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FontSize->18, CellTags->"Section 8.6"], Cell[TextData[{ "\tIf f is continuous on the interval ", Cell[BoxData[ \(b < x \[LessEqual] c\)]], ", then the improper integral of f over (b,c] is defined by \n\n\t\t", Cell[BoxData[ \(\[Integral]\_b\%c\ f \((x)\) \[DifferentialD]x\ = \ lim\+\(r\ \[Rule] \ \(b\^+\)\)\ \[Integral]\_r\%c\ f \((x)\) \[DifferentialD]x\)]], "\n\nprovided that the limit exists. Similarly, if f is continuous on \ the interval ", Cell[BoxData[ \(a \[LessEqual] x < b\)]], ", then the improper integral of f over [a,b) is defined by \n\n\t\t", Cell[BoxData[ \(\[Integral]\_a\%b\ f \((x)\) \[DifferentialD]x\ = \ lim\+\(r\ \[Rule] \ \(b\^-\)\)\ \[Integral]\_a\%r\ f \((x)\) \[DifferentialD]x\)]], "\n\nprovided that the limit exists. \n\n\tLet f be continuous for all \ values of x in the interval [a,c], except at the value x=b, where a < \ b < c. The ", StyleBox["Cauchy principal value", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " of f over [a,c] is defined by\n\t\n\t\t", Cell[BoxData[ \(\(\(P . V . \(\[Integral]\_a\%c\ f \((x)\) \[DifferentialD]x\)\)\(\ \)\(=\)\(\ \)\)\)]], Cell[BoxData[ \(lim\+\(r\ \[Rule] \ \(0\^+\)\)\ \([\[Integral]\_a\%\(b - r\)\ f \((x)\) \[DifferentialD]x + \[Integral]\_\(b + r\)\%c\ f \((x)\) \[DifferentialD]x\ ]\)\)]], "\n\t\nprovided that the limit exists. " }], "Text"], Cell[TextData[{ "For illustration purposes of comparison, the examples are also solved \ using ", StyleBox["Mathematica", FontSlant->"Italic"], "'s table of integrals." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["\n", FontWeight->"Bold"], StyleBox["Example 8.19, Page 333.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that ", Cell[BoxData[ \(P . V . \ \(\[Integral]\_\(-1\)\%8\( 1\/x\^\(1/3\)\) \[DifferentialD]x\) = 9\/2\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 8.19.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ "We must guide ", StyleBox["Mathematica", FontSlant->"Italic"], " and write an appropriate form of the anti-derivative." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(f[x_] = 1\/x\^\(1/3\);\)\ \), "\n", \(\(F[x_] = \(3\/2\) \((x\^2)\)\^\(1/3\);\)\ \), "\n", \(\(Print["\< F[x] = \>", F[x]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", \(F'\)[x], "\< = \>", PowerExpand[\(F'\)[x]]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\< f[x] = \>", f[x]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nThe integrand is discontinuous at x = 0 where it has a singularity. \ Indeed, ", StyleBox["Mathematica", FontSlant->"Italic"], " has a branch cut for ", Cell[BoxData[ \(w\ = \ z\^\(1/3\)\)]], " along the negative x-axis and maps these points onto the ray ", Cell[BoxData[ \(Arg[w] = \[Pi]\/3\)]], ". Hence the function ", Cell[BoxData[ \(f[x] = 1\/x\^\(1/3\)\)]], " does not have the graph as taught in elementary calculus. We must use \ the following function h[x]." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(h[x_ /; \ x > 0] := 1\/x\^\(1/3\);\)\ \), "\[IndentingNewLine]", \(\(h[ x_ /; \ x < 0] := \(-1\)\/Abs[x]\^\(1/3\);\)\ \), "\[IndentingNewLine]", \(\(g1 = FilledPlot[h[x], {x, \(-1\), 0}, Fills \[Rule] Cyan, DisplayFunction \[Rule] Identity];\)\ \), "\[IndentingNewLine]", \(\(g2 = FilledPlot[h[x], {x, 0, 1}, Fills \[Rule] Cyan, DisplayFunction \[Rule] Identity];\)\ \), "\[IndentingNewLine]", \(\(g3 = FilledPlot[h[x], {x, 1, 8}, Fills \[Rule] Red, DisplayFunction \[Rule] Identity];\)\ \), "\[IndentingNewLine]", \(\(Show[g1, g2, g3, PlotRange \[Rule] {\(-5\), 5}, AxesLabel \[Rule] {"\", "\"}, \[IndentingNewLine]Ticks \ \[Rule] {Range[\(-1\), 8, 1], Range[\(-10\), 10, 2]}, DisplayFunction \[Rule] $DisplayFunction];\)\ \), \ "\[IndentingNewLine]", \(\(Print["\< s = \>", f[x]];\)\ \)}], "Input"], Cell["\<\ Now we pursue the Cauchy principal value of the integral.