{
  "index": "1995-A-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "improper integral\n\\[\n\\int_{b}^{\\infty} \\left( \\sqrt{\\sqrt{x+a}-\\sqrt{x}} -\n\\sqrt{\\sqrt{x}-\\sqrt{x-b}} \\right)\\,dx\n\\]\nconverge?",
  "solution": "The integral converges iff $a=b$. The easiest proof uses\n``big-O'' notation and the fact that $(1+x)^{1/2} = 1 + x/2 +\nO(x^{2})$ for $|x|<1$. (Here $O(x^{2})$ means bounded by a constant\ntimes $x^{2}$.)\n\nSo\n\\begin{align*}\n\\sqrt{x+a}-\\sqrt{x} &= x^{1/2}(\\sqrt{1+a/x} - 1) \\\\\n&= x^{1/2}(1 + a/2x + O(x^{-2})),\n\\end{align*}\nhence\n\\[\n\\sqrt{\\sqrt{x+a} - \\sqrt{x}} = x^{1/4} (a/4x + O(x^{-2}))\n\\]\nand similarly\n\\[\n\\sqrt{\\sqrt{x} - \\sqrt{x-b}} = x^{1/4} (b/4x + O(x^{-2})).\n\\]\nHence the integral we're looking at is\n\\[\n\\int_{b}^{\\infty} x^{1/4} ((a-b)/4x + O(x^{-2}))\\,dx.\n\\]\nThe term $x^{1/4} O(x^{-2})$ is bounded by a constant times\n$x^{-7/4}$, whose integral converges. Thus we only have to decide\nwhether $x^{-3/4} (a-b)/4$ converges. But $x^{-3/4}$ has divergent\nintegral, so we get convergence if and only if $a=b$ (in which case\nthe integral telescopes anyway).",
  "vars": [
    "x"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "inputvar",
        "a": "paramone",
        "b": "paramtwo"
      },
      "question": "improper integral\n\\[\n\\int_{paramtwo}^{\\infty} \\left( \\sqrt{\\sqrt{inputvar+paramone}-\\sqrt{inputvar}} -\n\\sqrt{\\sqrt{inputvar}-\\sqrt{inputvar-paramtwo}} \\right)\\,dinputvar\n\\]\nconverge?",
      "solution": "The integral converges iff $paramone=paramtwo$. The easiest proof uses\n``big-O'' notation and the fact that $(1+inputvar)^{1/2} = 1 + inputvar/2 +\nO(inputvar^{2})$ for $|inputvar|<1$. (Here $O(inputvar^{2})$ means bounded by a constant\ntimes $inputvar^{2}$.)\n\nSo\n\\begin{align*}\n\\sqrt{inputvar+paramone}-\\sqrt{inputvar} &= inputvar^{1/2}(\\sqrt{1+paramone/inputvar} - 1) \\\\\n&= inputvar^{1/2}(1 + paramone/2inputvar + O(inputvar^{-2})),\n\\end{align*}\nhence\n\\[\n\\sqrt{\\sqrt{inputvar+paramone} - \\sqrt{inputvar}} = inputvar^{1/4} (paramone/4inputvar + O(inputvar^{-2}))\n\\]\nand similarly\n\\[\n\\sqrt{\\sqrt{inputvar} - \\sqrt{inputvar-paramtwo}} = inputvar^{1/4} (paramtwo/4inputvar + O(inputvar^{-2})).\n\\]\nHence the integral we're looking at is\n\\[\n\\int_{paramtwo}^{\\infty} inputvar^{1/4} ((paramone-paramtwo)/4inputvar + O(inputvar^{-2}))\\,dinputvar.\n\\]\nThe term $inputvar^{1/4} O(inputvar^{-2})$ is bounded by a constant times\n$inputvar^{-7/4}$, whose integral converges. Thus we only have to decide\nwhether $inputvar^{-3/4} (paramone-paramtwo)/4$ converges. But $inputvar^{-3/4}$ has divergent\nintegral, so we get convergence if and only if $paramone=paramtwo$ (in which case\nthe integral telescopes anyway)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "a": "lanterns",
        "b": "pinecones"
      },
      "question": "improper integral\n\\[\n\\int_{pinecones}^{\\infty} \\left( \\sqrt{\\sqrt{sunflower+lanterns}-\\sqrt{sunflower}} -\n\\sqrt{\\sqrt{sunflower}-\\sqrt{sunflower-pinecones}} \\right)\\,d sunflower\n\\]\nconverge?",
      "solution": "The integral converges iff $lanterns=pinecones$. The easiest proof uses\n``big-O'' notation and the fact that $(1+sunflower)^{1/2} = 1 + sunflower/2 +\nO(sunflower^{2})$ for $|sunflower|<1$. (Here $O(sunflower^{2})$ means bounded by a constant\ntimes $sunflower^{2}$.)\n\nSo\n\\begin{align*}\n\\sqrt{sunflower+lanterns}-\\sqrt{sunflower} &= sunflower^{1/2}(\\sqrt{1+lanterns/sunflower} - 1) \\\\\n&= sunflower^{1/2}(1 + lanterns/2sunflower + O(sunflower^{-2})),\n\\end{align*}\nhence\n\\[\n\\sqrt{\\sqrt{sunflower+lanterns} - \\sqrt{sunflower}} = sunflower^{1/4} (lanterns/4sunflower + O(sunflower^{-2}))\n\\]\nand similarly\n\\[\n\\sqrt{\\sqrt{sunflower} - \\sqrt{sunflower-pinecones}} = sunflower^{1/4} (pinecones/4sunflower + O(sunflower^{-2})).\n\\]\nHence the integral we're looking at is\n\\[\n\\int_{pinecones}^{\\infty} sunflower^{1/4} ((lanterns-pinecones)/4sunflower + O(sunflower^{-2}))\\,d sunflower.\n\\]\nThe term $sunflower^{1/4} O(sunflower^{-2})$ is bounded by a constant times\n$sunflower^{-7/4}$, whose integral converges. Thus we only have to decide\nwhether $sunflower^{-3/4} (lanterns-pinecones)/4$ converges. But $sunflower^{-3/4}$ has divergent\nintegral, so we get convergence if and only if $lanterns=pinecones$ (in which case\nthe integral telescopes anyway)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "a": "decremental",
