{
  "index": "1981-A-3",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Problem A-3\nFind\n\\[\n\\lim _{t \\rightarrow \\infty}\\left[e^{-t} \\int_{0}^{t} \\int_{0}^{t} \\frac{e^{x}-e^{y}}{x-y} d x d y\\right]\n\\]\nor show that the limit does not exist.",
  "solution": "A-3.\nLet \\( G(t) \\) be the double integral. Then\n\\[\n\\lim _{t \\rightarrow \\infty}\\left[G(t) / e^{t}\\right]=\\lim _{t \\rightarrow \\infty}\\left[G^{\\prime}(t) / e^{t}\\right]\n\\]\nby L'Hôpital's Rule. One finds that\n\\[\nG^{\\prime}(t)=\\int_{0}^{t} \\frac{e^{x}-e^{t}}{x-t} d x+\\int_{0}^{t} \\frac{e^{y}-e^{t}}{y-t} d y=2 \\int_{0}^{t} \\frac{e^{x}-e^{t}}{x-t} d x\n\\]\n\nThen using \\( e^{x}=e^{t}\\left[1+(x-t)+(x-t)^{2} / 2!+\\cdots\\right] \\), one sees that \\( e^{-t} G^{\\prime}(t) \\rightarrow \\infty \\) as \\( t \\rightarrow \\infty \\) since for sufficiently large \\( t \\),\n\\[\n\\frac{G^{\\prime}(t)}{2 e^{t}}=\\int_{0}^{t} \\frac{e^{x-t}-1}{x-t} d x=\\int_{0}^{t} \\frac{1-e^{-y}}{y} d y>\\int_{1}^{t} \\frac{1-e^{-y}}{y} d y>\\left(1-e^{-1}\\right) \\log t\n\\]",
  "vars": [
    "t",
    "x",
    "y",
    "G"
  ],
  "params": [],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "t": "timevar",
        "x": "xsymbol",
        "y": "ysymbol",
        "G": "doubleint"
      },
      "question": "Problem A-3\nFind\n\\[\n\\lim _{timevar \\rightarrow \\infty}\\left[e^{-timevar} \\int_{0}^{timevar} \\int_{0}^{timevar} \\frac{e^{xsymbol}-e^{ysymbol}}{xsymbol-ysymbol} d xsymbol d ysymbol\\right]\n\\]\nor show that the limit does not exist.",
      "solution": "A-3.\nLet \\( doubleint(timevar) \\) be the double integral. Then\n\\[\n\\lim _{timevar \\rightarrow \\infty}\\left[doubleint(timevar) / e^{timevar}\\right]=\\lim _{timevar \\rightarrow \\infty}\\left[doubleint^{\\prime}(timevar) / e^{timevar}\\right]\n\\]\nby L'Hôpital's Rule. One finds that\n\\[\ndoubleint^{\\prime}(timevar)=\\int_{0}^{timevar} \\frac{e^{xsymbol}-e^{timevar}}{xsymbol-timevar} d xsymbol+\\int_{0}^{timevar} \\frac{e^{ysymbol}-e^{timevar}}{ysymbol-timevar} d ysymbol=2 \\int_{0}^{timevar} \\frac{e^{xsymbol}-e^{timevar}}{xsymbol-timevar} d xsymbol\n\\]\n\nThen using \\( e^{xsymbol}=e^{timevar}\\left[1+(xsymbol-timevar)+(xsymbol-timevar)^{2} / 2!+\\cdots\\right] \\), one sees that \\( e^{-timevar} doubleint^{\\prime}(timevar) \\rightarrow \\infty \\) as \\( timevar \\rightarrow \\infty \\) since for sufficiently large \\( timevar \\),\n\\[\n\\frac{doubleint^{\\prime}(timevar)}{2 e^{timevar}}=\\int_{0}^{timevar} \\frac{e^{xsymbol-timevar}-1}{xsymbol-timevar} d xsymbol=\\int_{0}^{timevar} \\frac{1-e^{-ysymbol}}{ysymbol} d ysymbol>\\int_{1}^{timevar} \\frac{1-e^{-ysymbol}}{ysymbol} d ysymbol>\\left(1-e^{-1}\\right) \\log timevar\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "t": "marigolds",
