{
  "index": "2000-A-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Show that the improper integral\n\\[ \\lim_{B\\to\\infty}\\int_{0}^B \\sin(x) \\sin(x^2)\\,dx\\]\nconverges.",
  "solution": "To avoid some improper integrals at 0, we may as well replace the left endpoint of integration by some $\\epsilon > 0$. We now use integration by parts:\n\\begin{align*}\n\\int_\\epsilon^B \\sin x \\sin x^2\\,dx\n&= \\int_\\epsilon^B \\frac{\\sin x}{2x} \\sin x^2 (2x\\,dx) \\\\\n&= \\left. -\\frac{\\sin x}{2x} \\cos x^2 \\right|_\\epsilon^B \\\\\n&\\mbox{} + \\int_\\epsilon^B \\left( \\frac{\\cos x}{2x} - \\frac{\\sin x}{2x^2} \\right) \\cos x^2\\,dx.\n\\end{align*}\nNow $\\frac{\\sin x}{2x} \\cos x^2$ tends to 0 as $B \\to \\infty$,\nand the integral of $\\frac{\\sin x}{2x^2} \\cos x^2$ converges absolutely\nby comparison with $1/x^2$. Thus it suffices to note that\n\\begin{align*}\n\\int_\\epsilon^B \\frac{\\cos x}{2x} \\cos x^2\\,dx &=\n\\int_\\epsilon^B \\frac{\\cos x}{4x^2} \\cos x^2(2x\\,dx) \\\\\n&= \\left. \\frac{\\cos x}{4x^2} \\sin x^2 \\right|_\\epsilon^B \\\\\n&\\mbox{} - \\int_\\epsilon^B \\frac{2x\\cos x - \\sin x}{4x^3} \\sin x^2\\,dx,\n\\end{align*}\nand that the final integral converges absolutely by comparison to\n$1/x^3$.\n\nAn alternate approach is to first rewrite $\\sin x \\sin x^2$ as\n$\\frac{1}{2}(\\cos (x^2-x) - \\cos (x^2+x))$. Then\n\\begin{align*}\n\\int_\\epsilon^B \\cos(x^2+x)\\,dx &=\n- \\left. \\frac{\\sin (x^2+x)}{2x+1} \\right|_\\epsilon^B \\\\\n&\\mbox{} - \\int_\\epsilon^B \\frac{2\\sin(x^2+x)}{(2x+1)^2}\\,dx\n\\end{align*}\nconverges absolutely, and $\\int_0^B \\cos (x^2-x)$ can be\ntreated similarly.",
  "vars": [
    "x"
  ],
  "params": [
    "B",
    "\\\\epsilon"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "B": "upperlimit",
        "\\epsilon": "smallpos"
      },
      "question": "Show that the improper integral\n\\[ \\lim_{upperlimit\\to\\infty}\\int_{0}^{upperlimit} \\sin(variable) \\sin(variable^2)\\,d variable\\]\nconverges.",
      "solution": "To avoid some improper integrals at 0, we may as well replace the left endpoint of integration by some $smallpos > 0$. We now use integration by parts:\n\\begin{align*}\n\\int_{smallpos}^{upperlimit} \\sin variable \\sin variable^2\\,d variable\n&= \\int_{smallpos}^{upperlimit} \\frac{\\sin variable}{2variable} \\sin variable^2 (2variable\\,d variable) \\\\\n&= \\left. -\\frac{\\sin variable}{2variable} \\cos variable^2 \\right|_{smallpos}^{upperlimit} \\\\\n&\\mbox{} + \\int_{smallpos}^{upperlimit} \\left( \\frac{\\cos variable}{2variable} - \\frac{\\sin variable}{2variable^2} \\right) \\cos variable^2\\,d variable.\n\\end{align*}\nNow $\\frac{\\sin variable}{2variable} \\cos variable^2$ tends to 0 as $upperlimit \\to \\infty$, and the integral of $\\frac{\\sin variable}{2variable^2} \\cos variable^2$ converges absolutely by comparison with $1/variable^2$. Thus it suffices to note that\n\\begin{align*}\n\\int_{smallpos}^{upperlimit} \\frac{\\cos variable}{2variable} \\cos variable^2\\,d variable &= \\int_{smallpos}^{upperlimit} \\frac{\\cos variable}{4variable^2} \\cos variable^2(2variable\\,d variable) \\\\\n&= \\left. \\frac{\\cos variable}{4variable^2} \\sin variable^2 \\right|_{smallpos}^{upperlimit} \\\\\n&\\mbox{} - \\int_{smallpos}^{upperlimit} \\frac{2variable\\cos variable - \\sin variable}{4variable^3} \\sin variable^2\\,d variable,\n\\end{align*}\nand that the final integral converges absolutely by comparison to $1/variable^3$.\n\nAn alternate approach is to first rewrite $\\sin variable \\sin variable^2$ as $\\frac{1}{2}(\\cos (variable^2-variable) - \\cos (variable^2+variable))$. Then\n\\begin{align*}\n\\int_{smallpos}^{upperlimit} \\cos(variable^2+variable)\\,d variable &= - \\left. \\frac{\\sin (variable^2+variable)}{2variable+1} \\right|_{smallpos}^{upperlimit} \\\\\n&\\mbox{} - \\int_{smallpos}^{upperlimit} \\frac{2\\sin(variable^2+variable)}{(2variable+1)^2}\\,d variable\n\\end{align*}\nconverges absolutely, and $\\int_0^{upperlimit} \\cos (variable^2-variable)$ can be treated similarly."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "perimeter",