        "b": "ceilinglimit"
      },
      "question": "improper integral\n\\[\n\\int_{ceilinglimit}^{\\infty} \\left( \\sqrt{\\sqrt{constantvalue+decremental}-\\sqrt{constantvalue}} -\n\\sqrt{\\sqrt{constantvalue}-\\sqrt{constantvalue-ceilinglimit}} \\right)\\,dconstantvalue\n\\]\nconverge?",
      "solution": "The integral converges iff $decremental=ceilinglimit$. The easiest proof uses\n``big-O'' notation and the fact that $(1+constantvalue)^{1/2} = 1 + constantvalue/2 +\nO(constantvalue^{2})$ for $|constantvalue|<1$. (Here $O(constantvalue^{2})$ means bounded by a constant\ntimes $constantvalue^{2}$.)\n\nSo\n\\begin{align*}\n\\sqrt{constantvalue+decremental}-\\sqrt{constantvalue} &= constantvalue^{1/2}(\\sqrt{1+decremental/constantvalue} - 1) \\\\\n&= constantvalue^{1/2}(1 + decremental/2constantvalue + O(constantvalue^{-2})),\n\\end{align*}\nhence\n\\[\n\\sqrt{\\sqrt{constantvalue+decremental} - \\sqrt{constantvalue}} = constantvalue^{1/4} (decremental/4constantvalue + O(constantvalue^{-2}))\n\\]\nand similarly\n\\[\n\\sqrt{\\sqrt{constantvalue} - \\sqrt{constantvalue-ceilinglimit}} = constantvalue^{1/4} (ceilinglimit/4constantvalue + O(constantvalue^{-2})).\n\\]\nHence the integral we\\'re looking at is\n\\[\n\\int_{ceilinglimit}^{\\infty} constantvalue^{1/4} ((decremental-ceilinglimit)/4constantvalue + O(constantvalue^{-2}))\\,dconstantvalue.\n\\]\nThe term $constantvalue^{1/4} O(constantvalue^{-2})$ is bounded by a constant times\n$constantvalue^{-7/4}$, whose integral converges. Thus we only have to decide\nwhether $constantvalue^{-3/4} (decremental-ceilinglimit)/4$ converges. But $constantvalue^{-3/4}$ has divergent\nintegral, so we get convergence if and only if $decremental=ceilinglimit$ (in which case\nthe integral telescopes anyway)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "a": "hjgrksla",
        "b": "fpdqneiu"
      },
      "question": "improper integral\n\\[\n\\int_{fpdqneiu}^{\\infty} \\left( \\sqrt{\\sqrt{qzxwvtnp+hjgrksla}-\\sqrt{qzxwvtnp}} -\n\\sqrt{\\sqrt{qzxwvtnp}-\\sqrt{qzxwvtnp-fpdqneiu}} \\right)\\,d qzxwvtnp\n\\]\nconverge?",
      "solution": "The integral converges iff $hjgrksla=fpdqneiu$. The easiest proof uses\n``big-O'' notation and the fact that $(1+qzxwvtnp)^{1/2} = 1 + qzxwvtnp/2 +\nO(qzxwvtnp^{2})$ for $|qzxwvtnp|<1$. (Here $O(qzxwvtnp^{2})$ means bounded by a constant\ntimes $qzxwvtnp^{2}$.)\n\nSo\n\\begin{align*}\n\\sqrt{qzxwvtnp+hjgrksla}-\\sqrt{qzxwvtnp} &= qzxwvtnp^{1/2}(\\sqrt{1+hjgrksla/qzxwvtnp} - 1) \\\\\n&= qzxwvtnp^{1/2}(1 + hjgrksla/2qzxwvtnp + O(qzxwvtnp^{-2})),\n\\end{align*}\nhence\n\\[\n\\sqrt{\\sqrt{qzxwvtnp+hjgrksla} - \\sqrt{qzxwvtnp}} = qzxwvtnp^{1/4} (hjgrksla/4qzxwvtnp + O(qzxwvtnp^{-2}))\n\\]\nand similarly\n\\[\n\\sqrt{\\sqrt{qzxwvtnp} - \\sqrt{qzxwvtnp-fpdqneiu}} = qzxwvtnp^{1/4} (fpdqneiu/4qzxwvtnp + O(qzxwvtnp^{-2})).\n\\]\nHence the integral we're looking at is\n\\[\n\\int_{fpdqneiu}^{\\infty} qzxwvtnp^{1/4} ((hjgrksla-fpdqneiu)/4qzxwvtnp + O(qzxwvtnp^{-2}))\\,d qzxwvtnp.\n\\]\nThe term $qzxwvtnp^{1/4} O(qzxwvtnp^{-2})$ is bounded by a constant times\n$qzxwvtnp^{-7/4}$, whose integral converges. Thus we only have to decide\nwhether $qzxwvtnp^{-3/4} (hjgrksla-fpdqneiu)/4$ converges. But $qzxwvtnp^{-3/4}$ has divergent\nintegral, so we get convergence if and only if $hjgrksla=fpdqneiu$ (in which case\nthe integral telescopes anyway)."