        "x": "canisters",
        "y": "sandpiper",
        "G": "limestone"
      },
      "question": "Problem A-3\nFind\n\\[\n\\lim _{marigolds \\rightarrow \\infty}\\left[e^{-marigolds} \\int_{0}^{marigolds} \\int_{0}^{marigolds} \\frac{e^{canisters}-e^{sandpiper}}{canisters-sandpiper} d canisters d sandpiper\\right]\n\\]\nor show that the limit does not exist.",
      "solution": "A-3.\nLet \\( limestone(marigolds) \\) be the double integral. Then\n\\[\n\\lim _{marigolds \\rightarrow \\infty}\\left[limestone(marigolds) / e^{marigolds}\\right]=\\lim _{marigolds \\rightarrow \\infty}\\left[limestone^{\\prime}(marigolds) / e^{marigolds}\\right]\n\\]\nby L'Hôpital's Rule. One finds that\n\\[\nlimestone^{\\prime}(marigolds)=\\int_{0}^{marigolds} \\frac{e^{canisters}-e^{marigolds}}{canisters-marigolds} d canisters+\\int_{0}^{marigolds} \\frac{e^{sandpiper}-e^{marigolds}}{sandpiper-marigolds} d sandpiper=2 \\int_{0}^{marigolds} \\frac{e^{canisters}-e^{marigolds}}{canisters-marigolds} d canisters\n\\]\n\nThen using \\( e^{canisters}=e^{marigolds}\\left[1+(canisters-marigolds)+(canisters-marigolds)^{2} / 2!+\\cdots\\right] \\), one sees that \\( e^{-marigolds} limestone^{\\prime}(marigolds) \\rightarrow \\infty \\) as \\( marigolds \\rightarrow \\infty \\) since for sufficiently large \\( marigolds \\),\n\\[\n\\frac{limestone^{\\prime}(marigolds)}{2 e^{marigolds}}=\\int_{0}^{marigolds} \\frac{e^{canisters-marigolds}-1}{canisters-marigolds} d canisters=\\int_{0}^{marigolds} \\frac{1-e^{-sandpiper}}{sandpiper} d sandpiper>\\int_{1}^{marigolds} \\frac{1-e^{-sandpiper}}{sandpiper} d sandpiper>\\left(1-e^{-1}\\right) \\log marigolds\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "t": "timeless",
        "x": "knownvalue",
        "y": "fixedvalue",
        "G": "smallness"
      },
      "question": "Problem A-3\nFind\n\\[\n\\lim _{timeless \\rightarrow \\infty}\\left[e^{-timeless} \\int_{0}^{timeless} \\int_{0}^{timeless} \\frac{e^{knownvalue}-e^{fixedvalue}}{knownvalue-fixedvalue} d knownvalue d fixedvalue\\right]\n\\]\nor show that the limit does not exist.",
      "solution": "A-3.\nLet \\( smallness(timeless) \\) be the double integral. Then\n\\[\n\\lim _{timeless \\rightarrow \\infty}\\left[smallness(timeless) / e^{timeless}\\right]=\\lim _{timeless \\rightarrow \\infty}\\left[smallness^{\\prime}(timeless) / e^{timeless}\\right]\n\\]\nby L'Hôpital's Rule. One finds that\n\\[\nsmallness^{\\prime}(timeless)=\\int_{0}^{timeless} \\frac{e^{knownvalue}-e^{timeless}}{knownvalue-timeless} d knownvalue+\\int_{0}^{timeless} \\frac{e^{fixedvalue}-e^{timeless}}{fixedvalue-timeless} d fixedvalue=2 \\int_{0}^{timeless} \\frac{e^{knownvalue}-e^{timeless}}{knownvalue-timeless} d knownvalue\n\\]\nThen using \\( e^{knownvalue}=e^{timeless}\\left[1+(knownvalue-timeless)+(knownvalue-timeless)^{2} / 2!