        "B": "cylinder",
        "\\epsilon": "parabola"
      },
      "question": "Show that the improper integral\n\\[ \\lim_{cylinder\\to\\infty}\\int_{0}^{cylinder} \\sin(perimeter) \\sin(perimeter^2)\\,dperimeter\\]\nconverges.",
      "solution": "To avoid some improper integrals at 0, we may as well replace the left endpoint of integration by some $\\parabola > 0$. We now use integration by parts:\n\\begin{align*}\n\\int_\\parabola^{cylinder} \\sin perimeter \\sin perimeter^2\\,dperimeter\n&= \\int_\\parabola^{cylinder} \\frac{\\sin perimeter}{2perimeter} \\sin perimeter^2 (2perimeter\\,dperimeter) \\\\\n&= \\left. -\\frac{\\sin perimeter}{2perimeter} \\cos perimeter^2 \\right|_\\parabola^{cylinder} \\\\\n&\\mbox{} + \\int_\\parabola^{cylinder} \\left( \\frac{\\cos perimeter}{2perimeter} - \\frac{\\sin perimeter}{2perimeter^2} \\right) \\cos perimeter^2\\,dperimeter.\n\\end{align*}\nNow $\\frac{\\sin perimeter}{2perimeter} \\cos perimeter^2$ tends to 0 as $cylinder \\to \\infty$,\nand the integral of $\\frac{\\sin perimeter}{2perimeter^2} \\cos perimeter^2$ converges absolutely\nby comparison with $1/perimeter^2$. Thus it suffices to note that\n\\begin{align*}\n\\int_\\parabola^{cylinder} \\frac{\\cos perimeter}{2perimeter} \\cos perimeter^2\\,dperimeter &=\n\\int_\\parabola^{cylinder} \\frac{\\cos perimeter}{4perimeter^2} \\cos perimeter^2(2perimeter\\,dperimeter) \\\\\n&= \\left. \\frac{\\cos perimeter}{4perimeter^2} \\sin perimeter^2 \\right|_\\parabola^{cylinder} \\\\\n&\\mbox{} - \\int_\\parabola^{cylinder} \\frac{2perimeter\\cos perimeter - \\sin perimeter}{4perimeter^3} \\sin perimeter^2\\,dperimeter,\n\\end{align*}\nand that the final integral converges absolutely by comparison to\n$1/perimeter^3$.\n\nAn alternate approach is to first rewrite $\\sin perimeter \\sin perimeter^2$ as\n$\\frac{1}{2}(\\cos (perimeter^2-perimeter) - \\cos (perimeter^2+perimeter))$. Then\n\\begin{align*}\n\\int_\\parabola^{cylinder} \\cos(perimeter^2+perimeter)\\,dperimeter &=\n- \\left. \\frac{\\sin (perimeter^2+perimeter)}{2perimeter+1} \\right|_\\parabola^{cylinder} \\\\\n&\\mbox{} - \\int_\\parabola^{cylinder} \\frac{2\\sin(perimeter^2+perimeter)}{(2perimeter+1)^2}\\,dperimeter\n\\end{align*}\nconverges absolutely, and $\\int_0^{cylinder} \\cos (perimeter^2-perimeter)$ can be\ntreated similarly."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "B": "lowlimit",
        "\\epsilon": "megavalue"
      },
      "question": "Show that the improper integral\n\\[ \\lim_{lowlimit\\to\\infty}\\int_{0}^{lowlimit} \\sin(constantval) \\sin(constantval^2)\\,d constantval\\]\nconverges.",