    },
    "kernel_variant": {
      "question": "Let an integer $n\\ge 3$ and a \\emph{positive integer} $r\\ge 2$ be fixed and put\n\\[\n\\lambda:=\\frac1r\\in(0,1),\\qquad \\rho:=1-\\lambda .\n\\]\n\nFor $x>0$ introduce the $\\lambda$–fractional kernel\n\\[\nF_{r}(x):=\\sum_{k=1}^{n}w_{k}\\Bigl(\\,(x+a_{k})^{\\lambda}-x^{\\lambda}\\Bigr)^{\\lambda},\n\\]\nwhere  \n\n$\\bullet\\;a_{1},\\dots ,a_{n}>0$ are pairwise distinct,  \n\n$\\bullet\\;(w_{1},\\dots ,w_{n})\\in\\mathbf R^{\\,n}$ and $(w_{k})$ is not the zero–vector.\n\nFor a complex Mellin parameter $\\sigma$ put  \n\\[\nI(\\sigma):=\\int_{0}^{\\infty}x^{\\sigma-1}F_{r}(x)\\,dx\\tag{$\\star$}\n\\]\n(the integral is taken over the maximal open vertical strip on which it converges absolutely).\n\nThroughout, the symbol “$\\sim$’’ denotes a \\emph{complete} asymptotic expansion; as $x\\to\\infty$ the exponents drop by positive integers, whereas as $x\\to0^{+}$ they increase by integer multiples of $\\lambda$.\n\n(a)\\;Show that there exist computable constants $C_{j}(a),D_{j}(a)\\;(j=0,1,2,\\dots)$ such that, as $x\\to\\infty$ and $x\\to0^{+}$ respectively,\n\\[\nF_{r}(x)\\sim\\sum_{j=0}^{\\infty}S_{j}\\,x^{-\\lambda\\rho-j},\\qquad \n  S_{j}:=\\sum_{k=1}^{n}w_{k}C_{j}(a_{k}),\n\\]\n\\[\nF_{r}(x)\\sim\\sum_{j=0}^{\\infty}T_{j}\\,x^{j\\lambda},\\qquad\\ \n  T_{j}:=\\sum_{k=1}^{n}w_{k}D_{j}(a_{k}).\n\\]\n\nDefine  \n\\[\nm_{0}:=\\sup\\bigl\\{j\\ge 0\\;:\\;T_{0}=T_{1}=\\dots =T_{j-1}=0\\bigr\\},\n\\qquad   \nm_{\\infty}:=\\sup\\bigl\\{j\\ge 0\\;:\\;S_{0}=S_{1}=\\dots =S_{j-1}=0\\bigr\\},\n\\]\nand note that if $m_{0},m_{\\infty}<\\infty$ then necessarily\n$T_{m_{0}}\\neq 0$ and $S_{m_{\\infty}}\\neq 0$.\n\nProve that the integral $(\\star)$ converges absolutely if and only if\n\\[\n-\\;m_{0}\\lambda\\;<\\;\\operatorname{Re}\\sigma\\;<\\;\\lambda\\rho+m_{\\infty}.\\tag{$\\dagger$}\n\\]\n\n(b)\\;Analyse the two boundary lines.\n\n(i)\\;If $m_{0}<\\infty$, show that for every $\\sigma$ with\n$\\operatorname{Re}\\sigma=-m_{0}\\lambda$ the integral $(\\star)$ diverges (indeed it already diverges at the integrable end $x\\to 0^{+}$).  \n\n(ii)\\;If $m_{\\infty}<\\infty$, show that for every $\\sigma$ with\n$\\operatorname{Re}\\sigma=\\lambda\\rho+m_{\\infty}$ the integral $(\\star)$ diverges (already at $x\\to\\infty$).\n\n(When $m_{0}= \\infty$ or $m_{\\infty}= \\infty$ the corresponding boundary line is absent.)\n\n(c)\\;Assuming $(\\dagger)$, prove that $I(\\sigma)$ possesses a meromorphic continuation to the whole complex plane $\\mathbf C$.  All poles are simple and lie on the two arithmetic progressions  \n\\[\n\\sigma=-j\\lambda\\quad(j=0,1,2,\\dots),\\qquad \n\\sigma=\\lambda\\rho+j\\quad(j=0,1,2,\\dots).\n\\]\n\nLet  \n\\[\nj_{0}:=\\min\\{j\\ge 0:\\;T_{j}\\neq 0\\}\\quad(\\text{if such }j\\text{ exists}),\\qquad \nj_{1}:=\\min\\{j\\ge 0:\\;S_{j}\\neq 0\\}\\quad(\\text{if such }j\\text{ exists}).\n\\]\n\nShow that the left–most pole of $I(\\sigma)$ is  \n\\[\n\\sigma_{\\text{left}}\n   =\\begin{cases}\n        -\\,j_{0}\\lambda,&\\text{if }j_{0}\\text{ exists},\\\\[4pt]\n        \\lambda\\rho+j_{1},&\\text{otherwise,}\n     \\end{cases}\n\\]\nand compute the residues  \n\\[\n\\operatorname*{Res}_{\\sigma=-j\\lambda}I(\\sigma)=T_{j},\\qquad\n\\operatorname*{Res}_{\\sigma=\\lambda\\rho+j}I(\\sigma)=-\\,S_{j}.\n\\]",
