.\n\nm=2 (k=4).  Formula (*) gives  \n\n I₄(α)=\\dfrac{\\alpha^{-4}}{6}\\,\\psi^{(3)}\\!\\Bigl(\\frac1\\alpha\\Bigr).\n\nFor α=2, ψ^{(3)}(½)=π^{4}, hence  \n\n I₄(2)=\\frac{1}{16}\\cdot\\frac{π^{4}}{6}=\\dfrac{π^{4}}{96}.\n\nm=3 (k=6).  Formula (*) gives  \n\n I₆(α)=\\dfrac{\\alpha^{-6}}{120}\\,\\psi^{(5)}\\!\\Bigl(\\frac1\\alpha\\Bigr).\n\nFor α=2, ψ^{(5)}(½)=8π^{6}, hence  \n\n I₆(2)=\\frac{1}{64}\\cdot\\frac{8π^{6}}{120}=\\dfrac{π^{6}}{960}.\n\nBoth results are positive, in perfect agreement with the elementary formula\nI_{2m}(2)=(1-2^{-2m})ζ(2m).\n\n──────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.358462",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the problem generalises a 2-fold integral to an arbitrary k-fold one, forcing the solver to control a singularity on a codimension-k set and to justify term-wise integration in many variables.  \n\n• Additional structures: ζ(k) and β(s) (Riemann zeta and Dirichlet beta) appear; knowledge of their special values and interplay with Γ(s) is required.  \n\n• Parametric component: J(s) depends on a continuous parameter, demanding uniform convergence arguments and mastery of Mellin–type transforms.  \n\n• Interacting concepts: geometric-series expansion, Fubini/Tonelli theorems, Gamma integrals, special values of L-functions, and high-order differentiation all occur and must be combined coherently.  \n\n• Deeper theory: Part 3 asks for a non-trivial functional identity linking two ostensibly unrelated families of integrals via derivatives of β, pushing the solver into analytic number theory territory.  \n\nThe problem therefore requires substantially more sophisticated techniques and longer chains of reasoning than either the original exercise or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n(I) For every integer k ≥ 2 set  \n   Iₖ := ∬_{[0,1]ᵏ} \\frac{dx₁\\cdots dxₖ}{1-\\bigl(x₁x₂\\cdots xₖ\\bigr)^{\\,2}}\\!.\n\n(II) For every real number s>0 set  \n   J(s):=\\displaystyle\\int_{0}^{\\infty}\\frac{x^{\\,s-1}\\,dx}{e^{x}+e^{-x}}\\!.\n\nThroughout Γ denotes the Gamma–function, ζ the Riemann zeta–function and  \nβ(s)=\\displaystyle\\sum_{n=0}^{\\infty}\\frac{(-1)^{n}}{(2n+1)^{s}} the Dirichlet beta–function.\n\n1. (a) Prove that Iₖ converges for every k≥2.  \n (b) Show that  \n   Iₖ=\\displaystyle\\sum_{n=0}^{\\infty}\\frac1{(2n+1)^{k}}=(1-2^{-k})\\,\\zeta(k).  \n  Give explicit numerical values for k=3 and k=4.\n\n2. (a) Prove that J(s) converges for every s>0 and that  \n   J(s)=\\Gamma(s)\\,\\beta(s).  \n (b) Compute J(3) and J(5) in closed form.\n\n3. (Parametrised variant and higher-order polygamma)  \n Fix a real parameter α>0 and set  \n   Iₖ(α):=\\displaystyle\\iint_{[0,1]^{k}}\\frac{dx₁\\cdots dxₖ}{1-\\bigl(x₁x₂\\cdots xₖ\\bigr)^{\\alpha}}\\!.  \n\n (a) Show that Iₖ(α) converges for every k≥2.\n\n (b) Prove that  \n   Iₖ(α)=\\alpha^{-k}\\,\\zeta\\!\\bigl(k,\\tfrac1\\alpha\\bigr),  \n  where ζ(s,q)=\\displaystyle\\sum_{n=0}^{\\infty}\\frac1{(n+q)^{s}} is the Hurwitz zeta–function.\n\n (c) Deduce Part 1 (b) by putting α=2 and using the identity  \n   ζ(k,\\tfrac12)=(2^{k}-1)\\,\\zeta(k).