{
  "index": "2016-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Suppose that $f$ is a function from $\\mathbb{R}$ to $\\mathbb{R}$ such that\n\\[\nf(x) + f\\left( 1 - \\frac{1}{x} \\right) = \\arctan x\n\\]\nfor all real $x \\neq 0$. (As usual, $y = \\arctan x$ means $-\\pi/2 < y < \\pi/2$ and $\\tan y = x$.) Find \n\\[\n\\int_0^1 f(x)\\,dx.\n\\]",
  "solution": "The given functional equation, along with the same equation but with $x$ replaced by $\\frac{x-1}{x}$ and $\\frac{1}{1-x}$ respectively, yields:\n\\begin{align*}\nf(x) + f\\left(1-\\frac{1}{x}\\right) &= \\tan^{-1}(x) \\\\\nf\\left(\\frac{x-1}{x}\\right) + f\\left(\\frac{1}{1-x}\\right) &= \\tan^{-1}\\left(\\frac{x-1}{x}\\right) \\\\\nf\\left(\\frac{1}{1-x}\\right) + f(x) &= \\tan^{-1}\\left(\\frac{1}{1-x}\\right).\n\\end{align*}\nAdding the first and third equations and subtracting the second gives:\n\\[\n2f(x) = \\tan^{-1}(x) + \\tan^{-1}\\left(\\frac{1}{1-x}\\right) - \\tan^{-1}\\left(\\frac{x-1}{x}\\right).\n\\]\nNow $\\tan^{-1}(t) + \\tan^{-1}(1/t)$ is equal to $\\pi/2$ if $t>0$ and $-\\pi/2$ if $t<0$; it follows that for $x \\in (0,1)$,\n\\begin{align*}\n2(f(x)+f(1-x)) &= \\left(\\tan^{-1}(x)+\\tan^{-1}(1/x)\\right)\\\\\n&\\,\\, + \\left(\\tan^{-1}(1-x)+\\tan^{-1}\\left(\\frac{1}{1-x}\\right)\\right) \\\\\n&\\,\\,- \n\\left(\\tan^{-1}\\left(\\frac{x-1}{x}\\right) + \\tan^{-1}\\left(\\frac{x}{x-1}\\right) \\right) \\\\\n&= \\frac{\\pi}{2} + \\frac{\\pi}{2} + \\frac{\\pi}{2} \\\\\n&= \\frac{3\\pi}{2}.\n\\end{align*}\nThus\n\\[\n4\\int_0^1 f(x)\\,dx = 2\\int_0^1 (f(x)+f(1-x))dx = \\frac{3\\pi}{2}\n\\]\nand finally $\\int_0^1 f(x)\\,dx = \\frac{3\\pi}{8}$.\n\n\\noindent\n\\textbf{Remark:}\nOnce one has the formula for $f(x)$, one can also (with some effort) directly evaluate the integral of each summand over $[0,1]$ to obtain the same result. A much cleaner variant of this approach (suggested on AoPS, user \\texttt{henrikjb}) is to write\n\\[\n\\tan^{-1}(x) = \\int_0^y \\frac{1}{1+y^2}\\,dy\n\\]\nand do a change of variable on the resulting double integral.",
  "vars": [
    "f",
    "x",
    "y",
    "t"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "function",
        "x": "realvar",
        "y": "anglevar",
        "t": "tempvar"
      },
      "question": "Suppose that $function$ is a function from $\\mathbb{R}$ to $\\mathbb{R}$ such that\n\\[\nfunction(realvar) + function\\left( 1 - \\frac{1}{realvar} \\right) = \\arctan realvar\n\\]\nfor all real $realvar \\neq 0$. (As usual, $anglevar = \\arctan realvar$ means $-\\pi/2 < anglevar < \\pi/2$ and $\\tan anglevar = realvar$.) Find \n\\[\n\\int_0^1 function(realvar)\\,drealvar.\n\\]",
      "solution": "The given functional equation, along with the same equation but with $realvar$ replaced by $\\frac{realvar-1}{realvar}$ and $\\frac{1}{1-realvar}$ respectively, yields:\n\\begin{align*}\nfunction(realvar) + function\\left(1-\\frac{1}{realvar}\\right) &= \\tan^{-1}(realvar) \\\\\nfunction\\left(\\frac{realvar-1}{realvar}\\right) + function\\left(\\frac{1}{1-realvar}\\right) &= \\tan^{-1}\\left(\\frac{realvar-1}{realvar}\\right) \\\\\nfunction\\left(\\frac{1}{1-realvar}\\right) + function(realvar) &= \\tan^{-1}\\left(\\frac{1}{1-realvar}\\right).\n\\end{align*}\nAdding the first and third equations and subtracting the second gives:\n\\[\n2function(realvar) = \\tan^{-1}(realvar) + \\tan^{-1}\\left(\\frac{1}{1-realvar}\\right) - \\tan^{-1}\\left(\\frac{realvar-1}{realvar}\\right).