      "solution": "We abbreviate $\\lambda:=1/r$, $\\rho:=1-\\lambda$ and $F(x):=F_{r}(x)$.\n\n0.\\;Local coefficients.  \nFor $j\\ge 0$ define\n\\[\nC_{j}(a):=\\bigl[x^{-\\lambda\\rho-j}\\bigr]\\Bigl((x+a)^{\\lambda}-x^{\\lambda}\\Bigr)^{\\lambda}\\quad(x\\to\\infty),\n\\]\n\\[\nD_{j}(a):=\\bigl[x^{\\,j\\lambda}\\bigr]\\Bigl((x+a)^{\\lambda}-x^{\\lambda}\\Bigr)^{\\lambda}\\quad(x\\to0^{+}),\n\\]\nand put\n\\[\nS_{j}:=\\sum_{k=1}^{n}w_{k}\\,C_{j}(a_{k}),\\qquad\nT_{j}:=\\sum_{k=1}^{n}w_{k}\\,D_{j}(a_{k}).\\tag{0.1}\n\\]\n\nBecause $r\\in\\mathbf Z_{\\ge2}$, $\\lambda=1/r$; hence every exponent\n$k+\\ell\\lambda$ ($k,\\ell\\in\\mathbf N$) equals $j\\lambda$ for the\ninteger $j:=kr+\\ell$.  Consequently the “grid’’ that appears in the\nbinomial expansion near $x=0^{+}$ collapses to a \\emph{single}\narithmetic progression and the coefficients $D_{j}(a)$ are\nwell–defined.\n\n1.\\;Complete asymptotic expansions of $F$.\n\n(i)\\;$x\\to\\infty$.  \nWrite $(x+a)^{\\lambda}=x^{\\lambda}(1+a/x)^{\\lambda}$ and expand with the binomial series:\n\\[\n(x+a)^{\\lambda}-x^{\\lambda}=x^{\\lambda}\\bigl[(1+a/x)^{\\lambda}-1\\bigr]\n      =\\lambda a\\,x^{\\lambda-1}\\Bigl[1-\\frac{\\rho a}{2x}+O\\!\\bigl(x^{-2}\\bigr)\\Bigr].\n\\]\nBecause $\\lambda-1=-\\rho$, raising to the power $\\lambda$ gives\n\\[\n\\Bigl((x+a)^{\\lambda}-x^{\\lambda}\\Bigr)^{\\lambda}\n      =(\\lambda a)^{\\lambda}x^{-\\lambda\\rho}\n        \\Bigl[1-\\frac{\\rho a}{2x}+O\\!\\bigl(x^{-2}\\bigr)\\Bigr]^{\\lambda}\n      \\sim\\sum_{j=0}^{\\infty}C_{j}(a)\\,x^{-\\lambda\\rho-j}.\n\\]\nSumming over $k$ yields\n\\[\nF(x)\\sim\\sum_{j=0}^{\\infty}S_{j}\\,x^{-\\lambda\\rho-j}.\\tag{1.1}\n\\]\n\n(ii)\\;$x\\to0^{+}$.  \nWrite $(x+a)^{\\lambda}=a^{\\lambda}(1+x/a)^{\\lambda}$ and expand:\n\\[\n(x+a)^{\\lambda}-x^{\\lambda}=a^{\\lambda}\n      +\\lambda a^{\\lambda-1}x-x^{\\lambda}+O\\!\\bigl(x^{2}\\bigr).\n\\]\nRaising to the power $\\lambda$ and applying the binomial theorem twice gives the double sum\n\\[\n\\Bigl((x+a)^{\\lambda}-x^{\\lambda}\\Bigr)^{\\lambda}\n     =a^{\\lambda^{2}}\n        \\sum_{k,\\ell\\ge0}\\binom{\\lambda}{k+\\ell}\\binom{k+\\ell}{k}\n            \\bigl(\\lambda x/a\\bigr)^{k}\\Bigl(-x^{\\lambda}/a^{\\lambda}\\Bigr)^{\\ell}.\n\\]\nReplacing the pair $(k,\\ell)$ by $j:=kr+\\ell\\;(=0,1,2,\\dots)$ we can\nre–index the sum and obtain\n\\[\n\\Bigl((x+a)^{\\lambda}-x^{\\lambda}\\Bigr)^{\\lambda}\n      \\sim\\sum_{j=0}^{\\infty}D_{j}(a)\\,x^{j\\lambda}, \\qquad x\\to0^{+},\n\\]\nso that\n\\[\nF(x)\\sim\\sum_{j=0}^{\\infty}T_{j}\\,x^{j\\lambda}.\\tag{1.2}\n\\]\n\n2.\\;Absolute convergence of $I(\\sigma)$.\n\nDefine\n\\[\nm_{0}:=\\sup\\{j\\ge0:T_{0}=\\dots =T_{j-1}=0\\},\\qquad \nm_{\\infty}:=\\sup\\{j\\ge0:S_{0}=\\dots =S_{j-1}=0\\}.\n\\]\nIf $m_{0}<\\infty$ then $T_{m_{0}}\\neq0$, and similarly\n$S_{m_{\\infty}}\\neq0$ when $m_{\\infty}<\\infty$.\n\nA.\\;$x\\to\\infty$.  \nBy (1.1),\n\\[\nx^{\\sigma-1}F(x)\\sim S_{m_{\\infty}}\\,x^{\\sigma-1-\\lambda\\rho-m_{\\infty}}.\n\\]\nHence $\\int_{A}^{\\infty}$ converges absolutely iff\n\\[\n\\operatorname{Re}\\bigl(\\sigma-1-\\lambda\\rho-m_{\\infty}\\bigr)<-1\n\\quad\\Longleftrightarrow\\quad\n           \\operatorname{Re}\\sigma<\\lambda\\rho+m_{\\infty}.\\tag{2.1}\n\\]\n\nB.\\;$x\\to0^{+}$.  \nUsing (1.2),\n\\[\nx^{\\sigma-1}F(x)\\sim T_{m_{0}}\\,x^{\\sigma-1+m_{0}\\lambda},\n\\]\nso $\\int_{0}^{B}$ converges absolutely iff\n\\[\n\\operatorname{Re}\\bigl(\\sigma-1+m_{0}\\lambda\\bigr)>-1\n\\quad\\Longleftrightarrow\\quad\n\\operatorname{Re}\\sigma>-m_{0}\\lambda.\\tag{2.2}\n\\]\n\nC.\\;Combining (2.1)–(2.2) gives exactly the strip ($\\dagger$).\nConversely, violation of either inequality forces divergence, so ($\\dagger$) is necessary and sufficient.  This completes part (a).\n\n3.\\;Behaviour on the boundary lines (part (b)).\n\n(i)\\;Suppose $m_{0}<\\infty$ and $\\operatorname{Re}\\sigma=-m_{0}\\lambda$.  \nNear $x=0^{+}$,\n\\[\nx^{\\sigma-1}F(x)\\sim T_{m_{0}}\\,x^{-1},\\qquad T_{m_{0}}\\neq0,\n\\]\nso $\\int_{0}^{1}x^{-1}\\,dx$ diverges; therefore $(\\star)$ diverges.