+\\cdots\\right] \\), one sees that \\( e^{-timeless} smallness^{\\prime}(timeless) \\rightarrow \\infty \\) as \\( timeless \\rightarrow \\infty \\) since for sufficiently large \\( timeless \\),\n\\[\n\\frac{smallness^{\\prime}(timeless)}{2 e^{timeless}}=\\int_{0}^{timeless} \\frac{e^{knownvalue-timeless}-1}{knownvalue-timeless} d knownvalue=\\int_{0}^{timeless} \\frac{1-e^{-fixedvalue}}{fixedvalue} d fixedvalue>\\int_{1}^{timeless} \\frac{1-e^{-fixedvalue}}{fixedvalue} d fixedvalue>\\left(1-e^{-1}\\right) \\log timeless\n\\]"
    },
    "garbled_string": {
      "map": {
        "t": "qzxwvtnp",
        "x": "hjgrksla",
        "y": "mpdqlrzn",
        "G": "fjdkasoe"
      },
      "question": "Problem A-3\nFind\n\\[\n\\lim _{qzxwvtnp \\rightarrow \\infty}\\left[e^{-qzxwvtnp} \\int_{0}^{qzxwvtnp} \\int_{0}^{qzxwvtnp} \\frac{e^{hjgrksla}-e^{mpdqlrzn}}{hjgrksla-mpdqlrzn} d hjgrksla d mpdqlrzn\\right]\n\\]\nor show that the limit does not exist.",
      "solution": "A-3.\nLet \\( fjdkasoe(qzxwvtnp) \\) be the double integral. Then\n\\[\n\\lim _{qzxwvtnp \\rightarrow \\infty}\\left[fjdkasoe(qzxwvtnp) / e^{qzxwvtnp}\\right]=\\lim _{qzxwvtnp \\rightarrow \\infty}\\left[fjdkasoe^{\\prime}(qzxwvtnp) / e^{qzxwvtnp}\\right]\n\\]\nby L'H\\^opital's Rule. One finds that\n\\[\nfjdkasoe^{\\prime}(qzxwvtnp)=\\int_{0}^{qzxwvtnp} \\frac{e^{hjgrksla}-e^{qzxwvtnp}}{hjgrksla-qzxwvtnp} d hjgrksla+\\int_{0}^{qzxwvtnp} \\frac{e^{mpdqlrzn}-e^{qzxwvtnp}}{mpdqlrzn-qzxwvtnp} d mpdqlrzn=2 \\int_{0}^{qzxwvtnp} \\frac{e^{hjgrksla}-e^{qzxwvtnp}}{hjgrksla-qzxwvtnp} d hjgrksla\n\\]\n\nThen using \\( e^{hjgrksla}=e^{qzxwvtnp}\\left[1+(hjgrksla-qzxwvtnp)+(hjgrksla-qzxwvtnp)^{2} / 2!+\\cdots\\right] \\), one sees that \\( e^{-qzxwvtnp} fjdkasoe^{\\prime}(qzxwvtnp) \\rightarrow \\infty \\) as \\( qzxwvtnp \\rightarrow \\infty \\) since for sufficiently large \\( qzxwvtnp \\),\n\\[\n\\frac{fjdkasoe^{\\prime}(qzxwvtnp)}{2 e^{qzxwvtnp}}=\\int_{0}^{qzxwvtnp} \\frac{e^{hjgrksla-qzxwvtnp}-1}{hjgrksla-qzxwvtnp} d hjgrksla=\\int_{0}^{qzxwvtnp} \\frac{1-e^{-mpdqlrzn}}{mpdqlrzn} d mpdqlrzn>\\int_{1}^{qzxwvtnp} \\frac{1-e^{-mpdqlrzn}}{mpdqlrzn} d mpdqlrzn>\\left(1-e^{-1}\\right) \\log qzxwvtnp\n\\]"
    },
    "kernel_variant": {
      "question": "Let $a,b,c$ be fixed positive real numbers.  For $t>0$ define \n\\[\nK(t)\\;:=\\;\n        \\iint_{(0,t)^{2}}\n        \\frac{\\bigl(e^{ax}-e^{ay}\\bigr)\\,\n              \\bigl(1-\\cos b(x-y)\\bigr)\\,\n              e^{-c|x-y|}}\n             {x-y}\\,dx\\,dy .\n\\]\n\n(a)  Show that $K(t)<\\infty$ for every $t>0$ and that the limit \n\\[\nL\\;:=\\;\\lim_{t\\to\\infty} e^{-at}\\,K(t)\n\\]\nexists.\n\n(b)  Compute $L$ in closed form, proving that for all $(a,b,c)\\in(0,\\infty)^{3}$\n\\[\n\\boxed{\\;\n   L\\;=\\;\\frac{1}{a}\\,\n           \\ln\\!\\Bigl(\n                 \\frac{\\,1+(b/c)^{2}\\,}\n                      {\\,1+\\bigl(b/(a+c)\\bigr)^{2}\\,}\n               \\Bigr)\n   \\;}.\n\\]",