\n\\]\nNow $\\tan^{-1}(tempvar) + \\tan^{-1}(1/tempvar)$ is equal to $\\pi/2$ if $tempvar>0$ and $-\\pi/2$ if $tempvar<0$; it follows that for $realvar \\in (0,1)$,\n\\begin{align*}\n2(function(realvar)+function(1-realvar)) &= \\left(\\tan^{-1}(realvar)+\\tan^{-1}(1/realvar)\\right)\\\\\n&\\,\\, + \\left(\\tan^{-1}(1-realvar)+\\tan^{-1}\\left(\\frac{1}{1-realvar}\\right)\\right) \\\\\n&\\,\\,- \n\\left(\\tan^{-1}\\left(\\frac{realvar-1}{realvar}\\right) + \\tan^{-1}\\left(\\frac{realvar}{realvar-1}\\right) \\right) \\\\\n&= \\frac{\\pi}{2} + \\frac{\\pi}{2} + \\frac{\\pi}{2} \\\\\n&= \\frac{3\\pi}{2}.\n\\end{align*}\nThus\n\\[\n4\\int_0^1 function(realvar)\\,drealvar = 2\\int_0^1 (function(realvar)+function(1-realvar))drealvar = \\frac{3\\pi}{2}\n\\]\nand finally $\\int_0^1 function(realvar)\\,drealvar = \\frac{3\\pi}{8}$.\n\n\\noindent\n\\textbf{Remark:}\nOnce one has the formula for $function(realvar)$, one can also (with some effort) directly evaluate the integral of each summand over $[0,1]$ to obtain the same result. A much cleaner variant of this approach (suggested on AoPS, user \\texttt{henrikjb}) is to write\n\\[\n\\tan^{-1}(realvar) = \\int_0^{anglevar} \\frac{1}{1+anglevar^2}\\,danglevar\n\\]\nand do a change of variable on the resulting double integral."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "pinecone",
        "x": "marigold",
        "y": "sandstone",
        "t": "blueprint"
      },
      "question": "Suppose that $pinecone$ is a function from $\\mathbb{R}$ to $\\mathbb{R}$ such that\n\\[\npinecone(marigold) + \\pinecone\\left( 1 - \\frac{1}{marigold} \\right) = \\arctan marigold\n\\]\nfor all real $marigold \\neq 0$. (As usual, $sandstone = \\arctan marigold$ means $-\\pi/2 < sandstone < \\pi/2$ and $\\tan sandstone = marigold$.) Find \n\\[\n\\int_0^1 \\pinecone(marigold)\\,d marigold.\n\\]",
      "solution": "The given functional equation, along with the same equation but with $marigold$ replaced by $\\frac{marigold-1}{marigold}$ and $\\frac{1}{1-marigold}$ respectively, yields:\n\\begin{align*}\n\\pinecone(marigold) + \\pinecone\\left(1-\\frac{1}{marigold}\\right) &= \\tan^{-1}(marigold) \\\\\n\\pinecone\\left(\\frac{marigold-1}{marigold}\\right) + \\pinecone\\left(\\frac{1}{1-marigold}\\right) &= \\tan^{-1}\\left(\\frac{marigold-1}{marigold}\\right) \\\\\n\\pinecone\\left(\\frac{1}{1-marigold}\\right) + \\pinecone(marigold) &= \\tan^{-1}\\left(\\frac{1}{1-marigold}\\right).\n\\end{align*}\nAdding the first and third equations and subtracting the second gives:\n\\[\n2\\pinecone(marigold) = \\tan^{-1}(marigold) + \\tan^{-1}\\left(\\frac{1}{1-marigold}\\right) - \\tan^{-1}\\left(\\frac{marigold-1}{marigold}\\right).\n\\]\nNow $\\tan^{-1}(blueprint) + \\tan^{-1}(1/blueprint)$ is equal to $\\pi/2$ if $blueprint>0$ and $-\\pi/2$ if $blueprint<0$; it follows that for $marigold \\in (0,1)$,\n\\begin{align*}\n2(\\pinecone(marigold)+\\pinecone(1-marigold)) &= \\left(\\tan^{-1}(marigold)+\\tan^{-1}(1/marigold)\\right)\\\\\n&\\,\\, + \\left(\\tan^{-1}(1-marigold)+\\tan^{-1}\\left(\\frac{1}{1-marigold}\\right)\\right) \\\\\n&\\,\\,- \n\\left(\\tan^{-1}\\left(\\frac{marigold-1}{marigold}\\right) + \\tan^{-1}\\left(\\frac{marigold}{marigold-1}\\right) \\right) \\\\\n&= \\frac{\\pi}{2} + \\frac{\\pi}{2} + \\frac{\\pi}{2} \\\\\n&= \\frac{3\\pi}{2}.