\n\n(ii)\\;Suppose $m_{\\infty}<\\infty$ and $\\operatorname{Re}\\sigma=\\lambda\\rho+m_{\\infty}$.  \nAs $x\\to\\infty$,\n\\[\nx^{\\sigma-1}F(x)\\sim S_{m_{\\infty}}\\,x^{-1},\\qquad S_{m_{\\infty}}\\neq0,\n\\]\nand $\\int_{1}^{\\infty}x^{-1}\\,dx$ diverges; hence so does $(\\star)$.\nThus part (b) is established.\n\n4.\\;Meromorphic continuation to $\\mathbf C$ (part (c)).\n\nChoose integers $N_{0},N_{\\infty}\\ge1$ and smooth cut–off functions\n$\\varphi_{0},\\varphi_{\\infty}$ with\n\\[\n\\varphi_{0}\\equiv1\\ \\text{on }(0,1),\\quad\\operatorname{supp}\\varphi_{0}\\subset(0,2),\n\\]\n\\[\n\\varphi_{\\infty}\\equiv1\\ \\text{on }(1,\\infty),\\quad\\operatorname{supp}\\varphi_{\\infty}\\subset\\bigl(\\tfrac12,\\infty\\bigr).\n\\]\nDefine\n\\[\nG_{N_{0},N_{\\infty}}(x):=F(x)\n     -\\varphi_{0}(x)\\sum_{j=0}^{N_{0}-1}T_{j}x^{j\\lambda}\n     -\\varphi_{\\infty}(x)\\sum_{j=0}^{N_{\\infty}-1}S_{j}x^{-\\lambda\\rho-j}.\n\\tag{4.1}\n\\]\nBy (1.2)–(1.1) one has\n\\[\nG_{N_{0},N_{\\infty}}(x)=O\\!\\bigl(x^{N_{0}\\lambda}\\bigr)\\quad(x\\to0^{+}),\\qquad\nG_{N_{0},N_{\\infty}}(x)=O\\!\\bigl(x^{-\\lambda\\rho-N_{\\infty}}\\bigr)\\quad(x\\to\\infty).\n\\tag{4.2}\n\\]\nHence the Mellin transform\n\\[\n\\mathcal M_{N_{0},N_{\\infty}}(\\sigma)\n      :=\\int_{0}^{\\infty}x^{\\sigma-1}G_{N_{0},N_{\\infty}}(x)\\,dx\\tag{4.3}\n\\]\nis \\emph{holomorphic} in the vertical strip\n$-N_{0}\\lambda<\\operatorname{Re}\\sigma<\\lambda\\rho+N_{\\infty}$.\nBecause $N_{0},N_{\\infty}$ are arbitrary, the overlapping family of strips covers $\\mathbf C$, so the functions $\\mathcal M_{N_{0},N_{\\infty}}$ patch together to an \\emph{entire} function.\n\nReturning to $I(\\sigma)$ we obtain\n\\[\nI(\\sigma)=\\mathcal M_{N_{0},N_{\\infty}}(\\sigma)\n      +\\sum_{j=0}^{N_{0}-1}T_{j}\\,K_{0,j}(\\sigma)\n      +\\sum_{j=0}^{N_{\\infty}-1}S_{j}\\,K_{\\infty,j}(\\sigma),\\tag{4.4}\n\\]\nwhere\n\\[\nK_{0,j}(\\sigma):=\\int_{0}^{2}x^{\\sigma+j\\lambda-1}\\varphi_{0}(x)\\,dx,\n\\qquad\nK_{\\infty,j}(\\sigma):=\\int_{1/2}^{\\infty}x^{\\sigma-\\lambda\\rho-j-1}\\varphi_{\\infty}(x)\\,dx.\n\\]\n\n5.\\;Location of poles and computation of residues.\n\nBecause $\\varphi_{0}(0)=1$ and $\\varphi_{\\infty}(x)=1$ for $x\\ge1$, standard Mellin asymptotics give\n\\[\n\\operatorname*{Res}_{\\sigma=-j\\lambda}K_{0,j}(\\sigma)=1,\\qquad\n\\operatorname*{Res}_{\\sigma=\\lambda\\rho+j}K_{\\infty,j}(\\sigma)=-1.\n\\]\nEvery $K_{\\bullet,j}$ is otherwise holomorphic.  Since $\\mathcal M_{N_{0},N_{\\infty}}$ is entire, the only singularities of $I(\\sigma)$ are the simple poles\n\\[\n\\sigma=-j\\lambda\\quad(j=0,1,2,\\dots),\\qquad\n\\sigma=\\lambda\\rho+j\\quad(j=0,1,2,\\dots),\\tag{5.1}\n\\]\nwith residues\n\\[\n\\operatorname*{Res}_{\\sigma=-j\\lambda}I(\\sigma)=T_{j},\\qquad\n\\operatorname*{Res}_{\\sigma=\\lambda\\rho+j}I(\\sigma)=-\\,S_{j}.\\tag{5.2}\n\\]\n\n6.\\;The left–most pole.\n\nLet $j_{0},j_{1}$ be as in the problem statement.\n\n$\\bullet$ If $j_{0}$ exists, then $\\sigma=-j_{0}\\lambda$ lies strictly to the left of every pole in the second family, so it is the left–most pole with residue $T_{j_{0}}$.\n\n$\\bullet$ If no such $j_{0}$ exists, the first family is absent and the left–most pole is $\\sigma=\\lambda\\rho+j_{1}$ with residue $-\\,S_{j_{1}}$.\n\nThis completes the proof. \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.739736",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original one–dimensional, two-parameter integral, the enhanced variant\n\n• works with an arbitrary number n ≥ 3 of shifts and weights,  \n• introduces an extra Mellin parameter σ, turning the problem into the study of an entire family of integrals,  \n• forces simultaneous control of the integrand at both 0 and ∞,  \n• requires the identification of higher-order “moment” conditions (S_j, T_j) to describe cancellations,  \n• demands mastery of asymptotic expansions, Mellin transforms and meromorphic continuation to compute residues, and  \n• ties convergence questions to the analytic structure of I(σ), not just to simple comparison tests.