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%8\)\>\"", f[x], \*"\"\<\[DifferentialD]x = \!\(lim\+\(r \[Rule] 0\)\) ((\>\"", F[\(-r\)] - \ F[\(-1\)], "\<) + (\>", F[8] - \ F[r], "\<))\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%8\)\>\"", f[x], \*"\"\<\[DifferentialD]x = \!\(lim\+\(r \[Rule] 0\)\) (\>\"", F[\(-r\)] - \ F[\(-1\)] + F[8] - \ F[r], "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%8\)\>\"", f[x], "\<\[DifferentialD]x = \>", Limit[F[\(-r\)] - \ F[\(-1\)] + F[8] - \ F[r], r \[Rule] 0]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\n", StyleBox["Warning.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The following computations will not produce the principal value of the \ real integral. Indeed, ", StyleBox["Mathematica", FontSlant->"Italic"], " has a branch cut for ", Cell[BoxData[ \(w = \(g[z] = \(3\ z\^\(2/3\)\)\/2\)\)]], " along the negative x-axis and maps these points onto the ray ", Cell[BoxData[ \(Arg[w] = \(2\ \[Pi]\)\/3\)]], ", in particular ", Cell[BoxData[ \(g[\(-1\)] = \(-\(3\/4\)\) + \(3\ \[ImaginaryI]\ \@3\)\/4\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(g[ x_]\ = \[Integral]f[x] \[DifferentialD]x;\)\ \), "\n", \(\(val\ = \[Integral]\_\(-8\)\%1 f[x] \[DifferentialD]x;\)\ \), "\n", \(\(Print["\", f[x]];\)\ \), "\n", \(\(Print["\", f[x], "\< \[DifferentialD]x = \>", g[x]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", g[8]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \ g[\(-1\)], "\< = \>", ComplexExpand[g[\(-8\)]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\"];\)\ \), "\n\ ", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%8\)\>\"", f[x], "\< \[DifferentialD]x = g[8]-g[-1] = (\>", g[8]\ , "\<) - (\>", \ g[\(-1\)], "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%8\)\>\"", f[x], "\< \[DifferentialD]x = \>", g[8]\ - \ g[\(-1\)]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%8\)\>\"", f[x], "\< \[DifferentialD]x = \>", g[8]\ - ComplexExpand[g[\(-1\)]]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%8\)\>\"", f[x], "\< \[DifferentialD]x = \>", val];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Correction.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The following choice for the form for the indefinite integral will \ produce the desired computation." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(f[x_] = 1\/x\^\(1/3\);\)\ \), "\n", \(\(F[x_] = \(3\/2\) \((x\^2)\)\^\(1/3\);\)\ \), "\n", \(\(Print["\", f[x]];\)\ \), "\n", \(\(Print["\", f[x], "\< \[DifferentialD]x = \>", F[x]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", F[8], "\<, F[-1] = \>", \ F[\(-1\)]];\)\ \), "\n", \(\(Plot[F[x], {x, \(-1\), 8}, PlotRange \[Rule] {0, 6}, AxesLabel \[Rule] {"\", "\"}, \[IndentingNewLine]PlotStyle \ \[Rule] Magenta, Ticks \[Rule] {Range[\(-1\), 8, 1], Range[0, 6, 1]}];\)\ \), "\[IndentingNewLine]", \(\(Print["\< s = F[x] = \>", F[x]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%8\)\>\"", f[x], "\< \[DifferentialD]x = F[8]-F[-1] = (\>", F[8]\ , "\<) - (\>", \ F[\(-1\)], "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%8\)\>\"", f[x], "\< \[DifferentialD]x = \>", F[8]\ - \ F[\(-1\)]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Which is also the correct answer.\ \>", "Text"], Cell[TextData[{ "\n", StyleBox["Remark.