{
  "index": "1953-A-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "4. From the identity\n\\[\n\\begin{aligned}\n\\int_{0}^{\\pi / 2} \\log \\sin 2 x d x & =\\int_{0}^{\\pi / 2} \\log \\sin x d x \\\\\n& +\\int_{0}^{\\pi / 2} \\log \\cos x d x+\\int_{0}^{\\pi / 2} \\log 2 d x\n\\end{aligned}\n\\]\ndeduce the value of\n\\[\n\\int_{0}^{\\pi / 2} \\log \\sin x d x\n\\]",
  "solution": "Solution. Let\n\\[\nI=\\int_{0}^{\\pi / 2} \\log \\sin x d x\n\\]\n[This is, of course, an improper integral which we assume for the moment to exist.] Making the substitutions \\( x=\\pi / 2-u \\) and \\( x=\\pi-v \\) we see that\n\\[\nI=\\int_{0}^{\\pi / 2} \\log \\cos u d u=\\int_{\\pi / 2}^{\\pi} \\log \\sin v d v\n\\]\n\nTherefore\n\\[\n\\int_{0}^{\\pi} \\log \\sin v d v=\\int_{0}^{\\pi / 2} \\log \\sin v d v+\\int_{\\pi / 2}^{\\pi} \\log \\sin v d v=2 I .\n\\]\n\nMaking the substitution \\( v=2 w \\) we have also\n\\[\n\\int_{0}^{\\pi} \\log \\sin v d v=2 \\int_{0}^{\\pi / 2} \\log \\sin 2 w d w .\n\\]\n\nThen\n\\[\n\\begin{aligned}\nI=\\int_{0}^{\\pi / 2} \\log \\sin 2 w d w= & \\int_{0}^{\\pi / 2}(\\log 2) d w+\\int_{0}^{\\pi / 2} \\log \\sin w d w \\\\\n& +\\int_{0}^{\\pi / 2} \\log \\cos w d w \\\\\n= & \\frac{\\pi}{2} \\log 2+2 I\n\\end{aligned}\n\\]\n\nHence\n\\[\nI=-\\frac{\\pi}{2} \\log 2\n\\]\n\nTo justify these manipulations, note that\n\\[\n\\frac{2}{\\pi} x<\\sin x \\leq 1 \\text { for } 0<x \\leq \\frac{\\pi}{2}\n\\]\nand hence\n\\[\n\\log x-\\log \\frac{\\pi}{2}<\\log \\sin x \\leq 0\n\\]\n\nNow\n\\[\n\\int_{1}^{1} \\log x d x=-1-\\epsilon \\log \\epsilon+\\epsilon\n\\]\n\nSo\n\\[\n\\lim _{\\epsilon \\rightarrow 0} \\int_{t}^{1} \\log x d x=-1\n\\]\n\nIt follows that the improper integral\n\\[\n\\int_{0}^{\\pi / 2} \\log \\sin x d x=\\lim _{\\epsilon \\rightarrow 0} \\int_{\\epsilon}^{\\pi / 2} \\log \\sin x d x\n\\]\nexists. The several substitutions employed above can now be justified routinely.",
  "vars": [
    "x",
    "u",
    "v",
    "w",
    "t"
  ],
  "params": [
    "I",
    "\\\\epsilon"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "anglevar",
        "u": "rotatedvar",
        "v": "doubledvar",
        "w": "halfangle",
        "t": "limitvar",
        "I": "integralvalue",
        "\\epsilon": "epsilonpar"
      },
      "question": "4. From the identity\n\\[\n\\begin{aligned}\n\\int_{0}^{\\pi / 2} \\log \\sin 2 anglevar d anglevar & =\\int_{0}^{\\pi / 2} \\log \\sin anglevar d anglevar \\\\\n& +\\int_{0}^{\\pi / 2} \\log \\cos anglevar d anglevar+\\int_{0}^{\\pi / 2} \\log 2 d anglevar\n\\end{aligned}\n\\]\ndeduce the value of\n\\[\n\\int_{0}^{\\pi / 2} \\log \\sin anglevar d anglevar\n\\]",
      "solution": "Solution. Let\n\\[\nintegralvalue=\\int_{0}^{\\pi / 2} \\log \\sin anglevar d anglevar\n\\]\n[This is, of course, an improper integral which we assume for the moment to exist.] Making the substitutions \\( anglevar=\\pi / 2-rotatedvar \\) and \\( anglevar=\\pi-doubledvar \\) we see that\n\\[\nintegralvalue=\\int_{0}^{\\pi / 2} \\log \\cos rotatedvar d rotatedvar=\\int_{\\pi / 2}^{\\pi} \\log \\sin doubledvar d doubledvar\n\\]\n\nTherefore\n\\[\n\\int_{0}^{\\pi} \\log \\sin doubledvar d doubledvar=\\int_{0}^{\\pi / 2} \\log \\sin doubledvar d doubledvar+\\int_{\\pi / 2}^{\\pi} \\log \\sin doubledvar d doubledvar=2 integralvalue .