      "solution": "Throughout we fix $a,b,c>0$ and work with Lebesgue integration.  \nPut $s:=x-y$ and define \n\\[\nF(x,y)\\;:=\\;\n       \\frac{\\bigl(e^{ax}-e^{ay}\\bigr)\\,\n             \\bigl(1-\\cos bs\\bigr)\\,\n             e^{-c|s|}}\n            {s},\n\\qquad s\\neq 0 .\n\\]\n\n1.\\;Local integrability of the kernel  \n\nNear the diagonal $x=y$ we have the Taylor expansions\n\\[\ne^{ax}-e^{ay}=ae^{ax}s+O(s^{2}),\\qquad\n1-\\cos bs=\\tfrac{1}{2}b^{2}s^{2}+O(s^{4}),\\qquad\ne^{-c|s|}=1+O(|s|).\n\\]\nHence\n\\[\nF(x,y)=\\tfrac12\\,a b^{2}e^{ax}s^{2}+O(s^{3})=O(s^{2}),\n\\qquad s\\to 0.\n\\]\nBecause $s^{2}$ is integrable in a neighbourhood of $0$, the kernel is locally integrable on $(0,t)^{2}$ for every $t>0$.\n\n2.\\;A uniform $L^{1}$–majorant on the boundary  \n\nSince $|1-\\cos bs|\\le\\tfrac12 b^{2}s^{2}$ and $|e^{ax}-e^{ay}|\\le a e^{a\\max\\{x,y\\}}|s|$, \n\\[\n|F(x,y)|\\;\\le\\;\\frac{a b^{2}}{2}\\,e^{a(x\\vee y)}\\,|s|^{2}.\n\\tag{1}\n\\]\nIn particular, for the boundary $x=t$ we get the sharper estimate\n\\[\n|F(t,y)|\\;\\le\\;\\frac{a b^{2}}{2}\\,e^{at}\\,|t-y|^{2},\n\\qquad 0\\le y\\le t .\n\\tag{2}\n\\]\nBecause $\\int_{0}^{t}|t-y|^{2}\\,dy=\\tfrac13 t^{3}<\\infty$, the right–hand side of (2) is an $L^{1}$–majorant.  This will justify differentiation under the integral sign.\n\n3.\\;Differentiation of $K(t)$  \n\nBy Fubini–Tonelli and the majorant (2) the map $t\\mapsto K(t)$ is absolutely continuous and, for a.e.\\ $t>0$,\n\\[\nK'(t)=\\int_{0}^{t}F(t,y)\\,dy+\\int_{0}^{t}F(x,t)\\,dx.\n\\]\nBecause $F(x,y)=F(y,x)$, the two integrals coincide, hence\n\\[\nK'(t)=2\\int_{0}^{t}F(t,y)\\,dy.\n\\tag{3}\n\\]\nSubstituting $s=t-y$ ($y=t-s$) gives\n\\[\nF(t,y)=e^{at}\\,\n       \\frac{\\bigl(1-e^{-as}\\bigr)\n             \\bigl(1-\\cos bs\\bigr)\n             e^{-cs}}{s},\n\\]\nso that  \n\\[\nK'(t)=2e^{at}\\,J_{c}(t),\n\\qquad\nJ_{c}(t):=\\int_{0}^{t}\n           \\frac{\\bigl(1-e^{-as}\\bigr)\n                 \\bigl(1-\\cos bs\\bigr)\n                 e^{-cs}}{s}\\,ds.\n\\tag{4}\n\\]\n\n4.\\;Monotonicity and boundedness of $J_{c}$  \n\nThe integrand in (4) is non–negative.  Near $s=0$ we have\n\\[\n\\frac{\\bigl(1-e^{-as}\\bigr)\\bigl(1-\\cos bs\\bigr)e^{-cs}}{s}\n      =\\frac{\\bigl(a s+O(s^{2})\\bigr)\\bigl(\\tfrac12 b^{2}s^{2}+O(s^{4})\\bigr)}{s}\\,\n        (1+O(s))\n      =\\tfrac12 a b^{2}s^{2}+O(s^{3}),\n\\]\nwhich is integrable at $0$.  