\n\nThese additional layers move the task well beyond straightforward power-series matching and necessitate a coordinated use of real analysis, complex analysis and asymptotic techniques, making the variant substantially harder than both the original problem and the current kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let an integer $n\\ge 3$ and a rational number $r>1$ be fixed and set \n\n\\[\n\\lambda:=\\frac1r\\in(0,1),\\qquad \\rho:=1-\\lambda .\n\\]\n\nFor $x>0$ introduce the $\\lambda$–fractional kernel  \n\n\\[\nF_{r}(x):=\\sum_{k=1}^{n}w_{k}\\Bigl(\\,(x+a_{k})^{\\lambda}-x^{\\lambda}\\Bigr)^{\\lambda},\n\\]\n\nwhere  \n\n$\\bullet\\;a_{1},\\dots ,a_{n}>0$ are pairwise distinct,  \n\n$\\bullet\\;(w_{1},\\dots ,w_{n})\\in\\mathbf R^{\\,n}$ and $(w_{k})$ is not the zero–vector.\n\nFor a complex Mellin parameter $\\sigma$ put  \n\n\\[\nI(\\sigma):=\\int_{0}^{\\infty}x^{\\sigma-1}F_{r}(x)\\,dx\\tag{$\\star$}\n\\]\n(the integral is understood in the maximal open vertical strip on which it converges absolutely).\n\nThroughout, “$\\sim$’’ means a complete asymptotic expansion; as $x\\to\\infty$ the powers drop by integers, whereas as $x\\to0^{+}$ they increase by multiples of $\\lambda$.\n\n(a)\\; Show that there exist computable constants $C_{j}(a),D_{j}(a)\\;(j=0,1,2,\\dots)$ such that, as $x\\to\\infty$ and $x\\to0^{+}$ respectively,\n\\[\nF_{r}(x)\\sim\\sum_{j=0}^{\\infty}S_{j}x^{-\\lambda\\rho-j},\\qquad \n  S_{j}:=\\sum_{k}w_{k}C_{j}(a_{k}),\n\\]\n\\[\nF_{r}(x)\\sim\\sum_{j=0}^{\\infty}T_{j}x^{j\\lambda},\\qquad\\ \n  T_{j}:=\\sum_{k}w_{k}D_{j}(a_{k}).\n\\]\n\nDefine  \n\\[\nm_{0}:=\\sup\\bigl\\{j\\ge 0\\;:\\;T_{0}=T_{1}=\\dots =T_{j-1}=0\\bigr\\},\n\\qquad   \nm_{\\infty}:=\\sup\\bigl\\{j\\ge 0\\;:\\;S_{0}=S_{1}=\\dots =S_{j-1}=0\\bigr\\},\n\\]\nand note that if $m_{0},m_{\\infty}<\\infty$ then necessarily\n$T_{m_{0}}\\neq 0$ and $S_{m_{\\infty}}\\neq 0$.\n\nProve that the integral $(\\star)$ converges absolutely iff \n\\[\n-\\;m_{0}\\lambda\\;<\\;\\operatorname{Re}\\sigma\\;<\\;\\lambda\\rho+m_{\\infty}.\\tag{$\\dagger$}\n\\]\n\n(b)\\; Analyse the two boundary lines.\n\n(i) If $m_{0}<\\infty$, show that for every $\\sigma$ with \n$\\operatorname{Re}\\sigma=-m_{0}\\lambda$ the integral $(\\star)$ diverges (indeed it already diverges at the integrable end $x\\to0^{+}$).  \n\n(ii) If $m_{\\infty}<\\infty$, show that for every $\\sigma$ with \n$\\operatorname{Re}\\sigma=\\lambda\\rho+m_{\\infty}$ the integral $(\\star)$ diverges (already at $x\\to\\infty$).\n\n(When $m_{0}= \\infty$ or $m_{\\infty}= \\infty$ the corresponding boundary line is absent.)\n\n(c)\\; Assuming $(\\dagger)$, prove that $I(\\sigma)$ possesses a meromorphic continuation to the whole complex plane $\\mathbf C$. All poles are simple and lie on the two arithmetic progressions  \n\\[\n\\sigma=-j\\lambda\\quad(j=0,1,2,\\dots),\\qquad \n\\sigma=\\lambda\\rho+j\\quad(j=0,1,2,\\dots).\n\\]\n\nLet  \n\n\\[\nj_{0}:=\\min\\{j\\ge 0:\\;T_{j}\\neq 0\\}\\quad(\\text{if such }j\\text{ exists}),\\qquad \nj_{1}:=\\min\\{j\\ge 0:\\;S_{j}\\neq 0\\}\\quad(\\text{if such }j\\text{ exists}).\n\\]\n\nShow that the left–most pole of $I(\\sigma)$ is  \n\n\\[\n\\sigma_{\\text{left}}\n   =\\begin{cases}\n        -\\,j_{0}\\lambda,&\\text{if }j_{0}\\text{ exists},\\\\[4pt]\n        \\lambda\\rho+j_{1},&\\text{otherwise,}\n     \\end{cases}\n\\]\nand compute the residues  \n\\[\n\\operatorname*{Res}_{\\sigma=-j\\lambda}I(\\sigma)=T_{j},\\qquad\n\\operatorname*{Res}_{\\sigma=\\lambda\\rho+j}I(\\sigma)=S_{j}.\n\\]",