\n\\]\n\nMaking the substitution \\( doubledvar=2 halfangle \\) we have also\n\\[\n\\int_{0}^{\\pi} \\log \\sin doubledvar d doubledvar=2 \\int_{0}^{\\pi / 2} \\log \\sin 2 halfangle d halfangle .\n\\]\n\nThen\n\\[\n\\begin{aligned}\nintegralvalue=\\int_{0}^{\\pi / 2} \\log \\sin 2 halfangle d halfangle= & \\int_{0}^{\\pi / 2}(\\log 2) d halfangle+\\int_{0}^{\\pi / 2} \\log \\sin halfangle d halfangle \\\\\n& +\\int_{0}^{\\pi / 2} \\log \\cos halfangle d halfangle \\\\\n= & \\frac{\\pi}{2} \\log 2+2 integralvalue\n\\end{aligned}\n\\]\n\nHence\n\\[\nintegralvalue=-\\frac{\\pi}{2} \\log 2\n\\]\n\nTo justify these manipulations, note that\n\\[\n\\frac{2}{\\pi} anglevar<\\sin anglevar \\leq 1 \\text { for } 0<anglevar \\leq \\frac{\\pi}{2}\n\\]\nand hence\n\\[\n\\log anglevar-\\log \\frac{\\pi}{2}<\\log \\sin anglevar \\leq 0\n\\]\n\nNow\n\\[\n\\int_{epsilonpar}^{1} \\log anglevar d anglevar=-1-epsilonpar \\log epsilonpar+epsilonpar\n\\]\n\nSo\n\\[\n\\lim _{epsilonpar \\rightarrow 0} \\int_{limitvar}^{1} \\log anglevar d anglevar=-1\n\\]\n\nIt follows that the improper integral\n\\[\n\\int_{0}^{\\pi / 2} \\log \\sin anglevar d anglevar=\\lim _{epsilonpar \\rightarrow 0} \\int_{epsilonpar}^{\\pi / 2} \\log \\sin anglevar d anglevar\n\\]\nexists. The several substitutions employed above can now be justified routinely."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "parchment",
        "u": "snowflake",
        "v": "drumstick",
        "w": "feathered",
        "t": "rainstorm",
        "I": "windshield",
        "\\epsilon": "raspberry"
      },
      "question": "4. From the identity\n\\[\n\\begin{aligned}\n\\int_{0}^{\\pi / 2} \\log \\sin 2 parchment d parchment & =\\int_{0}^{\\pi / 2} \\log \\sin parchment d parchment \\\\\n& +\\int_{0}^{\\pi / 2} \\log \\cos parchment d parchment+\\int_{0}^{\\pi / 2} \\log 2 d parchment\n\\end{aligned}\n\\]\ndeduce the value of\n\\[\n\\int_{0}^{\\pi / 2} \\log \\sin parchment d parchment\n\\]\n",
      "solution": "Solution. Let\n\\[\nwindshield=\\int_{0}^{\\pi / 2} \\log \\sin parchment d parchment\n\\]\n[This is, of course, an improper integral which we assume for the moment to exist.] Making the substitutions \\( parchment=\\pi / 2-snowflake \\) and \\( parchment=\\pi-drumstick \\) we see that\n\\[\nwindshield=\\int_{0}^{\\pi / 2} \\log \\cos snowflake d snowflake=\\int_{\\pi / 2}^{\\pi} \\log \\sin drumstick d drumstick\n\\]\n\nTherefore\n\\[\n\\int_{0}^{\\pi} \\log \\sin drumstick d drumstick=\\int_{0}^{\\pi / 2} \\log \\sin drumstick d drumstick+\\int_{\\pi / 2}^{\\pi} \\log \\sin drumstick d drumstick=2 windshield .\n\\]\n\nMaking the substitution \\( drumstick=2 feathered \\) we have also\n\\[\n\\int_{0}^{\\pi} \\log \\sin drumstick d drumstick=2 \\int_{0}^{\\pi / 2} \\log \\sin 2 feathered d feathered .