      "solution": "To avoid some improper integrals at 0, we may as well replace the left endpoint of integration by some $megavalue > 0$. We now use integration by parts:\n\\begin{align*}\n\\int_{megavalue}^{lowlimit} \\sin constantval \\sin constantval^2\\,d constantval\n&= \\int_{megavalue}^{lowlimit} \\frac{\\sin constantval}{2 constantval} \\sin constantval^2 (2 constantval\\,d constantval) \\\\\n&= \\left. -\\frac{\\sin constantval}{2 constantval} \\cos constantval^2 \\right|_{megavalue}^{lowlimit} \\\\\n&\\mbox{} + \\int_{megavalue}^{lowlimit} \\left( \\frac{\\cos constantval}{2 constantval} - \\frac{\\sin constantval}{2 constantval^2} \\right) \\cos constantval^2\\,d constantval.\n\\end{align*}\nNow $\\frac{\\sin constantval}{2 constantval} \\cos constantval^2$ tends to 0 as $lowlimit \\to \\infty$, and the integral of $\\frac{\\sin constantval}{2 constantval^2} \\cos constantval^2$ converges absolutely by comparison with $1/constantval^2$. Thus it suffices to note that\n\\begin{align*}\n\\int_{megavalue}^{lowlimit} \\frac{\\cos constantval}{2 constantval} \\cos constantval^2\\,d constantval &= \\int_{megavalue}^{lowlimit} \\frac{\\cos constantval}{4 constantval^2} \\cos constantval^2(2 constantval\\,d constantval) \\\\\n&= \\left. \\frac{\\cos constantval}{4 constantval^2} \\sin constantval^2 \\right|_{megavalue}^{lowlimit} \\\\\n&\\mbox{} - \\int_{megavalue}^{lowlimit} \\frac{2 constantval\\cos constantval - \\sin constantval}{4 constantval^3} \\sin constantval^2\\,d constantval,\n\\end{align*}\nand that the final integral converges absolutely by comparison to $1/constantval^3$.\n\nAn alternate approach is to first rewrite $\\sin constantval \\sin constantval^2$ as $\\frac{1}{2}(\\cos (constantval^2-constantval) - \\cos (constantval^2+constantval))$. Then\n\\begin{align*}\n\\int_{megavalue}^{lowlimit} \\cos(constantval^2+constantval)\\,d constantval &= - \\left. \\frac{\\sin (constantval^2+constantval)}{2 constantval+1} \\right|_{megavalue}^{lowlimit} \\\\\n&\\mbox{} - \\int_{megavalue}^{lowlimit} \\frac{2\\sin(constantval^2+constantval)}{(2 constantval+1)^2}\\,d constantval\n\\end{align*}\nconverges absolutely, and $\\int_0^{lowlimit} \\cos (constantval^2-constantval)$ can be treated similarly."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "B": "mnlkjprs",
        "\\epsilon": "vcbnksld"
      },
      "question": "Show that the improper integral\n\\[ \\lim_{mnlkjprs\\to\\infty}\\int_{0}^{mnlkjprs} \\sin(qzxwvtnp) \\sin(qzxwvtnp^2)\\,dqzxwvtnp\\]\nconverges.",
      "solution": "To avoid some improper integrals at 0, we may as well replace the left endpoint of integration by some $vcbnksld > 0$. We now use integration by parts:\n\\begin{align*}\n\\int_{vcbnksld}^{mnlkjprs} \\sin qzxwvtnp \\sin qzxwvtnp^2\\,dqzxwvtnp\n&= \\int_{vcbnksld}^{mnlkjprs} \\frac{\\sin qzxwvtnp}{2qzxwvtnp} \\sin qzxwvtnp^2 \\bigl(2qzxwvtnp\\,dqzxwvtnp\\bigr) \\\\\n&= \\left. -\\frac{\\sin qzxwvtnp}{2qzxwvtnp} \\cos qzxwvtnp^2 \\right|_{vcbnksld}^{mnlkjprs} \\\\\n&\\mbox{} + \\int_{vcbnksld}^{mnlkjprs} \\left( \\frac{\\cos qzxwvtnp}{2qzxwvtnp} - \\frac{\\sin qzxwvtnp}{2qzxwvtnp^2} \\right) \\cos qzxwvtnp^2\\,dqzxwvtnp.\n\\end{align*}\nNow $\\frac{\\sin qzxwvtnp}{2qzxwvtnp} \\cos qzxwvtnp^2$ tends to 0 as $mnlkjprs \\to \\infty$, and the integral of $\\frac{\\sin qzxwvtnp}{2qzxwvtnp^2} \\cos qzxwvtnp^2$ converges absolutely by comparison with $1/qzxwvtnp^2$. Thus it suffices to note that\n\\begin{align*}\n\\int_{vcbnksld}^{mnlkjprs} \\frac{\\cos qzxwvtnp}{2qzxwvtnp} \\cos qzxwvtnp^2\\,dqzxwvtnp &=\n\\int_{vcbnksld}^{mnlkjprs} \\frac{\\cos qzxwvtnp}{4qzxwvtnp^2} \\cos qzxwvtnp^2\\bigl(2qzxwvtnp\\,dqzxwvtnp\\bigr) \\\\\n&= \\left. \\frac{\\cos qzxwvtnp}{4qzxwvtnp^2} \\sin qzxwvtnp^2 \\right|_{vcbnksld}^{mnlkjprs} \\\\\n&\\mbox{} - \\int_{vcbnksld}^{mnlkjprs} \\frac{2qzxwvtnp\\cos qzxwvtnp - \\sin qzxwvtnp}{4qzxwvtnp^3} \\sin qzxwvtnp^2\\,dqzxwvtnp,\n\\end{align*}\nand that the final integral converges absolutely by comparison to $1/qzxwvtnp^3$.