\n\n (d)  Denote by ψ^{(m)}(z)=\\dfrac{d^{m+1}}{dz^{m+1}}\\log\\Gamma(z) the m-th polygamma function.  \n  Show that for every even k=2m≥2 one has the exact formula  \n   I_{2m}(α)=\\dfrac{\\alpha^{-2m}}{(2m-1)!}\\;\\psi^{(2m-1)}\\!\\Bigl(\\frac1\\alpha\\Bigr). (*)\n\n  Using the known values  \n\n    ψ^{(3)}\\!\\bigl(\\tfrac12\\bigr)=\\pi^{4},  ψ^{(5)}\\!\\bigl(\\tfrac12\\bigr)=8\\pi^{6},  \n\n  obtain the explicit closed forms  \n\n    I₄(α)=\\dfrac{\\alpha^{-4}}{6}\\;\\psi^{(3)}\\!\\Bigl(\\dfrac1\\alpha\\Bigr),  \n    I₆(α)=\\dfrac{\\alpha^{-6}}{120}\\;\\psi^{(5)}\\!\\Bigl(\\dfrac1\\alpha\\Bigr),  \n\n  and verify in particular that  \n   I₄(2)=\\dfrac{\\pi^{4}}{96}, I₆(2)=\\dfrac{\\pi^{6}}{960}.",
      "solution": "Standard facts about the Gamma, zeta, beta, Hurwitz-zeta and polygamma functions are freely used.\n\n──────────────────────────────────\n1.  The basic integral Iₖ\n──────────────────────────────────\n1 (a) Convergence.  \nPut P=(x₁x₂⋯xₖ)².  Over the unit cube one has 0≤P≤1 and the only potential\nsingularity of the integrand 1/(1-P) occurs at the point (1,1,…,1) where P=1.\nWrite xᵢ=1-tᵢ with tᵢ∈[0,1].  Then\n\n 1-P=1-\\bigl(1-\\sum_{i=1}^{k}tᵢ+O(|t|^{2})\\bigr)=\\sum_{i=1}^{k}tᵢ+O(|t|^{2}).\n\nHence near the singular point the integrand behaves like 1/∑ tᵢ, whose\nk-dimensional integral over a small cube is finite as soon as k≥2.\nNo other singularities exist, so Iₖ converges for every k≥2.\n\n1 (b) Evaluation.  \nExpand 1/(1-P)=\\sum_{n=0}^{\\infty}P^{n}, interchange\nsum and integral (Tonelli) and compute\n\n Iₖ=\\sum_{n=0}^{\\infty}\\prod_{j=1}^{k}\\int_{0}^{1}x^{2n}\\,dx\n     =\\sum_{n=0}^{\\infty}\\Bigl(\\frac1{2n+1}\\Bigr)^{k}\n     =(1-2^{-k})\\,\\zeta(k).\n\nNumerical cases:\n\n k=3: I₃=(1-\\tfrac18)\\zeta(3)=\\tfrac78\\,\\zeta(3).  \n k=4: I₄=(1-\\tfrac1{16})\\zeta(4)=\\tfrac{15}{16}\\cdot\\frac{\\pi^{4}}{90}\n         =\\dfrac{\\pi^{4}}{96}.\n\n──────────────────────────────────\n2.  The integral  J(s)\n──────────────────────────────────\n2 (a) Convergence.  \nNear x=0 we have e^{x}+e^{-x}=2+O(x^{2}), so the integrand behaves like\nx^{s-1} and is integrable whenever s>0.\nFor x→∞ the denominator grows like e^{x}, so the integrand behaves like\nx^{s-1}e^{-x}, again integrable for every s.\n\nSeries representation:\n \\frac1{e^{x}+e^{-x}}\n     =\\frac{e^{-x}}{1+e^{-2x}}\n     =\\sum_{n=0}^{\\infty}(-1)^{n}\\,e^{-(2n+1)x},\nuniformly on [ε,∞) for any ε>0; thus\n\n J(s)=\\sum_{n=0}^{\\infty}(-1)^{n}\\int_{0}^{\\infty}x^{s-1}e^{-(2n+1)x}\\,dx\n     =\\Gamma(s)\\sum_{n=0}^{\\infty}\\frac{(-1)^{n}}{(2n+1)^{s}}\n     =\\Gamma(s)\\,\\beta(s).\n\n2 (b) Special values.\n β(3)=\\pi^{3}/32, β(5)=5\\pi^{5}/1536,\n\n Γ(3)=2, Γ(5)=24,\n\nso J(3)=\\pi^{3}/16 and J(5)=5\\pi^{5}/64.\n\n──────────────────────────────────\n3.  The parametrised integrals Iₖ(α)\n──────────────────────────────────\n3 (a) Convergence.  \nThe only critical point is (1,…,1).  Put xᵢ=1-tᵢ (tᵢ≥0);\nthen\n\n 1-\\bigl(x₁x₂\\cdots xₖ\\bigr)^{\\alpha}\n     =1-\\exp\\!