\n\\]\n\nThen\n\\[\n\\begin{aligned}\nwindshield=\\int_{0}^{\\pi / 2} \\log \\sin 2 feathered d feathered= & \\int_{0}^{\\pi / 2}(\\log 2) d feathered+\\int_{0}^{\\pi / 2} \\log \\sin feathered d feathered \\\\\n& +\\int_{0}^{\\pi / 2} \\log \\cos feathered d feathered \\\\\n= & \\frac{\\pi}{2} \\log 2+2 windshield\n\\end{aligned}\n\\]\n\nHence\n\\[\nwindshield=-\\frac{\\pi}{2} \\log 2\n\\]\n\nTo justify these manipulations, note that\n\\[\n\\frac{2}{\\pi} parchment<\\sin parchment \\leq 1 \\text { for } 0<parchment \\leq \\frac{\\pi}{2}\n\\]\nand hence\n\\[\n\\log parchment-\\log \\frac{\\pi}{2}<\\log \\sin parchment \\leq 0\n\\]\n\nNow\n\\[\n\\int_{1}^{1} \\log parchment d parchment=-1-raspberry \\log raspberry+raspberry\n\\]\n\nSo\n\\[\n\\lim _{raspberry \\rightarrow 0} \\int_{rainstorm}^{1} \\log parchment d parchment=-1\n\\]\n\nIt follows that the improper integral\n\\[\n\\int_{0}^{\\pi / 2} \\log \\sin parchment d parchment=\\lim _{raspberry \\rightarrow 0} \\int_{raspberry}^{\\pi / 2} \\log \\sin parchment d parchment\n\\]\nexists. The several substitutions employed above can now be justified routinely."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "steadfastconst",
        "u": "rigidparam",
        "v": "staticnumber",
        "w": "immutablesym",
        "t": "frozenvalue",
        "I": "variabletotal",
        "\\epsilon": "hugebound"
      },
      "question": "4. From the identity\n\\[\n\\begin{aligned}\n\\int_{0}^{\\pi / 2} \\log \\sin 2 steadfastconst d steadfastconst & =\\int_{0}^{\\pi / 2} \\log \\sin steadfastconst d steadfastconst \\\\\n& +\\int_{0}^{\\pi / 2} \\log \\cos steadfastconst d steadfastconst+\\int_{0}^{\\pi / 2} \\log 2 d steadfastconst\n\\end{aligned}\n\\]\ndeduce the value of\n\\[\n\\int_{0}^{\\pi / 2} \\log \\sin steadfastconst d steadfastconst\n\\]",
      "solution": "Solution. Let\n\\[\nvariabletotal=\\int_{0}^{\\pi / 2} \\log \\sin steadfastconst d steadfastconst\n\\]\n[This is, of course, an improper integral which we assume for the moment to exist.] Making the substitutions \\( steadfastconst=\\pi / 2-rigidparam \\) and \\( steadfastconst=\\pi-staticnumber \\) we see that\n\\[\nvariabletotal=\\int_{0}^{\\pi / 2} \\log \\cos rigidparam d rigidparam=\\int_{\\pi / 2}^{\\pi} \\log \\sin staticnumber d staticnumber\n\\]\nTherefore\n\\[\n\\int_{0}^{\\pi} \\log \\sin staticnumber d staticnumber=\\int_{0}^{\\pi / 2} \\log \\sin staticnumber d staticnumber+\\int_{\\pi / 2}^{\\pi} \\log \\sin staticnumber d staticnumber=2 variabletotal .\n\\]\nMaking the substitution \\( staticnumber=2 immutablesym \\) we have also\n\\[\n\\int_{0}^{\\pi} \\log \\sin staticnumber d staticnumber=2 \\int_{0}^{\\pi / 2} \\log \\sin 2 immutablesym d immutablesym .\n\\]\nThen\n\\[\n\\begin{aligned}\nvariabletotal=\\int_{0}^{\\pi / 2} \\log \\sin 2 immutablesym d immutablesym= & \\int_{0}^{\\pi / 2}(\\log 2) d immutablesym+\\int_{0}^{\\pi / 2} \\log \\sin immutablesym d immutablesym \\\\\n& +\\int_{0}^{\\pi / 2} \\log \\cos immutablesym d immutablesym \\\\\n= & \\frac{\\pi}{2} \\log 2+2 variabletotal\n\\end{aligned}\n\\]\nHence\n\\[\nvariabletotal=-\\frac{\\pi}{2} \\log 2\n\\]\nTo justify these manipulations, note that\n\\[\n\\frac{2}{\\pi} steadfastconst<\\sin steadfastconst \\leq 1 \\text { for } 0<steadfastconst \\leq \\frac{\\pi}{2}\n\\]\nand hence\n\\[\n\\log steadfastconst-\\log \\frac{\\pi}{2}<\\log \\sin steadfastconst \\leq 0\n\\]\nNow\n\\[\n\\int_{hugebound}^{1} \\log steadfastconst d steadfastconst=-1-hugebound \\log hugebound+hugebound\n\\]\nSo\n\\[\n\\lim _{hugebound \\rightarrow 0} \\int_{hugebound}^{1} \\log steadfastconst d steadfastconst=-1\n\\]\nIt follows that the improper integral\n\\[\n\\int_{0}^{\\pi / 2} \\log \\sin steadfastconst d steadfastconst=\\lim _{hugebound \\rightarrow 0} \\int_{hugebound}^{\\pi / 2} \\log \\sin steadfastconst d steadfastconst\n\\]\nexists. The several substitutions employed above can now be justified routinely."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "u": "hjgrksla",