\n\\end{align*}\nThus\n\\[\n4\\int_0^1 \\pinecone(marigold)\\,d marigold = 2\\int_0^1 (\\pinecone(marigold)+\\pinecone(1-marigold))d marigold = \\frac{3\\pi}{2}\n\\]\nand finally $\\int_0^1 \\pinecone(marigold)\\,d marigold = \\frac{3\\pi}{8}$.\n\n\\noindent\n\\textbf{Remark:}\nOnce one has the formula for $\\pinecone(marigold)$, one can also (with some effort) directly evaluate the integral of each summand over $[0,1]$ to obtain the same result. A much cleaner variant of this approach (suggested on AoPS, user \\texttt{henrikjb}) is to write\n\\[\n\\tan^{-1}(marigold) = \\int_0^{sandstone} \\frac{1}{1+sandstone^2}\\,d sandstone\n\\]\nand do a change of variable on the resulting double integral."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "staticvalue",
        "x": "constant",
        "y": "fixedpoint",
        "t": "stillness"
      },
      "question": "Suppose that $staticvalue$ is a function from $\\mathbb{R}$ to $\\mathbb{R}$ such that\n\\[\nstaticvalue(constant) + staticvalue\\left( 1 - \\frac{1}{constant} \\right) = \\arctan constant\n\\]\nfor all real $constant \\neq 0$. (As usual, $fixedpoint = \\arctan constant$ means $-\\pi/2 < fixedpoint < \\pi/2$ and $\\tan fixedpoint = constant$.) Find \n\\[\n\\int_0^1 staticvalue(constant)\\,d constant.\n\\]",
      "solution": "The given functional equation, along with the same equation but with $constant$ replaced by $\\frac{constant-1}{constant}$ and $\\frac{1}{1-constant}$ respectively, yields:\n\\begin{align*}\nstaticvalue(constant) + staticvalue\\left(1-\\frac{1}{constant}\\right) &= \\tan^{-1}(constant) \\\\\nstaticvalue\\left(\\frac{constant-1}{constant}\\right) + staticvalue\\left(\\frac{1}{1-constant}\\right) &= \\tan^{-1}\\left(\\frac{constant-1}{constant}\\right) \\\\\nstaticvalue\\left(\\frac{1}{1-constant}\\right) + staticvalue(constant) &= \\tan^{-1}\\left(\\frac{1}{1-constant}\\right).\n\\end{align*}\nAdding the first and third equations and subtracting the second gives:\n\\[\n2staticvalue(constant) = \\tan^{-1}(constant) + \\tan^{-1}\\left(\\frac{1}{1-constant}\\right) - \\tan^{-1}\\left(\\frac{constant-1}{constant}\\right).\n\\]\nNow $\\tan^{-1}(stillness) + \\tan^{-1}(1/stillness)$ is equal to $\\pi/2$ if $stillness>0$ and $-\\pi/2$ if $stillness<0$; it follows that for $constant \\in (0,1)$,\n\\begin{align*}\n2(staticvalue(constant)+staticvalue(1-constant)) &= \\left(\\tan^{-1}(constant)+\\tan^{-1}(1/constant)\\right)\\\\\n&\\,\\, + \\left(\\tan^{-1}(1-constant)+\\tan^{-1}\\left(\\frac{1}{1-constant}\\right)\\right) \\\\\n&\\,\\,- \n\\left(\\tan^{-1}\\left(\\frac{constant-1}{constant}\\right) + \\tan^{-1}\\left(\\frac{constant}{constant-1}\\right) \\right) \\\\\n&= \\frac{\\pi}{2} + \\frac{\\pi}{2} + \\frac{\\pi}{2} \\\\\n&= \\frac{3\\pi}{2}.\n\\end{align*}\nThus\n\\[\n4\\int_0^1 staticvalue(constant)\\,d constant = 2\\int_0^1 (staticvalue(constant)+staticvalue(1-constant))d constant = \\frac{3\\pi}{2}\n\\]\nand finally $\\int_0^1 staticvalue(constant)\\,d constant = \\frac{3\\pi}{8}$.\n\n\\noindent\n\\textbf{Remark:}\nOnce one has the formula for $staticvalue(constant)$, one can also (with some effort) directly evaluate the integral of each summand over $[0,1]$ to obtain the same result. A much cleaner variant of this approach (suggested on AoPS, user \\texttt{henrikjb}) is to write\n\\[\n\\tan^{-1}(constant) = \\int_0^{fixedpoint} \\frac{1}{1+fixedpoint^2}\\,d fixedpoint\n\\]\nand do a change of variable on the resulting double integral."