For large $s$ the factor $e^{-cs}$ enforces exponential decay.  Therefore \n\\[\nJ_{c}(\\infty):=\n\\int_{0}^{\\infty}\n\\frac{\\bigl(1-e^{-as}\\bigr)\\bigl(1-\\cos bs\\bigr)e^{-cs}}{s}\\,ds\n<\\infty .\n\\tag{5}\n\\]\nConsequently\n\\[\n|K'(t)|\\;\\le\\;2e^{at}J_{c}(\\infty),\\qquad\nK(t)=O(e^{at}).\n\\]\n\n5.\\;Cauchy’s form of l’Hospital’s rule  \n\nSet $f(t):=K(t)$ and $g(t):=e^{at}$.  Both are absolutely continuous, $g(t)\\to\\infty$, and $g'(t)=a e^{at}>0$.  From (3)–(5),\n\\[\n\\left|\\frac{f'(t)}{g'(t)}\\right|\n      =\\frac{2J_{c}(t)}{a}\\;\\le\\;\\frac{2}{a}J_{c}(\\infty)<\\infty .\n\\]\nSince $J_{c}(t)$ is non–decreasing and bounded, $\\displaystyle\\lim_{t\\to\\infty}J_{c}(t)=J_{c}(\\infty)$ exists.  Hence Cauchy’s form of l’Hospital’s rule yields\n\\[\n\\lim_{t\\to\\infty}\\frac{K(t)}{e^{at}}\n      =\\lim_{t\\to\\infty}\\frac{K'(t)}{a e^{at}}\n      =\\lim_{t\\to\\infty}\\frac{2J_{c}(t)}{a}\n      =\\frac{2}{a}J_{c}(\\infty).\n\\tag{6}\n\\]\nThus \n\\[\nL=\\frac{2}{a}J_{c}(\\infty).\n\\]\n\n6.\\;Evaluation of $J_{c}(\\infty)$  \n\nWrite $J_{c}=A-B$ with\n\\[\nA:=\\int_{0}^{\\infty}\n    \\frac{1-\\cos bs}{s}\\,e^{-cs}\\,ds,\n\\qquad\nB:=\\int_{0}^{\\infty}\n    \\frac{1-\\cos bs}{s}\\,e^{-(a+c)s}\\,ds.\n\\tag{7}\n\\]\n\nFor $\\lambda>0$ define\n\\[\n\\Phi(\\lambda):=\\int_{0}^{\\infty}\n               \\frac{1-\\cos bs}{s}\\,e^{-\\lambda s}\\,ds.\n\\tag{8}\n\\]\nDifferentiate under the integral with respect to $b$ (justified by dominated convergence, since $|s\\sin bs|\\le s$ and $\\int_{0}^{\\infty}s e^{-\\lambda s}\\,ds<\\infty$):\n\\[\n\\frac{\\partial\\Phi}{\\partial b}\n        =\\int_{0}^{\\infty}\\sin(bs)\\,e^{-\\lambda s}\\,ds\n        =\\frac{b}{\\lambda^{2}+b^{2}} .\n\\]\nIntegrating in $b$ and enforcing $\\Phi(0)=0$ gives\n\\[\n\\Phi(\\lambda)=\\frac12\\ln\\!\\bigl(1+(b/\\lambda)^{2}\\bigr).\n\\tag{9}\n\\]\nSince $c>0$ and $a+c>0$, monotone convergence implies\n\\[\nA=\\Phi(c),\\qquad B=\\Phi(a+c),\n\\]\nand therefore\n\\[\nJ_{c}(\\infty)=A-B\n             =\\frac12\n               \\ln\\!\\Bigl(\n                     \\frac{1+(b/c)^{2}}\n                          {1+\\bigl(b/(a+c)\\bigr)^{2}}\n               \\Bigr).\n\\tag{10}\n\\]\n\n7.\\;Final value of $L$  \n\nInsert (10) into (6):\n\\[\nL=\\frac{1}{a}\\,\n   \\ln\\!\\Bigl(\n         \\frac{1+(b/c)^{2}}\n              {1+\\bigl(b/(a+c)\\bigr)^{2}}\n     \\Bigr).\n\\]\nHence the limit exists and equals the claimed expression.\n\n8.\\;Consistency checks  \n\n(i) $c\\to\\infty$ (strong damping): numerator $\\to 1$, denominator $\\to 1$, so $L\\to 0$.  \n(ii) $c\\to 0^{+}$ (no damping): numerator $\\to\\infty$ while the denominator stays finite, so $L\\to\\infty$.  \n\nBoth behaviours are physically plausible, confirming the result.  \n$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.656587",