        "v": "mnbvcxzl",
        "w": "dsafghjk",
        "t": "poiuytre",
        "I": "lkjhgfds",
        "\\epsilon": "asdfghjk"
      },
      "question": "Problem:\n<<<\n4. From the identity\n\\[\n\\begin{aligned}\n\\int_{0}^{\\pi / 2} \\log \\sin 2 qzxwvtnp d qzxwvtnp & =\\int_{0}^{\\pi / 2} \\log \\sin qzxwvtnp d qzxwvtnp \\\\\n& +\\int_{0}^{\\pi / 2} \\log \\cos qzxwvtnp d qzxwvtnp+\\int_{0}^{\\pi / 2} \\log 2 d qzxwvtnp\n\\end{aligned}\n\\]\ndeduce the value of\n\\[\n\\int_{0}^{\\pi / 2} \\log \\sin qzxwvtnp d qzxwvtnp\n\\]\n>>>\n",
      "solution": "Solution:\n<<<\nSolution. Let\n\\[\nlkjhgfds=\\int_{0}^{\\pi / 2} \\log \\sin qzxwvtnp d qzxwvtnp\n\\]\n[This is, of course, an improper integral which we assume for the moment to exist.] Making the substitutions \\( qzxwvtnp=\\pi / 2-hjgrksla \\) and \\( qzxwvtnp=\\pi-mnbvcxzl \\) we see that\n\\[\nlkjhgfds=\\int_{0}^{\\pi / 2} \\log \\cos hjgrksla d hjgrksla=\\int_{\\pi / 2}^{\\pi} \\log \\sin mnbvcxzl d mnbvcxzl\n\\]\n\nTherefore\n\\[\n\\int_{0}^{\\pi} \\log \\sin mnbvcxzl d mnbvcxzl=\\int_{0}^{\\pi / 2} \\log \\sin mnbvcxzl d mnbvcxzl+\\int_{\\pi / 2}^{\\pi} \\log \\sin mnbvcxzl d mnbvcxzl=2 lkjhgfds .\n\\]\n\nMaking the substitution \\( mnbvcxzl=2 dsafghjk \\) we have also\n\\[\n\\int_{0}^{\\pi} \\log \\sin mnbvcxzl d mnbvcxzl=2 \\int_{0}^{\\pi / 2} \\log \\sin 2 dsafghjk d dsafghjk .\n\\]\n\nThen\n\\[\n\\begin{aligned}\nlkjhgfds=\\int_{0}^{\\pi / 2} \\log \\sin 2 dsafghjk d dsafghjk= & \\int_{0}^{\\pi / 2}(\\log 2) d dsafghjk+\\int_{0}^{\\pi / 2} \\log \\sin dsafghjk d dsafghjk \\\\\n& +\\int_{0}^{\\pi / 2} \\log \\cos dsafghjk d dsafghjk \\\\\n= & \\frac{\\pi}{2} \\log 2+2 lkjhgfds\n\\end{aligned}\n\\]\n\nHence\n\\[\nlkjhgfds=-\\frac{\\pi}{2} \\log 2\n\\]\n\nTo justify these manipulations, note that\n\\[\n\\frac{2}{\\pi} qzxwvtnp<\\sin qzxwvtnp \\leq 1 \\text { for } 0<qzxwvtnp \\leq \\frac{\\pi}{2}\n\\]\nand hence\n\\[\n\\log qzxwvtnp-\\log \\frac{\\pi}{2}<\\log \\sin qzxwvtnp \\leq 0\n\\]\n\nNow\n\\[\n\\int_{1}^{1} \\log qzxwvtnp d qzxwvtnp=-1-asdfghjk \\log asdfghjk+asdfghjk\n\\]\n\nSo\n\\[\n\\lim _{asdfghjk \\rightarrow 0} \\int_{poiuytre}^{1} \\log qzxwvtnp d qzxwvtnp=-1\n\\]\n\nIt follows that the improper integral\n\\[\n\\int_{0}^{\\pi / 2} \\log \\sin qzxwvtnp d qzxwvtnp=\\lim _{asdfghjk \\rightarrow 0} \\int_{asdfghjk}^{\\pi / 2} \\log \\sin qzxwvtnp d qzxwvtnp\n\\]\nexists. The several substitutions employed above can now be justified routinely.\n>>>\n"
    },
    "kernel_variant": {
      "question": "Let $\\log_{5}$ denote the logarithm to base $5$ and let  \n\\[\n\\Delta=\\Bigl\\{(x,y)\\in\\mathbb{R}^{2}\\mid 0\\le x\\le y\\le\\tfrac{\\pi}{2}\\Bigr\\}\n\\subset\\mathbb{R}^{2}.\n\\]\n\n(a)  Prove that the two–dimensional improper integral  \n\\[\nJ=\\iint_{\\Delta}\\log_{5}\\!\\bigl(\\sin(x+y)\\bigr)\\,dy\\,dx\n\\]\nis absolutely convergent, i.e.  \n\\[\n\\iint_{\\Delta}\\Bigl|\\log_{5}\\!\\bigl(\\sin(x+y)\\bigr)\\Bigr|\\,dy\\,dx<\\infty .\n\\]\n\n(b)  Evaluate $J$ in closed form.",