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The PrincipalValue option does not give and answer to this problem." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(int = Integrate[F[x], {x, \(-\[Infinity]\), \[Infinity]}, PrincipalValue \[Rule] True];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\)\>\"", F[x], "\< \[DifferentialD]x = \>", int];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Open ]], Cell[TextData[{ StyleBox["\nTheorem 8.5, Page 333.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Let ", Cell[BoxData[ \(f \((z)\) = \(P \((z)\)\)\/\(Q \((z)\)\)\)]], " where P and Q are polynomials with real coefficients of degree m and n, \ respectively, where ", Cell[BoxData[ \(n \[GreaterEqual] m + 2\)]], ". If Q has simple zeros at the points ", Cell[BoxData[ \(t\_1, t\_2, \ ... , t\_r\)]], " on the x-axis, then the Cauchy Principal Value (P.V.) of the integral is \ \n\n\t", Cell[BoxData[ \(\(\(P . V . \(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\(\( P \((x)\)\)\/\(Q \((x)\)\)\) \[DifferentialD]x\)\)\(\ \ \)\(=\)\(\ \)\)\)]], Cell[BoxData[ \(2 \[Pi]\ \[ImaginaryI]\ \(\[Sum]\+\(j = 1\)\%k Res[P\/Q, z\_j]\) + \[Pi]\ \[ImaginaryI]\ \(\[Sum]\+\(j = 1\)\%r Res[ P\/Q, t\_j]\)\)]], ", \n\nwhere ", Cell[BoxData[ \(z\_1, z\_2, \ ... , z\_k\)]], " are the poles of f that lie in the upper half-plane. " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[StyleBox["Proof of Theorem 8.5, see text Page 333.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["\nTheorem 8.6, Page 333.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Let P and Q be polynomials, of degree m and n, respectively, where ", Cell[BoxData[ \(n \[GreaterEqual] m + 1\)]], " and let Q have simple zeros at the points ", Cell[BoxData[ \(t\_1, t\_2, \ ... , t\_r\)]], " on the x-axis. If ", Cell[BoxData[ \(\[Alpha]\)]], " is a positive real number and if ", Cell[BoxData[ \(f \((z)\) = \(\[ExponentialE]\^\(\[ImaginaryI]\ \[Alpha]\ z\)\ P \ \((z)\)\)\/\(Q \((z)\)\)\)]], ", then we can compute the Cauchy Principal Value (P.V.) of the following \ integrals \n\n\t", Cell[BoxData[ \(\(\(P . V . \(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\(\( P \((x)\)\)\/\(Q \((x)\)\)\) cos \((\[Alpha]\ x)\) \[DifferentialD]x\)\)\(\ \)\(=\)\(\ \ \)\)\)]], Cell[BoxData[ \(\(-2\) \[Pi]\ \(\[Sum]\+\(j = 1\)\%k Im \((Res[f, z\_j])\)\) - \[Pi]\ \(\[Sum]\+\(j = 1\)\%r Im \((Res[f, t\_j])\)\)\)]], ", \n\t\n\tand \n\t\n\t", Cell[BoxData[ \(\(\(P . V . \(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\(\( P \((x)\)\)\/\(Q \((x)\)\)\) sin \((\[Alpha]\ x)\) \[DifferentialD]x\)\)\(\ \)\(=\)\(\ \ \)\)\)]], Cell[BoxData[ \(2 \[Pi]\ \ \(\[Sum]\+\(j = 1\)\%k Re \((Res[f, z\_j])\)\) + \[Pi]\ \ \(\[Sum]\+\(j = 1\)\%r Re \((Res[f, t\_j])\)\)\)]], ", \n\nwhere ", Cell[BoxData[ \(z\_1, z\_2, \ ... , z\_k\)]], " are the poles of f that lie in the upper half plane. " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[StyleBox["Proof of Theorem 8.6, see text Page 333.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "\n", StyleBox["Remark.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The formulas in these theorems give the Cauchy principal value of the \ integral, which pays special attention to the manner in which any limits are \ taken. They are similar to those in ", ButtonBox["Section 8.4", ButtonData:>{"ca0804.nb", None}, ButtonStyle->"Hyperlink"], " and ", ButtonBox["Section 8.5", ButtonData:>{"ca0805.nb", None}, ButtonStyle->"Hyperlink"], ", except here we add one-half of the value of each residue at the points \ ", Cell[BoxData[ \(t\_1, t\_2, \ ... , t\_r\)]], " on the x-axis." }], "Text"], Cell[TextData[{ StyleBox["\nExample 8.20, Page 334.