        "was_fixed": false,
        "difficulty_analysis": "1.  Additional structure: the original integrand \\(\\dfrac{e^{kx}-e^{ky}}{x-y}\\) is now multiplied by \\(\\sin(x-y)\\), introducing oscillatory behaviour that precludes naive Taylor-series estimates.  \n2.  Higher-level techniques: evaluation demands (i) differentiation under the integral sign, (ii) controlled use of L’Hôpital’s Rule on an \\(\\infty/\\infty\\) form, and (iii) a non-elementary Laplace–Fourier integral that links to the arctangent function.  \n3.  Interacting concepts: exponential growth, oscillatory kernels, and singular integrals all interplay; missing any one tool (e.g. the Laplace–Fourier identity) stalls the computation.  \n4.  Deeper insight: recognising that only the derivative behaves nicely after factoring out \\(e^{3t}\\), and that the sine kernel converts the problem into a known definite integral, is far less direct than in the original problem.  \n\nAll these layers make the enhanced variant significantly more challenging than both the source problem and the existing kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let $a,b,c$ be fixed positive real numbers.  For $t>0$ define \n\\[\nK(t)\\;:=\\;\n        \\iint_{(0,t)^{2}}\n        \\frac{\\bigl(e^{ax}-e^{ay}\\bigr)\\,\n              \\bigl(1-\\cos b(x-y)\\bigr)\\,\n              e^{-c|x-y|}}\n             {x-y}\\,dx\\,dy .\n\\]\n\n(a)  Show that $K(t)<\\infty$ for every $t>0$ and that the limit \n\\[\nL\\;:=\\;\\lim_{t\\to\\infty} e^{-at}\\,K(t)\n\\]\nexists.\n\n(b)  Compute $L$ in closed form, proving that for all $(a,b,c)\\in(0,\\infty)^{3}$\n\\[\n\\boxed{\\;\n   L\\;=\\;\\frac{1}{a}\\,\n           \\ln\\!\\Bigl(\n                 \\frac{\\,1+(b/c)^{2}\\,}\n                      {\\,1+\\bigl(b/(a+c)\\bigr)^{2}\\,}\n               \\Bigr)\n   \\;}.\n\\]",
      "solution": "Throughout we fix $a,b,c>0$ and work with Lebesgue integration.  \nPut $s:=x-y$ and define \n\\[\nF(x,y)\\;:=\\;\n       \\frac{\\bigl(e^{ax}-e^{ay}\\bigr)\\,\n             \\bigl(1-\\cos bs\\bigr)\\,\n             e^{-c|s|}}\n            {s},\n\\qquad s\\neq 0 .\n\\]\n\n1.\\;Local integrability of the kernel  \n\nNear the diagonal $x=y$ we have the Taylor expansions\n\\[\ne^{ax}-e^{ay}=ae^{ax}s+O(s^{2}),\\qquad\n1-\\cos bs=\\tfrac{1}{2}b^{2}s^{2}+O(s^{4}),\\qquad\ne^{-c|s|}=1+O(|s|).\n\\]\nHence\n\\[\nF(x,y)=\\tfrac12\\,a b^{2}e^{ax}s^{2}+O(s^{3})=O(s^{2}),\n\\qquad s\\to 0.\n\\]\nBecause $s^{2}$ is integrable in a neighbourhood of $0$, the kernel is locally integrable on $(0,t)^{2}$ for every $t>0$.\n\n2.\\;A uniform $L^{1}$–majorant on the boundary  \n\nSince $|1-\\cos bs|\\le\\tfrac12 b^{2}s^{2}$ and $|e^{ax}-e^{ay}|\\le a e^{a\\max\\{x,y\\}}|s|$, \n\\[\n|F(x,y)|\\;\\le\\;\\frac{a b^{2}}{2}\\,e^{a(x\\vee y)}\\,|s|^{2}.