      "solution": "Throughout $\\ln$ denotes the natural logarithm and $\\log_{5}t=\\dfrac{\\ln t}{\\ln 5}$.\n\n\\textbf{Step 0 – Absolute integrability}  \nPut $u=x+y$.  For every $u\\in[0,\\pi]$ the intersection of $\\Delta$ with the line\n$x+y=u$ is  \n\\[\n\\Bigl\\{\\,(x,y)\\in\\Delta\\mid x+y=u\\Bigr\\}\n     =\\Bigl\\{\\,(x,u-x)\\mid \\max\\!\\bigl(0,u-\\tfrac{\\pi}{2}\\bigr)\\le x\\le\\tfrac{u}{2}\\Bigr\\},\n\\]\nwhose length equals  \n\\[\nw(u)=\\frac12\n\\begin{cases}\nu, & 0\\le u\\le\\tfrac{\\pi}{2},\\\\\n\\pi-u,&\\tfrac{\\pi}{2}\\le u\\le\\pi .\n\\end{cases}\\tag{1}\n\\]\n\nApplying Tonelli’s theorem to the non–negative integrand\n$\\lvert\\log_{5}(\\sin(x+y))\\rvert$ gives  \n\\[\n\\iint_{\\Delta}\\Bigl|\\log_{5}\\!\\bigl(\\sin(x+y)\\bigr)\\Bigr|\\,dy\\,dx\n      =\\frac{1}{\\ln 5}\\int_{0}^{\\pi} w(u)\\,\\bigl|\\ln(\\sin u)\\bigr|\\,du .\n\\tag{2}\n\\]\n\nNear $u=0$ we have $\\sin u\\sim u$, hence $\\lvert\\ln(\\sin u)\\rvert\\le C_{0}\\lvert\\ln u\\rvert$\nfor $0<u<1$ and some absolute constant $C_{0}$.  Because $w(u)=u/2$ on\n$[0,\\pi/2]$, the contribution of $(0,\\varepsilon)$ to (2) is bounded by  \n\\[\n\\frac{C_{0}}{2\\ln 5}\\int_{0}^{\\varepsilon}u\\lvert\\ln u\\rvert\\,du\n      =\\frac{C_{0}}{4\\ln 5}\\,\\varepsilon^{2}\\bigl|\\ln\\varepsilon\\bigr|\n      \\xrightarrow[\\varepsilon\\downarrow0]{}0 .\n\\]\nAn analogous estimate at $u=\\pi$ proves integrability there as well, while on every\ncompact subinterval of $(0,\\pi)$ the integrand is continuous.  Hence the integral in (2) is finite, and $J$ is absolutely convergent.\n\n\\textbf{Step 1 – Reduction to one dimension}  \nBecause of absolute convergence we may change the order of integration:\n\\[\nJ=\\int_{0}^{\\pi}w(u)\\,\\log_{5}(\\sin u)\\,du.\\tag{3}\n\\]\nSince $w(u)=w(\\pi-u)$, the two halves of $[0,\\pi]$ contribute equally, so\n\\[\nJ= \\frac{1}{\\ln 5}\\,K,\\qquad\nK:=\\int_{0}^{\\pi/2}u\\,\\ln(\\sin u)\\,du .\\tag{4}\n\\]\n\n\\textbf{Step 2 – Fourier expansion and justified interchange of limit and integral}\n\nFor $0<u<\\pi$ the classical Fourier series  \n\\[\n\\ln\\bigl(2\\sin u\\bigr)=-\\sum_{n=1}^{\\infty}\\frac{\\cos(2nu)}{n}\\tag{5}\n\\]\nconverges pointwise.  Rewrite\n\\[\n\\ln(\\sin u)=-\\ln2-\\sum_{n=1}^{\\infty}\\frac{\\cos(2nu)}{n}.\\tag{6}\n\\]\n\nDefine the partial sums $S_{N}(u)=\\sum_{n=1}^{N}\\dfrac{\\cos(2nu)}{n}$.\nTo insert series (6) into (4) we must justify\n\\[\n\\int_{0}^{\\pi/2} u\\,S_{N}(u)\\,du \\longrightarrow\n      \\int_{0}^{\\pi/2}u\\left(\\sum_{n=1}^{\\infty}\\frac{\\cos(2nu)}{n}\\right)du\n\\qquad(N\\to\\infty).\\tag{7}\n\\]\n\nSplit the interval:\n\\[\n\\bigl(0,\\tfrac{\\pi}{2}\\bigr]= (0,\\delta]\\cup[\\delta,\\tfrac{\\pi}{2}]\n\\quad\\text{ with }\\quad 0<\\delta<\\tfrac{\\pi}{2}.\n\\]\n\n(i) On $[\\delta,\\pi/2]$ the Dirichlet kernel implies\n$\\lvert S_{N}(u)\\rvert\\le M_{\\delta}$ for all $u$ in this subinterval and\nevery $N$, where $M_{\\delta}$ is a finite constant depending on $\\delta$.\nConsequently $u\\lvert S_{N}(u)\\rvert\\le (\\pi/2)M_{\\delta}$, an integrable\nmajorant; dominated convergence applies.