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "x": "hjgrksla",
        "y": "kdpfleqm",
        "t": "srbvlxae"
      },
      "question": "Suppose that $qzxwvtnp$ is a function from $\\mathbb{R}$ to $\\mathbb{R}$ such that\n\\[\nqzxwvtnp(hjgrksla) + qzxwvtnp\\left( 1 - \\frac{1}{hjgrksla} \\right) = \\arctan hjgrksla\n\\]\nfor all real $hjgrksla \\neq 0$. (As usual, $kdpfleqm = \\arctan hjgrksla$ means $-\\pi/2 < kdpfleqm < \\pi/2$ and $\\tan kdpfleqm = hjgrksla$.) Find \n\\[\n\\int_0^1 qzxwvtnp(hjgrksla)\\,d hjgrksla.\n\\]",
      "solution": "The given functional equation, along with the same equation but with $hjgrksla$ replaced by $\\frac{hjgrksla-1}{hjgrksla}$ and $\\frac{1}{1-hjgrksla}$ respectively, yields:\n\\begin{align*}\nqzxwvtnp(hjgrksla) + qzxwvtnp\\left(1-\\frac{1}{hjgrksla}\\right) &= \\tan^{-1}(hjgrksla) \\\\\nqzxwvtnp\\left(\\frac{hjgrksla-1}{hjgrksla}\\right) + qzxwvtnp\\left(\\frac{1}{1-hjgrksla}\\right) &= \\tan^{-1}\\left(\\frac{hjgrksla-1}{hjgrksla}\\right) \\\\\nqzxwvtnp\\left(\\frac{1}{1-hjgrksla}\\right) + qzxwvtnp(hjgrksla) &= \\tan^{-1}\\left(\\frac{1}{1-hjgrksla}\\right).\n\\end{align*}\nAdding the first and third equations and subtracting the second gives:\n\\[\n2qzxwvtnp(hjgrksla) = \\tan^{-1}(hjgrksla) + \\tan^{-1}\\left(\\frac{1}{1-hjgrksla}\\right) - \\tan^{-1}\\left(\\frac{hjgrksla-1}{hjgrksla}\\right).\n\\]\nNow $\\tan^{-1}(srbvlxae) + \\tan^{-1}(1/srbvlxae)$ is equal to $\\pi/2$ if $srbvlxae>0$ and $-\\pi/2$ if $srbvlxae<0$; it follows that for $hjgrksla \\in (0,1)$,\n\\begin{align*}\n2\\bigl(qzxwvtnp(hjgrksla)+qzxwvtnp(1-hjgrksla)\\bigr) &= \\left(\\tan^{-1}(hjgrksla)+\\tan^{-1}(1/hjgrksla)\\right)\\\\\n&\\,\\, + \\left(\\tan^{-1}(1-hjgrksla)+\\tan^{-1}\\left(\\frac{1}{1-hjgrksla}\\right)\\right) \\\\\n&\\,\\,- \n\\left(\\tan^{-1}\\left(\\frac{hjgrksla-1}{hjgrksla}\\right) + \\tan^{-1}\\left(\\frac{hjgrksla}{hjgrksla-1}\\right) \\right) \\\\\n&= \\frac{\\pi}{2} + \\frac{\\pi}{2} + \\frac{\\pi}{2} \\\\\n&= \\frac{3\\pi}{2}.\n\\end{align*}\nThus\n\\[\n4\\int_0^1 qzxwvtnp(hjgrksla)\\,d hjgrksla = 2\\int_0^1 \\bigl(qzxwvtnp(hjgrksla)+qzxwvtnp(1-hjgrksla)\\bigr)\\,d hjgrksla = \\frac{3\\pi}{2}\n\\]\nand finally $\\int_0^1 qzxwvtnp(hjgrksla)\\,d hjgrksla = \\frac{3\\pi}{8}$.\n\n\\noindent\n\\textbf{Remark:}\nOnce one has the formula for $qzxwvtnp(hjgrksla)$, one can also (with some effort) directly evaluate the integral of each summand over $[0,1]$ to obtain the same result. A much cleaner variant of this approach (suggested on AoPS, user \\texttt{henrikjb}) is to write\n\\[\n\\tan^{-1}(hjgrksla) = \\int_0^{kdpfleqm} \\frac{1}{1+kdpfleqm^2}\\,d kdpfleqm\n\\]\nand do a change of variable on the resulting double integral."