\n\\tag{1}\n\\]\nIn particular, for the boundary $x=t$ we get the sharper estimate\n\\[\n|F(t,y)|\\;\\le\\;\\frac{a b^{2}}{2}\\,e^{at}\\,|t-y|^{2},\n\\qquad 0\\le y\\le t .\n\\tag{2}\n\\]\nBecause $\\int_{0}^{t}|t-y|^{2}\\,dy=\\tfrac13 t^{3}<\\infty$, the right–hand side of (2) is an $L^{1}$–majorant.  This will justify differentiation under the integral sign.\n\n3.\\;Differentiation of $K(t)$  \n\nBy Fubini–Tonelli and the majorant (2) the map $t\\mapsto K(t)$ is absolutely continuous and, for a.e.\\ $t>0$,\n\\[\nK'(t)=\\int_{0}^{t}F(t,y)\\,dy+\\int_{0}^{t}F(x,t)\\,dx.\n\\]\nBecause $F(x,y)=F(y,x)$, the two integrals coincide, hence\n\\[\nK'(t)=2\\int_{0}^{t}F(t,y)\\,dy.\n\\tag{3}\n\\]\nSubstituting $s=t-y$ ($y=t-s$) gives\n\\[\nF(t,y)=e^{at}\\,\n       \\frac{\\bigl(1-e^{-as}\\bigr)\n             \\bigl(1-\\cos bs\\bigr)\n             e^{-cs}}{s},\n\\]\nso that  \n\\[\nK'(t)=2e^{at}\\,J_{c}(t),\n\\qquad\nJ_{c}(t):=\\int_{0}^{t}\n           \\frac{\\bigl(1-e^{-as}\\bigr)\n                 \\bigl(1-\\cos bs\\bigr)\n                 e^{-cs}}{s}\\,ds.\n\\tag{4}\n\\]\n\n4.\\;Monotonicity and boundedness of $J_{c}$  \n\nThe integrand in (4) is non–negative.  Near $s=0$ we have\n\\[\n\\frac{\\bigl(1-e^{-as}\\bigr)\\bigl(1-\\cos bs\\bigr)e^{-cs}}{s}\n      =\\frac{\\bigl(a s+O(s^{2})\\bigr)\\bigl(\\tfrac12 b^{2}s^{2}+O(s^{4})\\bigr)}{s}\\,\n        (1+O(s))\n      =\\tfrac12 a b^{2}s^{2}+O(s^{3}),\n\\]\nwhich is integrable at $0$.  For large $s$ the factor $e^{-cs}$ enforces exponential decay.  Therefore \n\\[\nJ_{c}(\\infty):=\n\\int_{0}^{\\infty}\n\\frac{\\bigl(1-e^{-as}\\bigr)\\bigl(1-\\cos bs\\bigr)e^{-cs}}{s}\\,ds\n<\\infty .\n\\tag{5}\n\\]\nConsequently\n\\[\n|K'(t)|\\;\\le\\;2e^{at}J_{c}(\\infty),\\qquad\nK(t)=O(e^{at}).\n\\]\n\n5.\\;Cauchy’s form of l’Hospital’s rule  \n\nSet $f(t):=K(t)$ and $g(t):=e^{at}$.  Both are absolutely continuous, $g(t)\\to\\infty$, and $g'(t)=a e^{at}>0$.  From (3)–(5),\n\\[\n\\left|\\frac{f'(t)}{g'(t)}\\right|\n      =\\frac{2J_{c}(t)}{a}\\;\\le\\;\\frac{2}{a}J_{c}(\\infty)<\\infty .\n\\]\nSince $J_{c}(t)$ is non–decreasing and bounded, $\\displaystyle\\lim_{t\\to\\infty}J_{c}(t)=J_{c}(\\infty)$ exists.  Hence Cauchy’s form of l’Hospital’s rule yields\n\\[\n\\lim_{t\\to\\infty}\\frac{K(t)}{e^{at}}\n      =\\lim_{t\\to\\infty}\\frac{K'(t)}{a e^{at}}\n      =\\lim_{t\\to\\infty}\\frac{2J_{c}(t)}{a}\n      =\\frac{2}{a}J_{c}(\\infty).\n\\tag{6}\n\\]\nThus \n\\[\nL=\\frac{2}{a}J_{c}(\\infty).\n\\]\n\n6.\\;Evaluation of $J_{c}(\\infty)$  \n\nWrite $J_{c}=A-B$ with\n\\[\nA:=\\int_{0}^{\\infty}\n    \\frac{1-\\cos bs}{s}\\,e^{-cs}\\,ds,\n\\qquad\nB:=\\int_{0}^{\\infty}\n    \\frac{1-\\cos bs}{s}\\,e^{-(a+c)s}\\,ds.