\n\n(ii) On $(0,\\delta]$ one uses  \n\\[\n\\lvert S_{N}(u)\\rvert\\le C\\,|\\ln u| \\qquad\\bigl(0<u\\le\\delta,\\; N\\ge1\\bigr)\n\\]\nfor some absolute constant $C$.  Hence\n\\[\nu\\,\\lvert S_{N}(u)\\rvert\\le C\\,u\\,|\\ln u|,\n\\]\nand $u|\\ln u|$ is integrable at $0$.  By dominated convergence again,\nthe interchange in (7) is valid.\n\nThus\n\\[\nK=-\\ln2\\int_{0}^{\\pi/2}u\\,du\n    -\\sum_{n=1}^{\\infty}\\frac{1}{n}\\int_{0}^{\\pi/2}u\\,\\cos(2nu)\\,du .\\tag{8}\n\\]\n\n\\textbf{Step 3 – Evaluation of the oscillatory integrals}  \nFor $n\\ge1$ set\n\\[\nI_{n}:=\\int_{0}^{\\pi/2}u\\,\\cos(2nu)\\,du.\\tag{9}\n\\]\nAn integration by parts yields\n\\[\n\\begin{aligned}\nI_{n}&=\\Bigl[\\frac{u\\sin(2nu)}{2n}\\Bigr]_{0}^{\\pi/2}\n        -\\frac{1}{2n}\\int_{0}^{\\pi/2}\\sin(2nu)\\,du\n      =-\\frac{1-(-1)^{n}}{4n^{2}}\\\\\n      &=\n      \\begin{cases}\n      0,& n\\ \\text{even},\\\\[6pt]\n      -\\dfrac{1}{2n^{2}},& n\\ \\text{odd}.\n      \\end{cases}\n\\end{aligned}\\tag{10}\n\\]\n\n\\textbf{Step 4 – Summation}  \nInsert (10) into (8):\n\\[\n\\begin{aligned}\nK&=-\\frac{\\pi^{2}}{8}\\ln2\n     -\\sum_{k=0}^{\\infty}\\frac{1}{2k+1}\\Bigl(-\\frac{1}{2(2k+1)^{2}}\\Bigr)\\\\[4pt]\n  &= -\\frac{\\pi^{2}}{8}\\ln2\n     +\\frac12\\sum_{k=0}^{\\infty}\\frac{1}{(2k+1)^{3}}\n     =-\\frac{\\pi^{2}}{8}\\ln2+\\frac12\\Bigl(\\zeta(3)-\\frac18\\zeta(3)\\Bigr)\\\\[2pt]\n  &= -\\frac{\\pi^{2}}{8}\\ln2+\\frac{7}{16}\\,\\zeta(3).\n\\end{aligned}\\tag{11}\n\\]\n\n\\textbf{Step 5 – Final result}  \nCombining (4) and (11) we obtain\n\\[\n\\boxed{\\;\n   J=-\\frac{\\pi^{2}\\,\\ln2}{8\\,\\ln5}+\\frac{7\\,\\zeta(3)}{16\\,\\ln5}\n   \\;}\n\\quad\\bigl(\\;J\\approx-0.204927078\\bigr).\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.455191",
        "was_fixed": false,
        "difficulty_analysis": "1.  Dimension jump:  The problem is now two–dimensional, requiring skilful changes of variables and geometric reasoning about the integration region.\n\n2.  Additional structures:  A Fourier expansion of $\\ln(\\sin u)$ and non-trivial series evaluation appear.  The solution introduces Riemann’s zeta function at $3$, far beyond the elementary constants of the original problem.\n\n3.  Deeper theory:  Convergence of an infinite series under the integral sign, interchange of orders of integration, and careful justification of each step are necessary.  None of these issues arose in the one-dimensional kernel problem.\n\n4.  Multiple interacting concepts:  Geometry of the domain, real analysis (improper and double integrals), Fourier analysis, series acceleration, and special values of the zeta function all interplay.\n\n5.  Longer chain of reasoning:  The original integral collapses in a few lines; here one must  \n   • transform a domain,  \n   • identify a symmetry,  \n   • exploit a Fourier series,  \n   • evaluate a delicate alternating series, and  \n   • convert between logarithm bases.  \n\nAll these upgrades make the enhanced variant substantially harder while still revolving around the fundamental idea of integrating $\\log(\\sin)$–type functions."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\log_{5}$ denote the logarithm to base $5$ and let  \n\\[\n\\Delta=\\Bigl\\{(x,y)\\in\\mathbb{R}^{2}\\mid 0\\le x\\le y\\le\\tfrac{\\pi}{2}\\Bigr\\}\n\\subset\\mathbb{R}^{2}.\n\\]\n\n(a)  Prove that the two–dimensional improper integral  \n\\[\nJ=\\iint_{\\Delta}\\log_{5}\\!\\bigl(\\sin(x+y)\\bigr)\\,dy\\,dx\n\\]\nis absolutely convergent, i.e.  \n\\[\n\\iint_{\\Delta}\\Bigl|\\log_{5}\\!\\bigl(\\sin(x+y)\\bigr)\\Bigr|\\,dy\\,dx<\\infty .\n\\]\n\n(b)  Evaluate $J$ in closed form.",