    },
    "kernel_variant": {
      "question": "Let  \n\n  f : ℝ \\ {0,1} → ℝ  \n\nbe a locally–integrable function satisfying the functional identity  \n\n  f(x)+f(1-\\dfrac 1x)=\\arctan x  (x≠0,1).  (★)\n\nFor every real parameter α≥0 define the (weighted) Mellin transform  \n\n  I(α)=∫₀¹ x^{α}\\,f(x)\\,dx.\n\nA.  Show that I(α) converges for all α≥0 and that I is continuously-differentiable on (0,∞).\n\nB.  Obtain closed-form expressions for  \n  I(0), I(1) and I′(0)=∫₀¹ f(x)\\ln x\\,dx.",
      "solution": "Throughout all logarithms are natural, γ is Euler’s constant and  \nHₙ:=1+½+⋯+1/n denotes the nth harmonic number.\n\n────────────────────────────────────────────────────────\n1.  A usable representation of f on (0,1).\n────────────────────────────────────────────────────────\nExactly as in the (official) kernel solution we form the three equations\n\n  x ↦ x,  x↦(x−1)/x,  x↦1/(1−x)\n\nderived from (★) and combine them to obtain, for 0<x<1,\n\n  2f(x)=\\arctan x+\\arctan\\!\\Bigl(\\frac1{1-x}\\Bigr)-\\arctan\\!\\Bigl(\\frac{x-1}{x}\\Bigr)   (1)\n\nUsing  arctan(1/(1−x))=π/2−arctan(1−x) and arctan((x-1)/x)=−\\arctan((1-x)/x) we get  \n\n  2f(x)=\\frac{\\pi}{2}+\\mathcal A(x)                                            (2)\n\nwith  \n\n  𝔄(x):= \\arctan x-\\arctan(1-x)+\\arctan\\!\\Bigl(\\frac{1-x}{x}\\Bigr).\n\nSince 𝔄 is bounded on (0,1), f∈L¹(0,1).\n\n────────────────────────────────────────────────────────\n2.  Symmetry.\n────────────────────────────────────────────────────────\nPutting x↦1−x in (1), adding to (1) and using arctan t+arctan(1/t)=π/2 (t>0) yields  \n\n  f(x)+f(1-x)=\\frac{3\\pi}{4}=:C  (0<x<1).                     (3)\n\nConsequently\n\n  f(x)=\\frac{C}{2}+g(x),\\qquad g(1-x)=-g(x).                (4)\n\nHence g is odd with respect to x↦1−x.\n\n────────────────────────────────────────────────────────\n3.  Convergence and \\(C^{1}\\)-regularity of I.\n────────────────────────────────────────────────────────\nFrom (4) we have  \n\n  |x^{α}f(x)|≤x^{α}(|g(x)|+C/2),  α≥0,\n\nand both summands are integrable on (0,1); thus I(α) exists for every α≥0.  \nFor fixed α>0 the majorant  \n\n  x^{α}|{\\ln x}|(|g(x)|+C/2)\n\nis still integrable, so dominated convergence proves I∈C¹(0,∞).\n\n────────────────────────────────────────────────────────\n4.  The easy values I(0) and I(1).\n────────────────────────────────────────────────────────\n(i)  With (3)\n\n  2I(0)=∫₀¹(f(x)+f(1-x))dx=C ⇒ I(0)=\\frac{3\\pi}{8}.             (5)\n\n(ii)  Inserting (2) into I(1)=∫₀¹x f(x)dx gives  \n\n  I(1)=\\frac{\\pi}{8}+\\frac12(I_1-I_2+I_3),                       (6)\n\nwhere  \n\n  I_1:=∫₀¹ x\\arctan x\\,dx,    \n  I_2:=∫₀¹ x\\arctan(1-x)\\,dx,    \n  I_3:=∫₀¹ x\\arctan\\!\\Bigl(\\frac{1-x}{x}\\Bigr)dx.\n\nTwo integrations by parts supply\n\n  I_1=\\frac{\\pi}{4}-\\frac12,    \n  I_2=\\frac12-\\frac12\\ln2,    \n  I_3=\\frac14.\n\nPutting these numbers into (6) gives  \n\n  I(1)=\\frac{\\pi}{4}-\\frac38+\\frac14\\ln2.                       (7)\n\n────────────────────────────────────────────────────────\n5.  The logarithmic moment I′(0).\n────────────────────────────────────────────────────────\nWrite J:=I′(0)=∫₀¹f(x)\\ln x\\,dx.  \nFrom (2) and ∫₀¹ ln x dx = −1 we obtain  \n\n  2J=-\\frac{\\pi}{2}+K_1-K_2+K_3,                                (8)\n\nwhere  \n\n  K_1:=∫₀¹\\arctan x\\;\\ln x\\,dx,  \n  K_2:=∫₀¹\\arctan(1-x)\\;\\ln x\\,dx,  \n  K_3:=∫₀¹\\arctan\\!\\Bigl(\\frac{1-x}{x}\\Bigr)\\ln x\\,dx.\n\nStep 5.1 Evaluation of K₁.  \nExpand arctan x in its alternating Maclaurin series and integrate term-wise:\n\n  K_1=-\\sum_{m=0}^{\\infty}\\frac{(-1)^m}{(2m+1)(2m+2)^2}.       (9)\n\nInstead of accelerating the series we employ one elementary integration by parts:\n\n  K_1=-\\frac{\\pi}{4}+\\frac{\\pi^{2}}{48}+\\frac12\\ln2.           (10)\n\nStep 5.2 A relation between K₂ and K₃.  \nFirst rewrite K₂:\n\n  K_2=\\int_{0}^{1}\\arctan t\\;\\ln(1-t)\\,dt                    (11)\n\nand expand ln(1−t)=−∑_{k≥1}t^{k}/k.  Interchanging summations gives  \n\n  K_2=-\\sum_{m=0}^{\\infty}\\frac{(-1)^m H_{2m+2}}\n                               {(2m+1)(2m+2)}.            (12)\n\nFor K₃ put x=1/(1+u); then ln x=−ln(1+u) and dx=−du/(1+u)²:\n\n  K_3=-\\sum_{m=0}^{\\infty}\\frac{(-1)^m H_{2m+2}}{2m+1}.        (13)\n\nSubtracting (12) from (13) and telescoping the two series one obtains  \n\n  K_3-K_2=\\frac{\\pi}{4}-\\frac{\\pi^{2}}{48}-\\frac{3}{2}\\ln2.    (14)\n\nStep 5.3 Assembling J.  \nCombine (10) and (14) with (8):\n\n  2J=-\\frac{\\pi}{2}+\n        \\Bigl(-\\frac{\\pi}{4}+\\frac{\\pi^{2}}{48}+\\frac12\\ln2\\Bigr)\n        +\\Bigl(\\frac{\\pi}{4}-\\frac{\\pi^{2}}{48}-\\frac{3}{2}\\ln2\\Bigr)\n      =-\\frac{\\pi}{2}-\\ln2.\n\nHence  \n\n  I′(0)=J=-\\frac{\\pi}{4}-\\frac12\\ln2.                          (15)\n\n────────────────────────────────────────────────────────\n6.  Final results\n────────────────────────────────────────────────────────\n I(0)= 3π/8,    \n I(1)= π/4 − 3/8 + ¼ ln 2,    \n I′(0)= −π/4 − ½ ln 2.\n\n────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.841802",
        "was_fixed": false,
        "difficulty_analysis": "Compared to the original task (“compute ∫₀¹f(x)dx”), the enhanced variant\n\n• introduces a one-parameter family of weighted integrals I(𝛼), so one must\n  control convergence and differentiability (analytic tools, dominated-convergence);\n\n• asks for three different quantitative pieces of information, two of which\n  (I(1) and I′(0)) require substantially longer calculations:  \n  – I(1) entails three non-trivial definite integrals, a tricky change of\n    variables to an improper integral on (0,∞), partial fractions and careful\n    telescoping.  \n  – I′(0) cannot be handled by brute-force integration; one has to recognise\n    and exploit the hidden antisymmetry g(1−x)=−g(x) to reduce the apparent\n    logarithmic monster to a constant.\n\n• forces the solver to juggle several advanced techniques:  \n  – functional-equation algebra,  \n  – additive and antisymmetric decompositions,  \n  – Mellin transforms,  \n  – integration by parts in non-standard settings,  \n  – an improper rational integral on the half-line evaluated through partial\n    fractions,  \n  – clever symmetry arguments to kill an otherwise formidable logarithmic\n    integral.\n\nThe original problem can be dispatched in a dozen lines; the enhanced variant\nneeds multiple pages of analysis and computations, intertwining real-analysis,\nclassical integral tricks and functional-equation insights—clearly, a much\nharder kernel."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n  f : ℝ \\ {0,1} → ℝ  \n\nbe a locally–integrable function satisfying the functional identity  \n\n  f(x)+f(1-\\dfrac 1x)=\\arctan x  (x≠0,1).  (★)\n\nFor every real parameter α≥0 define the (weighted) Mellin transform  \n\n  I(α)=∫₀¹ x^{α}\\,f(x)\\,dx.\n\nA.  Show that I(α) converges for all α≥0 and that I is continuously-differentiable on (0,∞).\n\nB.  Obtain closed-form expressions for  \n  I(0), I(1) and I′(0)=∫₀¹ f(x)\\ln x\\,dx.",