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Evaluate ", Cell[BoxData[ \("P.V.\!\(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\)\!\(x\/\(x\^3 - \ 8\)\)\[DifferentialD]x"\)]], " by using complex analysis. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 8.20.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ "Enter the function and ", Cell[BoxData[ \(f[z] = z\/\(z\^3\ - \ 8\)\)], AspectRatioFixed->True], " and ", "locate the", " singularities." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[x, z, Z];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, g, Res1, Res2, solset, val];\)\ \), "\n", \(\(f[z_]\ = \ z\/\(z\^3\ - \ 8\);\)\ \), "\[IndentingNewLine]", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[Denominator[f[z]] \[Equal] 0];\)\ \), "\[IndentingNewLine]", \(\(solset\ = \ Sort[Solve[Denominator[f[z]] \[Equal] 0, z]];\)\ \ \), "\[IndentingNewLine]", \(\(Print[solset];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\n", \(\(Z\_1\ = \ ReplaceAll[z, solset\_\(\(\[LeftDoubleBracket]\)\(1\)\(\[RightDoubleBracket]\)\)];\ \)\ \ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(z\_1\) = \>\"", Z\_1];\)\ \), "\n", \(\(Z\_2\ = \ ComplexExpand[ ReplaceAll[z, solset\_\(\(\[LeftDoubleBracket]\)\(2\)\(\[RightDoubleBracket]\)\)\ ]];\)\ \ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(z\_2\) = \>\"", Z\_2];\)\ \), "\n", \(\(Z\_3\ = \ ComplexExpand[ ReplaceAll[z, solset\_\(\(\[LeftDoubleBracket]\)\(3\)\(\[RightDoubleBracket]\)\)\ ]];\)\ \ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(z\_3\) = \>\"", Z\_3];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->False], Cell["\<\ Which poles lie in the upper half plane ? Which poles lie on the real axis ?\ \ \>", "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(dots = Graphics[{{Blue, PointSize[0.03], Point[{Re[Z\_1], Im[Z\_1]}]}, {Red, PointSize[0.03], Point[{Re[Z\_2], Im[Z\_2]}]}, {Red, PointSize[0.03], Point[{Re[Z\_3], Im[Z\_3]}]}}];\)\ \), "\[IndentingNewLine]", \(\(centers = Graphics[{Text[\*"\"\<\!\(z\_1\)\>\"", {Re[Z\_1], Im[Z\_1] + 0.1}, {\(-1\), 0}], Text[\*"\"\<\!\(z\_2\)\>\"", {Re[Z\_2], Im[Z\_2] - 0.1}, {\(-1\), 0}], Text[\*"\"\<\!\(z\_3\)\>\"", {Re[Z\_3], Im[Z\_3] - 0.1}, {\(-1\), 0}]}];\)\ \), "\n", \(\(Show[dots, centers, PlotRange \[Rule] {{\(-2.2\), 2.2}, {\(-2.2\), 2.2}}, Ticks \[Rule] {Range[\(-2\), 2, 1], Range[\(-2\), 2, 1]}, \[IndentingNewLine]AspectRatio \[Rule] 1, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(z\_1\) = \>\"", Z\_1, "\<, Is Im(\>", Z\_1, "\<) > 0 ? \>", N[Im[Z\_1]] > 0];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_2\) = \>\"", Z\_2, "\<, Is Im(\>", Z\_2, "\<) > 0 ? \>", N[Im[Z\_2]] > 0];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(z\_3\) = \>\"", Z\_3, "\<, Is Im(\>", Z\_3, "\<) > 0 ? \>", N[Im[Z\_3]] > 0];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(z\_1\) = \>\"", Z\_1, "\<, Is Im(\>", Z\_1, "\<) = 0 ? \>", N[Im[Z\_1]] \[Equal] 0];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_2\) = \>\"", Z\_2, "\<, Is Im(\>", Z\_2, "\<) = 0 ? \>", N[Im[Z\_2]] \[Equal] 0];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_3\) = \>\"", Z\_3, "\<, Is Im(\>", Z\_3, "\<) = 0 ? \>", N[Im[Z\_3]] \[Equal] 0];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nCompute the residues at ", Cell[BoxData[ \(z\_1 = \(2\ \ and\ \ z\_3 = \(-1\) + \[ImaginaryI]\ \@3\)\)]], ", and use the residue calculus to compute the value of the integral." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Res1\ = \ Residue[f[z], {z, Z\_1}];\)\ \), "\n", \(\(Res3\ = \ Residue[f[z], {z, Z\_3}];\)\ \), "\[IndentingNewLine]", \(\(val\ = \ 2\ \[Pi]\ \[ImaginaryI]\ \ Res3 + \[Pi]\ \[ImaginaryI]\ \ Res1;\)\ \ \), "\[IndentingNewLine]", \(\(Print["\", Z\_1, "\<] = \>", Res1];\)\ \), "\n", \(\(Print["\", Z\_3, "\<] = \>", Res3];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Z\_3, "\<] = \>", ComplexExpand[Res3]];\)\ \), "\[IndentingNewLine]", \(\(dots = Graphics[{{Blue, PointSize[0.03], Point[{Re[Z\_1], Im[Z\_1]}]}, {Red, PointSize[0.03], Point[{Re[Z\_3], Im[Z\_3]}]}}];\)\ \), "\[IndentingNewLine]", \(\(centers = Graphics[{Text[\*"\"\<\!\(z\_1\)\>\"", {Re[Z\_1], Im[Z\_1] - 0.2}, {\(-1\), 0}], Text[\*"\"\<\!\(z\_3\)\>\"", {Re[Z\_3], Im[Z\_3]}, {\(-1.2\), 0}]}];\)\ \), "\[IndentingNewLine]", \(\(contour\ = \ Graphics[{{VioletRed, Circle[{0, 0}, 3, {0, \[Pi]}]}, {VioletRed, Circle[{Re[Z\_1], Im[Z\_1]}, 0.2, {0, \[Pi]}]}}];\)\ \), "\[IndentingNewLine]", \(\(segment = Graphics[{Thickness[0.02], Magenta, Line[{{\(-3\), 0}, {3, 0}}]}];\)\ \), "\n", \(\(Show[centers, contour, segment, dots, PlotRange \[Rule] {{\(-3.1\), 3.1}, {\(-0.05\), 3.05}}, Ticks \[Rule] {Range[\(-1\), 1, 1], Range[0, 2, 1]}, AspectRatio \[Rule] 1\/2, \[IndentingNewLine]AspectRatio \[Rule] 1, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[ContourIntegral]\+C\) (\>\"", f[z], "\<) \[DifferentialD]z = 2\[Pi]\[ImaginaryI](\>", ComplexExpand[Res3]\ , "\<) + \[Pi]\[ImaginaryI](\>", \ Res1\ , "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[ContourIntegral]\+C\) (\>\"", f[z], "\<) \[DifferentialD]z = \[Pi]\[ImaginaryI](\>", 2\ ComplexExpand[Res3]\ + \ Res1\ , "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[ContourIntegral]\+C\) (\>\"", f[z], "\<) \[DifferentialD]z = \>", ComplexExpand[val]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\""];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\"];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"\ ", f[x], "\< \[DifferentialD]x = \>", ComplexExpand[val]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"\ ", f[x], "\< \[DifferentialD]x = \>", Chop[N[val]]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Open ]], Cell[TextData[{ StyleBox["\nExample 8.21, Page 334.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Evaluate ", Cell[BoxData[ \("P.V.\!\(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\)\!\(x\/\(x\^3 - \ 8\)\)\[DifferentialD]x"\)]], " by using a computer algebra system. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 8.21.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ "Enter the function and ", Cell[BoxData[ \(f[z] = z\/\(z\^3\ - \ 8\)\)], AspectRatioFixed->True], " and ", "locate the", " singularities." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[x, z, Z];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, g, Res1, Res2, solset, val];\)\ \), "\n", \(\(f[z_]\ = \ z\/\(z\^3\ - \ 8\);\)\ \), "\[IndentingNewLine]", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->False], Cell[TextData[{ "\n", StyleBox["Warning.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]], " The indefinite integral of f[x] is g[x], but it ", StyleBox["cannot", FontColor->RGBColor[1, 0, 1]], " be used in any meaningful way to solve the problem at hand." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(g[ x_]\ = \[Integral]f[ x] \[DifferentialD]x;\)\ \), "\[IndentingNewLine]", \(\(lim1 = \ Limit[g[x], x \[Rule] \[Infinity]];\)\ \), "\[IndentingNewLine]", \(\(lim2 = Limit[g[x], x \[Rule] \(-\[Infinity]\)];\)\ \), "\n", \(\(Print["\", f[x], "\<\[DifferentialD]x = \>", g[x]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\< g[\[Infinity]] = \!