      "solution": "Throughout $\\ln$ denotes the natural logarithm and $\\log_{5}t=\\dfrac{\\ln t}{\\ln 5}$.\n\n\\textbf{Step 0 – Absolute integrability}  \nPut $u=x+y$.  For every $u\\in[0,\\pi]$ the intersection of $\\Delta$ with the line\n$x+y=u$ is  \n\\[\n\\Bigl\\{\\,(x,y)\\in\\Delta\\mid x+y=u\\Bigr\\}\n     =\\Bigl\\{\\,(x,u-x)\\mid \\max\\!\\bigl(0,u-\\tfrac{\\pi}{2}\\bigr)\\le x\\le\\tfrac{u}{2}\\Bigr\\},\n\\]\nwhose length equals  \n\\[\nw(u)=\\frac12\n\\begin{cases}\nu, & 0\\le u\\le\\tfrac{\\pi}{2},\\\\\n\\pi-u,&\\tfrac{\\pi}{2}\\le u\\le\\pi .\n\\end{cases}\\tag{1}\n\\]\n\nApplying Tonelli’s theorem to the non–negative integrand\n$\\lvert\\log_{5}(\\sin(x+y))\\rvert$ gives  \n\\[\n\\iint_{\\Delta}\\Bigl|\\log_{5}\\!\\bigl(\\sin(x+y)\\bigr)\\Bigr|\\,dy\\,dx\n      =\\frac{1}{\\ln 5}\\int_{0}^{\\pi} w(u)\\,\\bigl|\\ln(\\sin u)\\bigr|\\,du .\n\\tag{2}\n\\]\n\nNear $u=0$ we have $\\sin u\\sim u$, hence $\\lvert\\ln(\\sin u)\\rvert\\le C_{0}\\lvert\\ln u\\rvert$\nfor $0<u<1$ and some absolute constant $C_{0}$.  Because $w(u)=u/2$ on\n$[0,\\pi/2]$, the contribution of $(0,\\varepsilon)$ to (2) is bounded by  \n\\[\n\\frac{C_{0}}{2\\ln 5}\\int_{0}^{\\varepsilon}u\\lvert\\ln u\\rvert\\,du\n      =\\frac{C_{0}}{4\\ln 5}\\,\\varepsilon^{2}\\bigl|\\ln\\varepsilon\\bigr|\n      \\xrightarrow[\\varepsilon\\downarrow0]{}0 .\n\\]\nAn analogous estimate at $u=\\pi$ proves integrability there as well, while on every\ncompact subinterval of $(0,\\pi)$ the integrand is continuous.  Hence the integral in (2) is finite, and $J$ is absolutely convergent.\n\n\\textbf{Step 1 – Reduction to one dimension}  \nBecause of absolute convergence we may change the order of integration:\n\\[\nJ=\\int_{0}^{\\pi}w(u)\\,\\log_{5}(\\sin u)\\,du.\\tag{3}\n\\]\nSince $w(u)=w(\\pi-u)$, the two halves of $[0,\\pi]$ contribute equally, so\n\\[\nJ= \\frac{1}{\\ln 5}\\,K,\\qquad\nK:=\\int_{0}^{\\pi/2}u\\,\\ln(\\sin u)\\,du .\\tag{4}\n\\]\n\n\\textbf{Step 2 – Fourier expansion and justified interchange of limit and integral}\n\nFor $0<u<\\pi$ the classical Fourier series  \n\\[\n\\ln\\bigl(2\\sin u\\bigr)=-\\sum_{n=1}^{\\infty}\\frac{\\cos(2nu)}{n}\\tag{5}\n\\]\nconverges pointwise.  Rewrite\n\\[\n\\ln(\\sin u)=-\\ln2-\\sum_{n=1}^{\\infty}\\frac{\\cos(2nu)}{n}.\\tag{6}\n\\]\n\nDefine the partial sums $S_{N}(u)=\\sum_{n=1}^{N}\\dfrac{\\cos(2nu)}{n}$.\nTo insert series (6) into (4) we must justify\n\\[\n\\int_{0}^{\\pi/2} u\\,S_{N}(u)\\,du \\longrightarrow\n      \\int_{0}^{\\pi/2}u\\left(\\sum_{n=1}^{\\infty}\\frac{\\cos(2nu)}{n}\\right)du\n\\qquad(N\\to\\infty).\\tag{7}\n\\]\n\nSplit the interval:\n\\[\n\\bigl(0,\\tfrac{\\pi}{2}\\bigr]= (0,\\delta]\\cup[\\delta,\\tfrac{\\pi}{2}]\n\\quad\\text{ with }\\quad 0<\\delta<\\tfrac{\\pi}{2}.\n\\]\n\n(i) On $[\\delta,\\pi/2]$ the Dirichlet kernel implies\n$\\lvert S_{N}(u)\\rvert\\le M_{\\delta}$ for all $u$ in this subinterval and\nevery $N$, where $M_{\\delta}$ is a finite constant depending on $\\delta$.