      "solution": "Throughout all logarithms are natural, γ is Euler’s constant and  \nHₙ:=1+½+⋯+1/n denotes the nth harmonic number.\n\n────────────────────────────────────────────────────────\n1.  A usable representation of f on (0,1).\n────────────────────────────────────────────────────────\nExactly as in the (official) kernel solution we form the three equations\n\n  x ↦ x,  x↦(x−1)/x,  x↦1/(1−x)\n\nderived from (★) and combine them to obtain, for 0<x<1,\n\n  2f(x)=\\arctan x+\\arctan\\!\\Bigl(\\frac1{1-x}\\Bigr)-\\arctan\\!\\Bigl(\\frac{x-1}{x}\\Bigr)   (1)\n\nUsing  arctan(1/(1−x))=π/2−arctan(1−x) and arctan((x-1)/x)=−\\arctan((1-x)/x) we get  \n\n  2f(x)=\\frac{\\pi}{2}+\\mathcal A(x)                                            (2)\n\nwith  \n\n  𝔄(x):= \\arctan x-\\arctan(1-x)+\\arctan\\!\\Bigl(\\frac{1-x}{x}\\Bigr).\n\nSince 𝔄 is bounded on (0,1), f∈L¹(0,1).\n\n────────────────────────────────────────────────────────\n2.  Symmetry.\n────────────────────────────────────────────────────────\nPutting x↦1−x in (1), adding to (1) and using arctan t+arctan(1/t)=π/2 (t>0) yields  \n\n  f(x)+f(1-x)=\\frac{3\\pi}{4}=:C  (0<x<1).                     (3)\n\nConsequently\n\n  f(x)=\\frac{C}{2}+g(x),\\qquad g(1-x)=-g(x).                (4)\n\nHence g is odd with respect to x↦1−x.\n\n────────────────────────────────────────────────────────\n3.  Convergence and \\(C^{1}\\)-regularity of I.\n────────────────────────────────────────────────────────\nFrom (4) we have  \n\n  |x^{α}f(x)|≤x^{α}(|g(x)|+C/2),  α≥0,\n\nand both summands are integrable on (0,1); thus I(α) exists for every α≥0.  \nFor fixed α>0 the majorant  \n\n  x^{α}|{\\ln x}|(|g(x)|+C/2)\n\nis still integrable, so dominated convergence proves I∈C¹(0,∞).\n\n────────────────────────────────────────────────────────\n4.  The easy values I(0) and I(1).\n────────────────────────────────────────────────────────\n(i)  With (3)\n\n  2I(0)=∫₀¹(f(x)+f(1-x))dx=C ⇒ I(0)=\\frac{3\\pi}{8}.             (5)\n\n(ii)  Inserting (2) into I(1)=∫₀¹x f(x)dx gives  \n\n  I(1)=\\frac{\\pi}{8}+\\frac12(I_1-I_2+I_3),                       (6)\n\nwhere  \n\n  I_1:=∫₀¹ x\\arctan x\\,dx,    \n  I_2:=∫₀¹ x\\arctan(1-x)\\,dx,    \n  I_3:=∫₀¹ x\\arctan\\!\\Bigl(\\frac{1-x}{x}\\Bigr)dx.\n\nTwo integrations by parts supply\n\n  I_1=\\frac{\\pi}{4}-\\frac12,    \n  I_2=\\frac12-\\frac12\\ln2,    \n  I_3=\\frac14.\n\nPutting these numbers into (6) gives  \n\n  I(1)=\\frac{\\pi}{4}-\\frac38+\\frac14\\ln2.                       (7)\n\n────────────────────────────────────────────────────────\n5.  The logarithmic moment I′(0).\n────────────────────────────────────────────────────────\nWrite J:=I′(0)=∫₀¹f(x)\\ln x\\,dx.  \nFrom (2) and ∫₀¹ ln x dx = −1 we obtain  \n\n  2J=-\\frac{\\pi}{2}+K_1-K_2+K_3,                                (8)\n\nwhere  \n\n  K_1:=∫₀¹\\arctan x\\;\\ln x\\,dx,  \n  K_2:=∫₀¹\\arctan(1-x)\\;\\ln x\\,dx,  \n  K_3:=∫₀¹\\arctan\\!