\\bigl(\\alpha\\sum_{i}\\log(1-tᵢ)\\bigr)\n     =\\alpha\\sum_{i}tᵢ+O(|t|^{2}).\n\nThus the integrand behaves like 1/∑ tᵢ, which is k-integrable for k≥2.\n\n3 (b) Evaluation through the Hurwitz zeta.  \nExpand\n\n \\frac1{1-P^{\\alpha}}=\\sum_{n=0}^{\\infty}P^{\\alpha n}, P=x₁x₂\\cdots xₖ,\n\ninterchange sum and integral and obtain\n\n Iₖ(α)=\\sum_{n=0}^{\\infty}\\Bigl(\\frac1{\\alpha n+1}\\Bigr)^{k}\n         =\\alpha^{-k}\\sum_{n=0}^{\\infty}\\frac1{(n+\\frac1\\alpha)^{k}}\n         =\\alpha^{-k}\\,\\zeta\\!\\bigl(k,\\tfrac1\\alpha\\bigr).\n\n3 (c) Specialisation α=2.  \nBecause ζ(k,\\tfrac12)=(2^{k}-1)\\,ζ(k) we recover\n\n Iₖ(2)=2^{-k}(2^{k}-1)\\,ζ(k)=(1-2^{-k})\\,ζ(k)\n\nin agreement with Part 1 (b).\n\n3 (d) Even orders and the polygamma function.  \nRecall the higher-order\npolygamma: ψ^{(m)}(z)=\\dfrac{d^{m+1}}{dz^{m+1}}\\log\\Gamma(z).\nFor n≥0 one has  \n\n ψ^{(n)}(z)=(-1)^{\\,n+1}n!\\sum_{r=0}^{\\infty}\\frac1{(r+z)^{\\,n+1}}.\n\nTaking n=2m-1 (m≥1) gives  \n\n ζ(2m,z)=\\sum_{r=0}^{\\infty}\\frac1{(r+z)^{2m}}\n         =\\frac{1}{(2m-1)!}\\,\\psi^{(2m-1)}(z).\n\nInserting z=1/α into part (b) yields the exact identity\n\n I_{2m}(α)=\\frac{\\alpha^{-2m}}{(2m-1)!}\\,\n             \\psi^{(2m-1)}\\!\\Bigl(\\frac1\\alpha\\Bigr)\\!,  m≥1. (*)\n\nExplicit cases.\n\nm=2 (k=4).  Formula (*) gives  \n\n I₄(α)=\\dfrac{\\alpha^{-4}}{6}\\,\\psi^{(3)}\\!\\Bigl(\\frac1\\alpha\\Bigr).\n\nFor α=2, ψ^{(3)}(½)=π^{4}, hence  \n\n I₄(2)=\\frac{1}{16}\\cdot\\frac{π^{4}}{6}=\\dfrac{π^{4}}{96}.\n\nm=3 (k=6).  Formula (*) gives  \n\n I₆(α)=\\dfrac{\\alpha^{-6}}{120}\\,\\psi^{(5)}\\!\\Bigl(\\frac1\\alpha\\Bigr).\n\nFor α=2, ψ^{(5)}(½)=8π^{6}, hence  \n\n I₆(2)=\\frac{1}{64}\\cdot\\frac{8π^{6}}{120}=\\dfrac{π^{6}}{960}.\n\nBoth results are positive, in perfect agreement with the elementary formula\nI_{2m}(2)=(1-2^{-2m})ζ(2m).\n\n──────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.311520",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: the problem generalises a 2-fold integral to an arbitrary k-fold one, forcing the solver to control a singularity on a codimension-k set and to justify term-wise integration in many variables.  \n\n• Additional structures: ζ(k) and β(s) (Riemann zeta and Dirichlet beta) appear; knowledge of their special values and interplay with Γ(s) is required.  \n\n• Parametric component: J(s) depends on a continuous parameter, demanding uniform convergence arguments and mastery of Mellin–type transforms.  \n\n• Interacting concepts: geometric-series expansion, Fubini/Tonelli theorems, Gamma integrals, special values of L-functions, and high-order differentiation all occur and must be combined coherently.  \n\n• Deeper theory: Part 3 asks for a non-trivial functional identity linking two ostensibly unrelated families of integrals via derivatives of β, pushing the solver into analytic number theory territory.  \n\nThe problem therefore requires substantially more sophisticated techniques and longer chains of reasoning than either the original exercise or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}