\n\nAn alternate approach is to first rewrite $\\sin qzxwvtnp \\sin qzxwvtnp^2$ as $\\frac{1}{2}(\\cos (qzxwvtnp^2-qzxwvtnp) - \\cos (qzxwvtnp^2+qzxwvtnp))$. Then\n\\begin{align*}\n\\int_{vcbnksld}^{mnlkjprs} \\cos(qzxwvtnp^2+qzxwvtnp)\\,dqzxwvtnp &=\n- \\left. \\frac{\\sin (qzxwvtnp^2+qzxwvtnp)}{2qzxwvtnp+1} \\right|_{vcbnksld}^{mnlkjprs} \\\\\n&\\mbox{} - \\int_{vcbnksld}^{mnlkjprs} \\frac{2\\sin(qzxwvtnp^2+qzxwvtnp)}{(2qzxwvtnp+1)^2}\\,dqzxwvtnp\n\\end{align*}\nconverges absolutely, and $\\int_{0}^{mnlkjprs} \\cos (qzxwvtnp^2-qzxwvtnp)$ can be treated similarly."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nQ(x,y)=x^{2}+2xy+2y^{2},\\qquad (x,y)\\in\\mathbb R^{2}.\n\\]  \nProve that the oscillatory improper double integral taken through the nested squares $[-R,R]^{2}$  \n\\[\n\\mathbf I\n=\\lim_{R\\to\\infty}\n\\iint_{[-R,R]^{2}}\n\\cos\\!\\bigl(\\pi\\,Q(x,y)\\bigr)\\,\n\\cos(x+y)\\,dx\\,dy\n\\]  \nexists and compute its exact value.\n\nHints.  \n(i)  Diagonalise the quadratic form $Q$ and use  \n\\[\n\\cos\\bigl(\\pi Q\\bigr)\\cos(x+y)\n=\\tfrac12\\sum_{\\pm}\\cos\\bigl(\\pi Q\\pm(x+y)\\bigr).\n\\]  \n(ii)  Evaluate  \n\\[\n\\iint_{\\mathbb R^{2}}\n\\exp\\!\\bigl(i\\,[\\pi Q(x,y)\\pm(x+y)]\\bigr)\\,dx\\,dy\n\\]  \nwith the two–dimensional Fresnel formula.  \n(iii)  To justify replacing the bounded squares by the whole plane,\nintroduce a smooth cut–off that localises to the exterior of the square\nand integrate {\\em twice} by parts with the standard non-stationary-phase\noperator.",
      "solution": "Throughout write  \n\\[\nX=\\begin{pmatrix}x\\\\y\\end{pmatrix},\n\\qquad \nA=\\begin{pmatrix}1&1\\\\ 1&2\\end{pmatrix},\n\\qquad\nQ(x,y)=X^{\\!\\top}AX .\n\\]\nThe symmetric matrix $A$ is positive definite,  \n\\[\n\\det A=1,\n\\qquad\nA^{-1}=\\begin{pmatrix}2&-1\\\\ -1&1\\end{pmatrix}.\n\\]\n\n--------------------------------------------------------------------\n1.\\;Half–angle decomposition  \n\n\\[\n\\cos\\!\\bigl(\\pi Q(x,y)\\bigr)\\cos(x+y)\n=\\tfrac12\\sum_{\\sigma=\\pm1}\n\\cos\\!\\bigl(\\pi Q(x,y)+\\sigma(x+y)\\bigr).\n\\]\nPut  \n\\[\nI_{\\sigma}(R)=\\iint_{[-R,R]^{2}}\n\\cos\\!\\bigl(\\pi Q(x,y)+\\sigma(x+y)\\bigr)\\,dx\\,dy,\n\\qquad\n\\mathbf I=\\tfrac12\\lim_{R\\to\\infty}\\bigl(I_{+}(R)+I_{-}(R)\\bigr).\n\\]\nIt suffices to treat $\\sigma=+1$.  \nDefine the phase  \n\\[\n\\Phi(x,y)=\\pi Q(x,y)+(x+y),\n\\qquad\nJ(R)=\\iint_{[-R,R]^{2}}e^{\\,i\\Phi(x,y)}\\,dx\\,dy,\n\\qquad\nI_{+}(R)=\\Re J(R).\n\\]\n\n--------------------------------------------------------------------\n2.\\;Whole–plane Fresnel evaluation  \n\nCompleting the square gives  \n\\[\n\\Phi(x,y)=\\pi\\bigl(X^{\\!\\top}AX+2B^{\\!\\top}X\\bigr),\n\\qquad\nB=\\frac1{2\\pi}\\begin{pmatrix}1\\\\1\\end{pmatrix}.\n\\]\nThe two–dimensional Fresnel identity for a nonsingular symmetric matrix\n$A$ states  \n\\[\n\\int_{\\mathbb R^{2}}\ne^{\\,i\\pi\\,(X^{\\!\\top}AX+2B^{\\!\\top}X)}\\,dX\n=\ne^{\\,i\\pi\\,\\operatorname{sgn}(A)/4}\\,\n|\\det A|^{-1/2}\\,\ne^{-\\,i\\pi\\,B^{\\!\\top}A^{-1}B}.\n\\tag{2.1}\n\\]\nBecause $A$ is positive definite, $\\operatorname{sgn}(A)=2$; hence  \n\\[\nJ_{\\infty}:=\\iint_{\\mathbb R^{2}}e^{\\,i\\Phi}\\,dx\\,dy\n=i\\,e^{-\\,i/(4\\pi)},\\qquad\n\\Re J_{\\infty}=\\sin\\!\\Bigl(\\tfrac1{4\\pi}\\Bigr).\n\\tag{2.2}\n\\]\nReplacing $B$ by $-B$ produces an identical real part, so\n\\[\n\\Re\\iint_{\\mathbb R^{2}}\ne^{\\,i[\\pi Q(x,y)\\pm(x+y)]}\\,dx\\,dy\n=\\sin\\!\\Bigl(\\tfrac1{4\\pi}\\Bigr).\n\\]\n\nIt remains to justify  \n\\[\n\\lim_{R\\to\\infty}J(R)=J_{\\infty}.\n\\tag{2.3}\n\\]\n\n--------------------------------------------------------------------\n3.\\;A quantitative non–stationary-phase lemma  \n\nSet  \n\\[\n\\nabla\\Phi(x,y)=2\\pi AX+\\begin{pmatrix}1\\\\1\\end{pmatrix}.