\nConsequently $u\\lvert S_{N}(u)\\rvert\\le (\\pi/2)M_{\\delta}$, an integrable\nmajorant; dominated convergence applies.\n\n(ii) On $(0,\\delta]$ one uses  \n\\[\n\\lvert S_{N}(u)\\rvert\\le C\\,|\\ln u| \\qquad\\bigl(0<u\\le\\delta,\\; N\\ge1\\bigr)\n\\]\nfor some absolute constant $C$.  Hence\n\\[\nu\\,\\lvert S_{N}(u)\\rvert\\le C\\,u\\,|\\ln u|,\n\\]\nand $u|\\ln u|$ is integrable at $0$.  By dominated convergence again,\nthe interchange in (7) is valid.\n\nThus\n\\[\nK=-\\ln2\\int_{0}^{\\pi/2}u\\,du\n    -\\sum_{n=1}^{\\infty}\\frac{1}{n}\\int_{0}^{\\pi/2}u\\,\\cos(2nu)\\,du .\\tag{8}\n\\]\n\n\\textbf{Step 3 – Evaluation of the oscillatory integrals}  \nFor $n\\ge1$ set\n\\[\nI_{n}:=\\int_{0}^{\\pi/2}u\\,\\cos(2nu)\\,du.\\tag{9}\n\\]\nAn integration by parts yields\n\\[\n\\begin{aligned}\nI_{n}&=\\Bigl[\\frac{u\\sin(2nu)}{2n}\\Bigr]_{0}^{\\pi/2}\n        -\\frac{1}{2n}\\int_{0}^{\\pi/2}\\sin(2nu)\\,du\n      =-\\frac{1-(-1)^{n}}{4n^{2}}\\\\\n      &=\n      \\begin{cases}\n      0,& n\\ \\text{even},\\\\[6pt]\n      -\\dfrac{1}{2n^{2}},& n\\ \\text{odd}.\n      \\end{cases}\n\\end{aligned}\\tag{10}\n\\]\n\n\\textbf{Step 4 – Summation}  \nInsert (10) into (8):\n\\[\n\\begin{aligned}\nK&=-\\frac{\\pi^{2}}{8}\\ln2\n     -\\sum_{k=0}^{\\infty}\\frac{1}{2k+1}\\Bigl(-\\frac{1}{2(2k+1)^{2}}\\Bigr)\\\\[4pt]\n  &= -\\frac{\\pi^{2}}{8}\\ln2\n     +\\frac12\\sum_{k=0}^{\\infty}\\frac{1}{(2k+1)^{3}}\n     =-\\frac{\\pi^{2}}{8}\\ln2+\\frac12\\Bigl(\\zeta(3)-\\frac18\\zeta(3)\\Bigr)\\\\[2pt]\n  &= -\\frac{\\pi^{2}}{8}\\ln2+\\frac{7}{16}\\,\\zeta(3).\n\\end{aligned}\\tag{11}\n\\]\n\n\\textbf{Step 5 – Final result}  \nCombining (4) and (11) we obtain\n\\[\n\\boxed{\\;\n   J=-\\frac{\\pi^{2}\\,\\ln2}{8\\,\\ln5}+\\frac{7\\,\\zeta(3)}{16\\,\\ln5}\n   \\;}\n\\quad\\bigl(\\;J\\approx-0.204927078\\bigr).\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.388866",
        "was_fixed": false,
        "difficulty_analysis": "1.  Dimension jump:  The problem is now two–dimensional, requiring skilful changes of variables and geometric reasoning about the integration region.\n\n2.  Additional structures:  A Fourier expansion of $\\ln(\\sin u)$ and non-trivial series evaluation appear.  The solution introduces Riemann’s zeta function at $3$, far beyond the elementary constants of the original problem.\n\n3.  Deeper theory:  Convergence of an infinite series under the integral sign, interchange of orders of integration, and careful justification of each step are necessary.  None of these issues arose in the one-dimensional kernel problem.\n\n4.  Multiple interacting concepts:  Geometry of the domain, real analysis (improper and double integrals), Fourier analysis, series acceleration, and special values of the zeta function all interplay.\n\n5.  Longer chain of reasoning:  The original integral collapses in a few lines; here one must  \n   • transform a domain,  \n   • identify a symmetry,  \n   • exploit a Fourier series,  \n   • evaluate a delicate alternating series, and  \n   • convert between logarithm bases.  \n\nAll these upgrades make the enhanced variant substantially harder while still revolving around the fundamental idea of integrating $\\log(\\sin)$–type functions."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}