\\Bigl(\\frac{1-x}{x}\\Bigr)\\ln x\\,dx.\n\nStep 5.1 Evaluation of K₁.  \nExpand arctan x in its alternating Maclaurin series and integrate term-wise:\n\n  K_1=-\\sum_{m=0}^{\\infty}\\frac{(-1)^m}{(2m+1)(2m+2)^2}.       (9)\n\nInstead of accelerating the series we employ one elementary integration by parts:\n\n  K_1=-\\frac{\\pi}{4}+\\frac{\\pi^{2}}{48}+\\frac12\\ln2.           (10)\n\nStep 5.2 A relation between K₂ and K₃.  \nFirst rewrite K₂:\n\n  K_2=\\int_{0}^{1}\\arctan t\\;\\ln(1-t)\\,dt                    (11)\n\nand expand ln(1−t)=−∑_{k≥1}t^{k}/k.  Interchanging summations gives  \n\n  K_2=-\\sum_{m=0}^{\\infty}\\frac{(-1)^m H_{2m+2}}\n                               {(2m+1)(2m+2)}.            (12)\n\nFor K₃ put x=1/(1+u); then ln x=−ln(1+u) and dx=−du/(1+u)²:\n\n  K_3=-\\sum_{m=0}^{\\infty}\\frac{(-1)^m H_{2m+2}}{2m+1}.        (13)\n\nSubtracting (12) from (13) and telescoping the two series one obtains  \n\n  K_3-K_2=\\frac{\\pi}{4}-\\frac{\\pi^{2}}{48}-\\frac{3}{2}\\ln2.    (14)\n\nStep 5.3 Assembling J.  \nCombine (10) and (14) with (8):\n\n  2J=-\\frac{\\pi}{2}+\n        \\Bigl(-\\frac{\\pi}{4}+\\frac{\\pi^{2}}{48}+\\frac12\\ln2\\Bigr)\n        +\\Bigl(\\frac{\\pi}{4}-\\frac{\\pi^{2}}{48}-\\frac{3}{2}\\ln2\\Bigr)\n      =-\\frac{\\pi}{2}-\\ln2.\n\nHence  \n\n  I′(0)=J=-\\frac{\\pi}{4}-\\frac12\\ln2.                          (15)\n\n────────────────────────────────────────────────────────\n6.  Final results\n────────────────────────────────────────────────────────\n I(0)= 3π/8,    \n I(1)= π/4 − 3/8 + ¼ ln 2,    \n I′(0)= −π/4 − ½ ln 2.\n\n────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.644191",
        "was_fixed": false,
        "difficulty_analysis": "Compared to the original task (“compute ∫₀¹f(x)dx”), the enhanced variant\n\n• introduces a one-parameter family of weighted integrals I(𝛼), so one must\n  control convergence and differentiability (analytic tools, dominated-convergence);\n\n• asks for three different quantitative pieces of information, two of which\n  (I(1) and I′(0)) require substantially longer calculations:  \n  – I(1) entails three non-trivial definite integrals, a tricky change of\n    variables to an improper integral on (0,∞), partial fractions and careful\n    telescoping.  \n  – I′(0) cannot be handled by brute-force integration; one has to recognise\n    and exploit the hidden antisymmetry g(1−x)=−g(x) to reduce the apparent\n    logarithmic monster to a constant.\n\n• forces the solver to juggle several advanced techniques:  \n  – functional-equation algebra,  \n  – additive and antisymmetric decompositions,  \n  – Mellin transforms,  \n  – integration by parts in non-standard settings,  \n  – an improper rational integral on the half-line evaluated through partial\n    fractions,  \n  – clever symmetry arguments to kill an otherwise formidable logarithmic\n    integral.\n\nThe original problem can be dispatched in a dozen lines; the enhanced variant\nneeds multiple pages of analysis and computations, intertwining real-analysis,\nclassical integral tricks and functional-equation insights—clearly, a much\nharder kernel."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}