\n\\]\nSince $A$ is positive definite, there exist constants $c_{0},R_{0}>0$\nsuch that  \n\\[\n|\\nabla\\Phi(X)|\\ge c_{0}\\,\\lVert X\\rVert\n\\quad\\text{for all }\\lVert X\\rVert\\ge R_{0}.\n\\tag{3.1}\n\\]\nIntroduce the first–order differential operator  \n\\[\nLf\n=\\frac{1}{i}\\,\n\\frac{\\nabla\\Phi}{|\\nabla\\Phi|^{2}}\\!\\cdot\\!\\nabla f,\n\\qquad\\text{so that}\\qquad\nL\\bigl(e^{\\,i\\Phi}\\bigr)=e^{\\,i\\Phi}.\n\\tag{3.2}\n\\]\nIts formal adjoint is  \n\\[\nL^{\\!*}g\n=-\\frac{1}{i}\\,\\nabla\\!\\cdot\\!\\Bigl(\ng\\,\\frac{\\nabla\\Phi}{|\\nabla\\Phi|^{2}}\\Bigr).\n\\tag{3.3}\n\\]\n\nLemma 3.1 (two–dimensional non-stationary phase).  \nThere exists $C>0$ such that  \n\\[\n\\Bigl|\\iint_{\\lVert X\\rVert\\ge R}\\!e^{\\,i\\Phi(X)}\\,dX\\Bigr|\n\\le \\frac{C}{R},\n\\qquad R\\ge R_{0}.\n\\tag{3.4}\n\\]\n\nProof.  \nLet $\\chi\\in C^{\\infty}_{c}(\\mathbb R)$ satisfy $0\\le\\chi\\le1$,\n$\\chi(\\rho)=0$ for $\\rho\\le1$ and $\\chi(\\rho)=1$ for $\\rho\\ge2$,\nand put $\\chi_{R}(X)=\\chi(\\lVert X\\rVert/R)$.  \nThen $\\chi_{R}=1$ on $\\{\\,\\lVert X\\rVert\\ge2R\\,\\}$ and\n$\\operatorname{supp}\\nabla\\chi_{R}\\subset\\{\\,R\\le\\lVert X\\rVert\\le2R\\,\\}$.\nDefine  \n\\[\nI_{R}:=\\iint_{\\mathbb R^{2}}\\!e^{\\,i\\Phi(X)}\\,\\chi_{R}(X)\\,dX.\n\\]\nBecause  \n\\[\nI_{R}=\\iint e^{\\,i\\Phi}\\,L^{\\!*}\\!\\bigl(\\chi_{R}\\bigr)\\,dX\n\\quad\\text{by \\eqref{3.2} and integration by parts},\n\\]\nwe obtain  \n\\[\n|I_{R}|\n\\le\\iint\\bigl|L^{\\!*}(\\chi_{R})\\bigr|\\,dX.\n\\tag{3.5}\n\\]\nWrite $V(X)=\\nabla\\Phi/|\\nabla\\Phi|^{2}$; then  \n\\[\nL^{\\!*}(\\chi_{R})\n=-\\frac{1}{i}\\Bigl(V\\!\\cdot\\!\\nabla\\chi_{R}\n+(\\nabla\\!\\cdot\\!V)\\,\\chi_{R}\\Bigr).\n\\]\nBy \\eqref{3.1} we have $|V(X)|\\le C_{1}\\lVert X\\rVert^{-1}$ and\n$|\\nabla\\!\\cdot\\!V(X)|\\le C_{1}\\lVert X\\rVert^{-2}$ for\n$\\lVert X\\rVert\\ge R_{0}$, whereas\n$|\\nabla\\chi_{R}|\\le C_{2}/R$ and\n$\\operatorname{supp}\\nabla\\chi_{R}\\subset\\{R\\le\\lVert X\\rVert\\le2R\\}$.\nConsequently  \n\\[\n|L^{\\!*}(\\chi_{R})|\n\\le\\frac{C}{R\\,\\lVert X\\rVert}\n+\\frac{C}{\\lVert X\\rVert^{2}}\n\\le\\frac{C}{R^{2}}\n\\quad\\text{on }\\operatorname{supp}(\\chi_{R}).\n\\]\nBecause this support has area $\\mathcal O(R^{2})$, inequality\n\\eqref{3.5} yields $|I_{R}|\\le C$, which is {\\em not} yet sufficient.\n\nApply $L^{\\!*}$ a second time:\n\\[\nI_{R}\n=\\iint e^{\\,i\\Phi}\\,L^{\\!*}\\!\\bigl(\\chi_{R}\\bigr)\n=\\iint e^{\\,i\\Phi}\\,L^{\\!*}\\!\\bigl(L^{\\!*}\\chi_{R}\\bigr)\\,dX.\n\\]\nNow \\eqref{3.1} implies $|L^{\\!*}L^{\\!*}(\\chi_{R})|\\le C/R^{3}$ on\n$\\operatorname{supp}(\\chi_{R})$, hence\n\\[\n|I_{R}|\\le C\\frac{1}{R^{3}}\\cdot R^{2}\\le\\frac{C}{R},\n\\]\nproving \\eqref{3.4}. ∎\n\n--------------------------------------------------------------------\n4.\\;Passage from squares to the whole plane  \n\nObserve that  \n\\[\n[-R,R]^{2}\\subset\\{\\,\\lVert X\\rVert\\le R\\sqrt2\\,\\},\n\\quad\n\\{\\,\\lVert X\\rVert\\le R\\,\\}\\subset[-R,R]^{2},\n\\]\nso  \n\\[\n\\iint_{\\mathbb R^{2}\\setminus[-R,R]^{2}}\\!\ne^{\\,i\\Phi}\\,dX\n=\\mathcal O\\!\\bigl(R^{-1}\\bigr)\n\\quad\\text{by \\eqref{3.4}.}\n\\tag{4.1}\n\\]\nTherefore  \n\\[\n\\lim_{R\\to\\infty}J(R)\n=\\lim_{R\\to\\infty}\\Bigl(\n\\iint_{\\mathbb R^{2}}e^{\\,i\\Phi}\n-\\iint_{\\mathbb R^{2}\\setminus[-R,R]^{2}}e^{\\,i\\Phi}\\Bigr)\n=J_{\\infty},\n\\]\nestablishing \\eqref{2.3}.\n\n--------------------------------------------------------------------\n5.\\;Completion of the proof  \n\nTaking real parts and invoking \\eqref{2.2}–\\eqref{2.3},\n\\[\n\\lim_{R\\to\\infty}I_{+}(R)=\\Re J_{\\infty}\n=\\sin\\!\\Bigl(\\tfrac1{4\\pi}\\Bigr).\n\\]\nExactly the same computation with $\\sigma=-1$ gives the identical\nlimit.  Consequently  \n\\[\n\\mathbf I\n=\\tfrac12\\Bigl(\\sin\\!\\bigl(\\tfrac1{4\\pi}\\bigr)\n+\\sin\\!\\bigl(\\tfrac1{4\\pi}\\bigr)\\Bigr)\n=\\sin\\!\\Bigl(\\frac1{4\\pi}\\Bigr).\n\\]\n\n--------------------------------------------------------------------\nAnswer  \n\n\\[\n\\boxed{\\displaystyle\n\\mathbf I=\\sin\\!\\Bigl(\\frac1{4\\pi}\\Bigr)}\n\\]\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.769272",