\n\\tag{7}\n\\]\n\nFor $\\lambda>0$ define\n\\[\n\\Phi(\\lambda):=\\int_{0}^{\\infty}\n               \\frac{1-\\cos bs}{s}\\,e^{-\\lambda s}\\,ds.\n\\tag{8}\n\\]\nDifferentiate under the integral with respect to $b$ (justified by dominated convergence, since $|s\\sin bs|\\le s$ and $\\int_{0}^{\\infty}s e^{-\\lambda s}\\,ds<\\infty$):\n\\[\n\\frac{\\partial\\Phi}{\\partial b}\n        =\\int_{0}^{\\infty}\\sin(bs)\\,e^{-\\lambda s}\\,ds\n        =\\frac{b}{\\lambda^{2}+b^{2}} .\n\\]\nIntegrating in $b$ and enforcing $\\Phi(0)=0$ gives\n\\[\n\\Phi(\\lambda)=\\frac12\\ln\\!\\bigl(1+(b/\\lambda)^{2}\\bigr).\n\\tag{9}\n\\]\nSince $c>0$ and $a+c>0$, monotone convergence implies\n\\[\nA=\\Phi(c),\\qquad B=\\Phi(a+c),\n\\]\nand therefore\n\\[\nJ_{c}(\\infty)=A-B\n             =\\frac12\n               \\ln\\!\\Bigl(\n                     \\frac{1+(b/c)^{2}}\n                          {1+\\bigl(b/(a+c)\\bigr)^{2}}\n               \\Bigr).\n\\tag{10}\n\\]\n\n7.\\;Final value of $L$  \n\nInsert (10) into (6):\n\\[\nL=\\frac{1}{a}\\,\n   \\ln\\!\\Bigl(\n         \\frac{1+(b/c)^{2}}\n              {1+\\bigl(b/(a+c)\\bigr)^{2}}\n     \\Bigr).\n\\]\nHence the limit exists and equals the claimed expression.\n\n8.\\;Consistency checks  \n\n(i) $c\\to\\infty$ (strong damping): numerator $\\to 1$, denominator $\\to 1$, so $L\\to 0$.  \n(ii) $c\\to 0^{+}$ (no damping): numerator $\\to\\infty$ while the denominator stays finite, so $L\\to\\infty$.  \n\nBoth behaviours are physically plausible, confirming the result.  \n$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.518095",
        "was_fixed": false,
        "difficulty_analysis": "1.  Additional structure: the original integrand \\(\\dfrac{e^{kx}-e^{ky}}{x-y}\\) is now multiplied by \\(\\sin(x-y)\\), introducing oscillatory behaviour that precludes naive Taylor-series estimates.  \n2.  Higher-level techniques: evaluation demands (i) differentiation under the integral sign, (ii) controlled use of L’Hôpital’s Rule on an \\(\\infty/\\infty\\) form, and (iii) a non-elementary Laplace–Fourier integral that links to the arctangent function.  \n3.  Interacting concepts: exponential growth, oscillatory kernels, and singular integrals all interplay; missing any one tool (e.g. the Laplace–Fourier identity) stalls the computation.  \n4.  Deeper insight: recognising that only the derivative behaves nicely after factoring out \\(e^{3t}\\), and that the sine kernel converts the problem into a known definite integral, is far less direct than in the original problem.  \n\nAll these layers make the enhanced variant significantly more challenging than both the source problem and the existing kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}