\(lim\+\(x \[Rule] \[Infinity]\)\) \ (\>\"", g[x], "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\< g[\[Infinity]] = \>", lim1];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", g[x], "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\", lim2];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\< The following answer is suspect\>"];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", lim1 - lim2];\)\ \), "\n", \(\(Print["\", ComplexExpand[lim1 - lim2]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", N[lim1 - lim2]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Warning.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]], " The following computation will ", StyleBox["not", FontColor->RGBColor[1, 0, 1]], " produce the correct answer." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(val\ = \[Integral]\_\(-\[Infinity]\)\%\[Infinity] f[x] \[DifferentialD]x;\)\ \), "\[IndentingNewLine]", \(\(Print["\< The following answer is suspect\>"];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\)\>\"", f[x], "\< \[DifferentialD]x = \>", val];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\)\>\"", f[x], "\< \[DifferentialD]x = \>", N[val]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\n", StyleBox["Correction.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The indefinite integral of g[x] can be written in an algebraic form that \ will produce the correct answer for real functions." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(g[ x_]\ = \[Integral]f[ x] \[DifferentialD]x;\)\ \), "\[IndentingNewLine]", \(\(Print["\", f[x], "\<\[DifferentialD]x = \>", g[x]];\)\ \), "\[IndentingNewLine]", \(\(g[x_] = ArcTan[\(1 + x\)\/\@3]\/\(2\ \@3\) + 1\/12\ Log[\((\(-2\) + x)\)\^2\/\(4 + 2\ x + x\^2\)];\)\ \), "\ \[IndentingNewLine]", \(\(lim1 = \ Limit[g[x], x \[Rule] \[Infinity]];\)\ \), "\[IndentingNewLine]", \(\(lim2 = Limit[g[x], x \[Rule] \(-\[Infinity]\)];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", f[x], "\<\[DifferentialD]x = \>", g[x]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\< g[\[Infinity]] = \!\(lim\+\(x \[Rule] \[Infinity]\)\) \ (\>\"", g[x], "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\< g[\[Infinity]] = \>", lim1];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", g[x], "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\", lim2];\)\ \), "\n", \(\(Plot[g[x], {x, \(-20\), 20}, PlotRange \[Rule] {\(-0.5\), 0.5}, AxesLabel \[Rule] {"\", "\"}, \[IndentingNewLine]PlotStyle \ \[Rule] Magenta, Ticks \[Rule] {Range[\(-20\), 20, 10], Range[\(-\[Pi]\)\/\(4\ \@3\), \[Pi]\/\(4\ \@3\), \[Pi]\/\(4\ \ \@3\)]}];\)\ \), "\[IndentingNewLine]", \(Print["\", g[x]]\), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"\ ", f[x], "\< \[DifferentialD]x = (\>", lim1, "\<) - 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Cell[StyleData["Subsubtitle", "Printout"], CellMargins->{{2, 10}, {8, 10}}, FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Section"], CellDingbat->"\[FilledSquare]", CellMargins->{{25, Inherited}, {8, 24}}, CellGroupingRules->{"SectionGrouping", 30}, PageBreakBelow->False, CounterIncrements->"Section", CounterAssignments->{{"Subsection", 0}, {"Subsubsection", 0}}, FontFamily->"Helvetica", FontSize->16, FontWeight->"Bold"], Cell[StyleData["Section", "Presentation"], CellMargins->{{40, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->24], Cell[StyleData["Section", "Condensed"], CellMargins->{{18, Inherited}, {6, 12}}, FontSize->12], Cell[StyleData["Section", "Printout"], CellMargins->{{13, 