        "was_fixed": false,
        "difficulty_analysis": "• Dimensional upgrade: the problem now involves a two–dimensional improper integral instead of a single integral, immediately increasing the technical load (multiple limits, Fubini issues, decay estimates in $\\mathbb R^{2}$).  \n• Coupled phases: the quadratic form $x^{2}+2xy+2y^{2}$ contains cross–terms; it cannot be separated into a product of one–dimensional factors and requires linear–algebraic diagonalisation.  \n• Mixed oscillations: the linear phase $\\cos(x+y)$ interacts with the quadratic Fresnel–type phase, forcing the use of complex Gaussian integrals and the general multi-dimensional Fresnel formula (or stationary-phase arguments) rather than elementary one-dimensional tricks.  \n• Conditional convergence: unlike the original kernel problem, absolute convergence is false.  One must establish decay via successive integrations by parts and manage the two interacting variables to justify exchanging limits and integrals.  \n• Exact evaluation: the problem does not only ask for convergence but for a closed-form value, which requires completing the square in several variables, computing determinants and inverses of matrices, and manipulating complex exponentials—techniques well beyond the original setting.\n\nAll these additions demand a broader toolkit—linear algebra of quadratic forms, multi-variable oscillatory integrals, and rigorous justification of conditional convergence—making the enhanced variant substantially more sophisticated than either the original or the first kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nQ(x,y)=x^{2}+2xy+2y^{2},\\qquad (x,y)\\in\\mathbb R^{2}.\n\\]  \nProve that the oscillatory improper double integral taken through the nested squares $[-R,R]^{2}$  \n\\[\n\\mathbf I\n=\\lim_{R\\to\\infty}\n\\iint_{[-R,R]^{2}}\n\\cos\\!\\bigl(\\pi\\,Q(x,y)\\bigr)\\,\n\\cos(x+y)\\,dx\\,dy\n\\]  \nexists and compute its exact value.\n\nHints.  \n(i)  Diagonalise the quadratic form $Q$ and use  \n\\[\n\\cos\\bigl(\\pi Q\\bigr)\\cos(x+y)\n=\\tfrac12\\sum_{\\pm}\\cos\\bigl(\\pi Q\\pm(x+y)\\bigr).\n\\]  \n(ii)  Evaluate  \n\\[\n\\iint_{\\mathbb R^{2}}\n\\exp\\!\\bigl(i\\,[\\pi Q(x,y)\\pm(x+y)]\\bigr)\\,dx\\,dy\n\\]  \nwith the two–dimensional Fresnel formula.  \n(iii)  To justify replacing the bounded squares by the whole plane,\nintroduce a smooth cut–off that localises to the exterior of the square\nand integrate {\\em twice} by parts with the standard non-stationary-phase\noperator.",
      "solution": "Throughout write  \n\\[\nX=\\begin{pmatrix}x\\\\y\\end{pmatrix},\n\\qquad \nA=\\begin{pmatrix}1&1\\\\ 1&2\\end{pmatrix},\n\\qquad\nQ(x,y)=X^{\\!\\top}AX .\n\\]\nThe symmetric matrix $A$ is positive definite,  \n\\[\n\\det A=1,\n\\qquad\nA^{-1}=\\begin{pmatrix}2&-1\\\\ -1&1\\end{pmatrix}.\n\\]\n\n--------------------------------------------------------------------\n1.\\;Half–angle decomposition  \n\n\\[\n\\cos\\!\\bigl(\\pi Q(x,y)\\bigr)\\cos(x+y)\n=\\tfrac12\\sum_{\\sigma=\\pm1}\n\\cos\\!\\bigl(\\pi Q(x,y)+\\sigma(x+y)\\bigr).\n\\]\nPut  \n\\[\nI_{\\sigma}(R)=\\iint_{[-R,R]^{2}}\n\\cos\\!\\bigl(\\pi Q(x,y)+\\sigma(x+y)\\bigr)\\,dx\\,dy,\n\\qquad\n\\mathbf I=\\tfrac12\\lim_{R\\to\\infty}\\bigl(I_{+}(R)+I_{-}(R)\\bigr).