      "solution": "We abbreviate $\\lambda:=1/r$, $\\rho:=1-\\lambda$ and $F(x):=F_{r}(x)$.  \n\n0.\\;Local coefficients.  \nFor $j\\ge 0$ define  \n\\[\nC_{j}(a):=\\bigl[x^{-\\lambda\\rho-j}\\bigr]\\Bigl((x+a)^{\\lambda}-x^{\\lambda}\\Bigr)^{\\lambda}\\quad(x\\to\\infty),\n\\]\n\\[\nD_{j}(a):=\\bigl[x^{\\,j\\lambda}\\bigr]\\Bigl((x+a)^{\\lambda}-x^{\\lambda}\\Bigr)^{\\lambda}\\quad(x\\to0^{+}),\n\\]\nand put  \n\\[\nS_{j}:=\\sum_{k=1}^{n}w_{k}\\,C_{j}(a_{k}),\\qquad\nT_{j}:=\\sum_{k=1}^{n}w_{k}\\,D_{j}(a_{k}).\\tag{0.1}\n\\]\n\n1.\\;Complete asymptotic expansions of $F$.\n\n(i)\\;$x\\to\\infty$.  \nWrite $(x+a)^{\\lambda}=x^{\\lambda}(1+a/x)^{\\lambda}$ and use the binomial series:\n\\[\n(x+a)^{\\lambda}-x^{\\lambda}=x^{\\lambda}\\bigl[(1+a/x)^{\\lambda}-1\\bigr]\n      =\\lambda a\\,x^{\\lambda-1}\\Bigl[1-\\frac{\\rho a}{2x}+O(x^{-2})\\Bigr].\n\\]\nBecause $\\lambda-1=-\\rho$, raising to the power $\\lambda$ gives\n\\[\n\\bigl((x+a)^{\\lambda}-x^{\\lambda}\\bigr)^{\\lambda}\n      =(\\lambda a)^{\\lambda}x^{-\\lambda\\rho}\n        \\Bigl[1-\\frac{\\rho a}{2x}+O(x^{-2})\\Bigr]^{\\lambda}\n      \\sim\\sum_{j=0}^{\\infty}C_{j}(a)\\,x^{-\\lambda\\rho-j}.\n\\]\nSumming over $k$ yields  \n\\[\nF(x)\\sim\\sum_{j=0}^{\\infty}S_{j}\\,x^{-\\lambda\\rho-j}.\\tag{1.1}\n\\]\n\n(ii)\\;$x\\to0^{+}$.  \nWrite $(x+a)^{\\lambda}=a^{\\lambda}(1+x/a)^{\\lambda}$ and expand:\n\\[\n(x+a)^{\\lambda}-x^{\\lambda}=a^{\\lambda}\n      +\\lambda a^{\\lambda-1}x-x^{\\lambda}+O(x^{2}).\n\\]\nSince $0<\\lambda<1$, one has $x=o(x^{\\lambda})$ near $0$, so the dominant pieces are $a^{\\lambda}$ and $-x^{\\lambda}$.  Factoring $a^{\\lambda}$ and expanding again,\n\\[\n\\bigl((x+a)^{\\lambda}-x^{\\lambda}\\bigr)^{\\lambda}\n     =a^{\\lambda^{2}}\n        \\Bigl[1-\\bigl(x/a\\bigr)^{\\lambda}\n                 +\\lambda(x/a)+O(x^{2})\\Bigr]^{\\lambda}\n     \\sim a^{\\lambda^{2}}\\sum_{j=0}^{\\infty}E_{j}(a)\\,x^{j\\lambda},\n\\]\nso that  \n\\[\nF(x)\\sim\\sum_{j=0}^{\\infty}T_{j}\\,x^{j\\lambda}.\\tag{1.2}\n\\]\n\n2.\\;Absolute convergence of $I(\\sigma)$.\n\nDefine  \n\\[\nm_{0}:=\\sup\\{j\\ge0:T_{0}=\\dots =T_{j-1}=0\\},\\qquad \nm_{\\infty}:=\\sup\\{j\\ge0:S_{0}=\\dots =S_{j-1}=0\\}.\n\\]\nIf $m_{0}<\\infty$ then $T_{m_{0}}\\neq0$, and similarly\n$S_{m_{\\infty}}\\neq0$ when $m_{\\infty}<\\infty$.\n\nA.\\;$x\\to\\infty$.  \nBy (1.1),\n\\[\nx^{\\sigma-1}F(x)\\sim S_{m_{\\infty}}\\,x^{\\sigma-1-\\lambda\\rho-m_{\\infty}}.\n\\]\nHence $\\int_{A}^{\\infty}$ converges absolutely iff  \n\\[\n\\operatorname{Re}(\\sigma-1-\\lambda\\rho-m_{\\infty})<-1\\quad\\Longleftrightarrow\\quad\n           \\operatorname{Re}\\sigma<\\lambda\\rho+m_{\\infty}.\\tag{2.1}\n\\]\n\nB.\\;$x\\to0^{+}$.  \nUsing (1.2),\n\\[\nx^{\\sigma-1}F(x)\\sim T_{m_{0}}\\,x^{\\sigma-1+m_{0}\\lambda},\n\\]\nso $\\int_{0}^{B}$ converges absolutely iff  \n\\[\n\\operatorname{Re}(\\sigma-1+m_{0}\\lambda)>-1\n\\quad\\Longleftrightarrow\\quad \n\\operatorname{Re}\\sigma>-m_{0}\\lambda.\\tag{2.2}\n\\]\n\nC.\\;Combining (2.1)–(2.2) gives precisely the strip ($\\dagger$).\nConversely, violation of either inequality forces divergence, so ($\\dagger$) is necessary and sufficient.  This completes part (a).\n\n3.\\;Behaviour on the boundary lines (part (b)).\n\n(i)\\;Suppose $m_{0}<\\infty$ and $\\operatorname{Re}\\sigma=-m_{0}\\lambda$.  \nNear $x=0^{+}$,\n\\[\nx^{\\sigma-1}F(x)\\sim T_{m_{0}}\\,x^{-1},\\qquad T_{m_{0}}\\neq0,\n\\]\nso $\\int_{0}^{1}x^{-1}\\,dx$ diverges; therefore $(\\star)$ diverges.\n\n(ii)\\;Suppose $m_{\\infty}<\\infty$ and $\\operatorname{Re}\\sigma=\\lambda\\rho+m_{\\infty}$.  \nAs $x\\to\\infty$,\n\\[\nx^{\\sigma-1}F(x)\\sim S_{m_{\\infty}}\\,x^{-1},\\qquad S_{m_{\\infty}}\\neq0,\n\\]\nand $\\int_{1}^{\\infty}x^{-1}\\,dx$ diverges; hence so does $(\\star)$.\nThus part (b) is established.\n\n4.\\;Meromorphic continuation to $\\mathbf C$ (part (c)).\n\nPick integers $N_{0},N_{\\infty}\\ge1$ and smooth cut–off functions \n$\\varphi_{0},\\varphi_{\\infty}$ with  \n\\[\n\\varphi_{0}\\equiv1\\ \\text{on }(0,1),\\quad\\operatorname{supp}\\varphi_{0}\\subset(0,2),\n\\]\n\\[\n\\varphi_{\\infty}\\equiv1\\ \\text{on }(1,\\infty),\\quad\\operatorname{supp}\\varphi_{\\infty}\\subset\\bigl(\\tfrac12,\\infty\\bigr).\n\\]\nDefine\n\\[\nG_{N_{0},N_{\\infty}}(x):=F(x)\n     -\\varphi_{0}(x)\\sum_{j=0}^{N_{0}-1}T_{j}x^{j\\lambda}\n     -\\varphi_{\\infty}(x)\\sum_{j=0}^{N_{\\infty}-1}S_{j}x^{-\\lambda\\rho-j}.\n\\tag{4.1}\n\\]\nBy (1.2)–(1.1) one has  \n\\[\nG_{N_{0},N_{\\infty}}(x)=O\\bigl(x^{N_{0}\\lambda}\\bigr)\\quad(x\\to0^{+}),\\qquad\nG_{N_{0},N_{\\infty}}(x)=O\\bigl(x^{-\\lambda\\rho-N_{\\infty}}\\bigr)\\quad(x\\to\\infty).\n\\tag{4.2}\n\\]\nHence the Mellin transform  \n\\[\n\\mathcal M_{N_{0},N_{\\infty}}(\\sigma)\n      :=\\int_{0}^{\\infty}x^{\\sigma-1}G_{N_{0},N_{\\infty}}(x)\\,dx\\tag{4.3}\n\\]\nis an entire function in the vertical strip\n$-N_{0}\\lambda<\\operatorname{Re}\\sigma<\\lambda\\rho+N_{\\infty}$.\nBecause $N_{0},N_{\\infty}$ are arbitrary, $\\mathcal M_{N_{0},N_{\\infty}}$\nextends to an entire function on $\\mathbf C$.\n\nA direct calculation gives\n\\[\nI(\\sigma)=\\mathcal M_{N_{0},N_{\\infty}}(\\sigma)\n      +\\sum_{j=0}^{N_{0}-1}\\frac{T_{j}}{\\sigma+j\\lambda}\n      +\\sum_{j=0}^{N_{\\infty}-1}\\frac{S_{j}}{\\sigma-(\\lambda\\rho+j)}.\n\\tag{4.4}\n\\]\nIncreasing $N_{0}$ or $N_{\\infty}$ alters only the entire part, so\n(4.4) furnishes a meromorphic continuation of $I$ to $\\mathbf C$ whose\npoles are exactly\n\\[\n\\sigma=-j\\lambda\\quad(j=0,1,2,\\dots),\\qquad\n\\sigma=\\lambda\\rho+j\\quad(j=0,1,2,\\dots),\\tag{4.5}\n\\]\nall simple.  Part (c) is proved.\n\n5.\\;Residues and the left–most pole.\n\nFrom (4.4)\n\\[\n\\operatorname*{Res}_{\\sigma=-j\\lambda}I(\\sigma)=T_{j},\\qquad\n\\operatorname*{Res}_{\\sigma=\\lambda\\rho+j}I(\\sigma)=S_{j}.\\tag{5.1}\n\\]\nLet $j_{0},j_{1}$ be as in the problem statement.  If $j_{0}$ exists,\nthen $\\sigma=-j_{0}\\lambda$ lies strictly to the left of every pole in\nthe second family and is therefore the left–most pole, with residue\n$T_{j_{0}}$.  If no such $j_{0}$ exists, the first family is absent and\nthe left–most pole is $\\sigma=\\lambda\\rho+j_{1}$ with residue\n$S_{j_{1}}$. \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.572279",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original one–dimensional, two-parameter integral, the enhanced variant\n\n• works with an arbitrary number n ≥ 3 of shifts and weights,  \n• introduces an extra Mellin parameter σ, turning the problem into the study of an entire family of integrals,  \n• forces simultaneous control of the integrand at both 0 and ∞,  \n• requires the identification of higher-order “moment” conditions (S_j, T_j) to describe cancellations,  \n• demands mastery of asymptotic expansions, Mellin transforms and meromorphic continuation to compute residues, and  \n• ties convergence questions to the analytic structure of I(σ), not just to simple comparison tests.\n\nThese additional layers move the task well beyond straightforward power-series matching and necessitate a coordinated use of real analysis, complex analysis and asymptotic techniques, making the variant substantially harder than both the original problem and the current kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}