0}, {7, 22}}, FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 20}}, CellGroupingRules->{"SectionGrouping", 40}, PageBreakBelow->False, CounterIncrements->"Subsection", CounterAssignments->{{"Subsubsection", 0}}, FontSize->14, FontWeight->"Bold"], Cell[StyleData["Subsection", "Presentation"], CellMargins->{{36, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->22], Cell[StyleData["Subsection", "Condensed"], CellMargins->{{16, Inherited}, {6, 12}}, FontSize->12], Cell[StyleData["Subsection", "Printout"], CellMargins->{{9, 0}, {7, 22}}, FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsubsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 18}}, CellGroupingRules->{"SectionGrouping", 50}, PageBreakBelow->False, CounterIncrements->"Subsubsection", FontWeight->"Bold"], Cell[StyleData["Subsubsection", "Presentation"], CellMargins->{{34, 10}, {11, 26}}, LineSpacing->{1, 0}, FontSize->18], Cell[StyleData["Subsubsection", "Condensed"], CellMargins->{{17, Inherited}, {6, 12}}, FontSize->10], Cell[StyleData["Subsubsection", "Printout"], CellMargins->{{9, 0}, {7, 14}}, FontSize->11] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Body Text", "Section"], Cell[CellGroupData[{ Cell[StyleData["Text"], CellMargins->{{12, 10}, {7, 7}}, LineSpacing->{1, 3}, CounterIncrements->"Text"], Cell[StyleData["Text", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}], Cell[StyleData["Text", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}], Cell[StyleData["Text", "Printout"], CellMargins->{{2, 2}, {6, 6}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SmallText"], CellMargins->{{12, 10}, {6, 6}}, LineSpacing->{1, 3}, CounterIncrements->"SmallText", FontFamily->"Helvetica", FontSize->9], Cell[StyleData["SmallText", "Presentation"], CellMargins->{{24, 10}, {8, 8}}, LineSpacing->{1, 5}, FontSize->12], Cell[StyleData["SmallText", "Condensed"], CellMargins->{{8, 10}, {5, 5}}, LineSpacing->{1, 2}, FontSize->9], Cell[StyleData["SmallText", "Printout"], CellMargins->{{2, 2}, {5, 5}}, FontSize->7] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles 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If a cell's FormatType matches the name of one of the styles \ defined below, then that style is applied between the cell's style and its \ own options.\ \>", "Text"], Cell[StyleData["CellExpression"], PageWidth->Infinity, CellMargins->{{6, Inherited}, {Inherited, Inherited}}, ShowCellLabel->False, ShowSpecialCharacters->False, AllowInlineCells->False, AutoItalicWords->{}, StyleMenuListing->None, FontFamily->"Courier", Background->GrayLevel[1]], Cell[StyleData["InputForm"], AllowInlineCells->False, StyleMenuListing->None, FontFamily->"Courier"], Cell[StyleData["OutputForm"], PageWidth->Infinity, TextAlignment->Left, LineSpacing->{1, -5}, StyleMenuListing->None, FontFamily->"Courier"], Cell[StyleData["StandardForm"], LineSpacing->{1.25, 0}, StyleMenuListing->None, FontFamily->"Courier"], Cell[StyleData["TraditionalForm"], LineSpacing->{1.25, 0}, SingleLetterItalics->True, TraditionalFunctionNotation->True, DelimiterMatching->None, StyleMenuListing->None], Cell["\<\ The style defined below is mixed in to any cell that is in an \ inline cell within another.\ \>", "Text"], Cell[StyleData["InlineCell"], TextAlignment->Left, ScriptLevel->1, StyleMenuListing->None], Cell[StyleData["InlineCellEditing"], StyleMenuListing->None, Background->RGBColor[1, 0.749996, 0.8]] }, Closed]] }, Open ]] }] ] (******************************************************************* Cached data follows. 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