\n\\]\nIt suffices to treat $\\sigma=+1$.  \nDefine the phase  \n\\[\n\\Phi(x,y)=\\pi Q(x,y)+(x+y),\n\\qquad\nJ(R)=\\iint_{[-R,R]^{2}}e^{\\,i\\Phi(x,y)}\\,dx\\,dy,\n\\qquad\nI_{+}(R)=\\Re J(R).\n\\]\n\n--------------------------------------------------------------------\n2.\\;Whole–plane Fresnel evaluation  \n\nCompleting the square gives  \n\\[\n\\Phi(x,y)=\\pi\\bigl(X^{\\!\\top}AX+2B^{\\!\\top}X\\bigr),\n\\qquad\nB=\\frac1{2\\pi}\\begin{pmatrix}1\\\\1\\end{pmatrix}.\n\\]\nThe two–dimensional Fresnel identity for a nonsingular symmetric matrix\n$A$ states  \n\\[\n\\int_{\\mathbb R^{2}}\ne^{\\,i\\pi\\,(X^{\\!\\top}AX+2B^{\\!\\top}X)}\\,dX\n=\ne^{\\,i\\pi\\,\\operatorname{sgn}(A)/4}\\,\n|\\det A|^{-1/2}\\,\ne^{-\\,i\\pi\\,B^{\\!\\top}A^{-1}B}.\n\\tag{2.1}\n\\]\nBecause $A$ is positive definite, $\\operatorname{sgn}(A)=2$; hence  \n\\[\nJ_{\\infty}:=\\iint_{\\mathbb R^{2}}e^{\\,i\\Phi}\\,dx\\,dy\n=i\\,e^{-\\,i/(4\\pi)},\\qquad\n\\Re J_{\\infty}=\\sin\\!\\Bigl(\\tfrac1{4\\pi}\\Bigr).\n\\tag{2.2}\n\\]\nReplacing $B$ by $-B$ produces an identical real part, so\n\\[\n\\Re\\iint_{\\mathbb R^{2}}\ne^{\\,i[\\pi Q(x,y)\\pm(x+y)]}\\,dx\\,dy\n=\\sin\\!\\Bigl(\\tfrac1{4\\pi}\\Bigr).\n\\]\n\nIt remains to justify  \n\\[\n\\lim_{R\\to\\infty}J(R)=J_{\\infty}.\n\\tag{2.3}\n\\]\n\n--------------------------------------------------------------------\n3.\\;A quantitative non–stationary-phase lemma  \n\nSet  \n\\[\n\\nabla\\Phi(x,y)=2\\pi AX+\\begin{pmatrix}1\\\\1\\end{pmatrix}.\n\\]\nSince $A$ is positive definite, there exist constants $c_{0},R_{0}>0$\nsuch that  \n\\[\n|\\nabla\\Phi(X)|\\ge c_{0}\\,\\lVert X\\rVert\n\\quad\\text{for all }\\lVert X\\rVert\\ge R_{0}.\n\\tag{3.1}\n\\]\nIntroduce the first–order differential operator  \n\\[\nLf\n=\\frac{1}{i}\\,\n\\frac{\\nabla\\Phi}{|\\nabla\\Phi|^{2}}\\!\\cdot\\!\\nabla f,\n\\qquad\\text{so that}\\qquad\nL\\bigl(e^{\\,i\\Phi}\\bigr)=e^{\\,i\\Phi}.\n\\tag{3.2}\n\\]\nIts formal adjoint is  \n\\[\nL^{\\!*}g\n=-\\frac{1}{i}\\,\\nabla\\!\\cdot\\!\\Bigl(\ng\\,\\frac{\\nabla\\Phi}{|\\nabla\\Phi|^{2}}\\Bigr).\n\\tag{3.3}\n\\]\n\nLemma 3.1 (two–dimensional non-stationary phase).  \nThere exists $C>0$ such that  \n\\[\n\\Bigl|\\iint_{\\lVert X\\rVert\\ge R}\\!e^{\\,i\\Phi(X)}\\,dX\\Bigr|\n\\le \\frac{C}{R},\n\\qquad R\\ge R_{0}.\n\\tag{3.4}\n\\]\n\nProof.  \nLet $\\chi\\in C^{\\infty}_{c}(\\mathbb R)$ satisfy $0\\le\\chi\\le1$,\n$\\chi(\\rho)=0$ for $\\rho\\le1$ and $\\chi(\\rho)=1$ for $\\rho\\ge2$,\nand put $\\chi_{R}(X)=\\chi(\\lVert X\\rVert/R)$.  \nThen $\\chi_{R}=1$ on $\\{\\,\\lVert X\\rVert\\ge2R\\,\\}$ and\n$\\operatorname{supp}\\nabla\\chi_{R}\\subset\\{\\,R\\le\\lVert X\\rVert\\le2R\\,\\}$.\nDefine  \n\\[\nI_{R}:=\\iint_{\\mathbb R^{2}}\\!e^{\\,i\\Phi(X)}\\,\\chi_{R}(X)\\,dX.\n\\]\nBecause  \n\\[\nI_{R}=\\iint e^{\\,i\\Phi}\\,L^{\\!*}\\!\\bigl(\\chi_{R}\\bigr)\\,dX\n\\quad\\text{by \\eqref{3.2} and integration by parts},\n\\]\nwe obtain  \n\\[\n|I_{R}|\n\\le\\iint\\bigl|L^{\\!*}(\\chi_{R})\\bigr|\\,dX.\n\\tag{3.5}\n\\]\nWrite $V(X)=\\nabla\\Phi/|\\nabla\\Phi|^{2}$; then  \n\\[\nL^{\\!*}(\\chi_{R})\n=-\\frac{1}{i}\\Bigl(V\\!\\cdot\\!\\nabla\\chi_{R}\n+(\\nabla\\!\\cdot\\!V)\\,\\chi_{R}\\Bigr).\n\\]\nBy \\eqref{3.1} we have $|V(X)|\\le C_{1}\\lVert X\\rVert^{-1}$ and\n$|\\nabla\\!\\cdot\\!V(X)|\\le C_{1}\\lVert X\\rVert^{-2}$ for\n$\\lVert X\\rVert\\ge R_{0}$, whereas\n$|\\nabla\\chi_{R}|\\le C_{2}/R$ and\n$\\operatorname{supp}\\nabla\\chi_{R}\\subset\\{R\\le\\lVert X\\rVert\\le2R\\}$.\nConsequently  \n\\[\n|L^{\\!*}(\\chi_{R})|\n\\le\\frac{C}{R\\,\\lVert X\\rVert}\n+\\frac{C}{\\lVert X\\rVert^{2}}\n\\le\\frac{C}{R^{2}}\n\\quad\\text{on }\\operatorname{supp}(\\chi_{R}).\n\\]\nBecause this support has area $\\mathcal O(R^{2})$, inequality\n\\eqref{3.5} yields $|I_{R}|\\le C$, which is {\\em not} yet sufficient.\n\nApply $L^{\\!*}$ a second time:\n\\[\nI_{R}\n=\\iint e^{\\,i\\Phi}\\,L^{\\!*}\\!\\bigl(\\chi_{R}\\bigr)\n=\\iint e^{\\,i\\Phi}\\,L^{\\!*}\\!\\bigl(L^{\\!*}\\chi_{R}\\bigr)\\,dX.\n\\]\nNow \\eqref{3.1} implies $|L^{\\!*}L^{\\!*}(\\chi_{R})|\\le C/R^{3}$ on\n$\\operatorname{supp}(\\chi_{R})$, hence\n\\[\n|I_{R}|\\le C\\frac{1}{R^{3}}\\cdot R^{2}\\le\\frac{C}{R},\n\\]\nproving \\eqref{3.4}. ∎\n\n--------------------------------------------------------------------\n4.\\;Passage from squares to the whole plane  \n\nObserve that  \n\\[\n[-R,R]^{2}\\subset\\{\\,\\lVert X\\rVert\\le R\\sqrt2\\,\\},\n\\quad\n\\{\\,\\lVert X\\rVert\\le R\\,\\}\\subset[-R,R]^{2},\n\\]\nso  \n\\[\n\\iint_{\\mathbb R^{2}\\setminus[-R,R]^{2}}\\!\ne^{\\,i\\Phi}\\,dX\n=\\mathcal O\\!\\bigl(R^{-1}\\bigr)\n\\quad\\text{by \\eqref{3.4}.}\n\\tag{4.1}\n\\]\nTherefore  \n\\[\n\\lim_{R\\to\\infty}J(R)\n=\\lim_{R\\to\\infty}\\Bigl(\n\\iint_{\\mathbb R^{2}}e^{\\,i\\Phi}\n-\\iint_{\\mathbb R^{2}\\setminus[-R,R]^{2}}e^{\\,i\\Phi}\\Bigr)\n=J_{\\infty},\n\\]\nestablishing \\eqref{2.3}.\n\n--------------------------------------------------------------------\n5.\\;Completion of the proof  \n\nTaking real parts and invoking \\eqref{2.2}–\\eqref{2.3},\n\\[\n\\lim_{R\\to\\infty}I_{+}(R)=\\Re J_{\\infty}\n=\\sin\\!\\Bigl(\\tfrac1{4\\pi}\\Bigr).\n\\]\nExactly the same computation with $\\sigma=-1$ gives the identical\nlimit.  Consequently  \n\\[\n\\mathbf I\n=\\tfrac12\\Bigl(\\sin\\!\\bigl(\\tfrac1{4\\pi}\\bigr)\n+\\sin\\!\\bigl(\\tfrac1{4\\pi}\\bigr)\\Bigr)\n=\\sin\\!\\Bigl(\\frac1{4\\pi}\\Bigr).\n\\]\n\n--------------------------------------------------------------------\nAnswer  \n\n\\[\n\\boxed{\\displaystyle\n\\mathbf I=\\sin\\!\\Bigl(\\frac1{4\\pi}\\Bigr)}\n\\]\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.589360",
        "was_fixed": false,
        "difficulty_analysis": "• Dimensional upgrade: the problem now involves a two–dimensional improper integral instead of a single integral, immediately increasing the technical load (multiple limits, Fubini issues, decay estimates in $\\mathbb R^{2}$).  \n• Coupled phases: the quadratic form $x^{2}+2xy+2y^{2}$ contains cross–terms; it cannot be separated into a product of one–dimensional factors and requires linear–algebraic diagonalisation.  \n• Mixed oscillations: the linear phase $\\cos(x+y)$ interacts with the quadratic Fresnel–type phase, forcing the use of complex Gaussian integrals and the general multi-dimensional Fresnel formula (or stationary-phase arguments) rather than elementary one-dimensional tricks.  \n• Conditional convergence: unlike the original kernel problem, absolute convergence is false.  One must establish decay via successive integrations by parts and manage the two interacting variables to justify exchanging limits and integrals.  \n• Exact evaluation: the problem does not only ask for convergence but for a closed-form value, which requires completing the square in several variables, computing determinants and inverses of matrices, and manipulating complex exponentials—techniques well beyond the original setting.\n\nAll these additions demand a broader toolkit—linear algebra of quadratic forms, multi-variable oscillatory integrals, and rigorous justification of conditional convergence—making the enhanced variant substantially more sophisticated than either the original or the first kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}