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To introduce \ the integral of a complex function, we start by defining what is meant by the \ integral of a complex-valued function of a ", StyleBox["real", FontSlant->"Italic"], " variable. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Definition 6.1 (definite integral of a complex integrand), Page \ 201.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Let ", Cell[BoxData[ \(f[t] = u[t] + \[ImaginaryI]\ v[t]\)]], " where u and v are real-valued functions of the real variable t for ", Cell[BoxData[ \(a \[LessEqual] t \[LessEqual] b\)]], " . Then\n\n\t", Cell[BoxData[ \(\[Integral]\_a\%b f[ t] \[DifferentialD]t\ = \ \[Integral]\_a\%b\((u[ t] + \[ImaginaryI]\ v[ t])\) \(\(\[DifferentialD]t\)\(\ \)\)\)]], Cell[BoxData[ \(\(\(=\)\(\ \)\(\[Integral]\_a\%b u[ t] \[DifferentialD]t + \[ImaginaryI]\ \(\[Integral]\_a\%b v[ t] \[DifferentialD]t\)\)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n\n\tWe generally evaluate integrals of this type by finding the \ antiderivatives of u and v and evaluating the definite integrals on the \ right side of the equation. That is, if ", Cell[BoxData[ \(\(U'\)[t]\ = \ u[t]\)]], " and ", Cell[BoxData[ \(\(V'\)[t]\ = \ v[t]\)]], ", then we write \n\n\t", Cell[BoxData[ \(\[Integral]\_a\%b f[ t] \[DifferentialD]t\ = \ \(\[Integral]\_a\%b u[ t] \[DifferentialD]t + \[ImaginaryI]\ \(\[Integral]\_a\%b v[ t] \[DifferentialD]t\) = \ U[b] - U[a]\ + \ \[ImaginaryI] \((V[b] - V[a])\)\)\)]] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 6.1, Page 202.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that ", Cell[BoxData[ \(\[Integral]\_0\%1\(\((t - \[ImaginaryI])\)\^3\) \[DifferentialD]t = \ \(-\(5\/4\)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.1.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ StyleBox["Method (i).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " Expand the integrand f[t] into its real and imaginary parts, then \ integrate u[t] and v[t]." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[t, u, v, w];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, Int, ImInt, ImVal, ReInt, ReVal, Val];\)\ \), "\[IndentingNewLine]", \(\(w = \((t\ - \ \[ImaginaryI])\)\^3;\)\ \), "\n", \(\(f[t_]\ = \ ComplexExpand[w];\)\ \), "\n", \(\(u[t_]\ = \ ComplexExpand[Re[f[t]]];\)\ \), "\n", \(\(v[t_]\ = \ ComplexExpand[Im[f[t]]];\)\ \), "\[IndentingNewLine]", \(\(U[ t_]\ = \ \[Integral]u[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(V[t_]\ = \ \[Integral]v[t] \[DifferentialD]t;\)\ \), "\n", \(\(Int\ = \ U[t]\ + \ \[ImaginaryI]\ V[t];\)\ \), "\[IndentingNewLine]", \(\(ReVal\ = \ \ \[Integral]\_0\%1 u[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(ImVal\ = \ \ \[Integral]\_0\%1 v[t] \[DifferentialD]t;\)\ \), "\n", \(\(Val\ = \ ReVal\ + \ \[ImaginaryI]\ ImVal;\)\ \), "\n", \(\(Print["\", f[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", u[t]];\)\ \), "\n", \(\(Print["\", v[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", U[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", V[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\<\[Integral]u[t]\[DifferentialD]t + \[ImaginaryI] \ \[Integral]v[t]\[DifferentialD]t = (\>", U[t], "\<) + \[ImaginaryI](\>", V[t], "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)\>\"", w, \*"\"\<\[DifferentialD]t = \!\(\[Integral]\_0\%1\)u[t]\ \[DifferentialD]t + \[ImaginaryI] \ \!\(\[Integral]\_0\%1\)v[t]\[DifferentialD]t\>\""];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)\>\"", w, "\<\[DifferentialD]t = (U[1]-U[0]) + \[ImaginaryI](V[1]-V[0])\>"];\ \)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)\>\"", w, "\<\[DifferentialD]t = (\>", ReVal, "\<) + \[ImaginaryI](\>", ImVal, "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)\>\"", w, "\<\[DifferentialD]t = \>", Val];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Method (ii).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " Compute the integral directly using a complex function for the \ integrand." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[t];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, F, val];\)\ \), "\[IndentingNewLine]", \(\(f[ t_]\ = \ \((t\ - \ \[ImaginaryI])\)\^3;\)\ \), "\ \[IndentingNewLine]", \(\(F[ t_]\ = \ \[Integral]f[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(val\ = \ \[Integral]\_0\%1 f[t] \[DifferentialD]t;\)\ \), "\n", \(\(Print["\", f[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", f[t], "\<\[DifferentialD]t = \>", F[t]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)\>\"", f[t], "\<\[DifferentialD]t = F[1] - F[0]\>"];\)\ \), "\ \[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)\>\"", f[t], "\<\[DifferentialD]t = (\>", F[1]\ , "\<) - (\>", F[0], "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)\>\"", f[t], "\<\[DifferentialD]t = \>", val];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 6.2, Page 202.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(\[ExponentialE]\^\(t + \[ImaginaryI]\ \ t\)\) \[DifferentialD]t = \(1\/2\) \((\[ExponentialE]\^\(\[Pi]/2\) - 1)\) + \(\(\(\[ImaginaryI]\)\(\ \)\)\/2\) \ \((\[ExponentialE]\^\(\[Pi]/2\) + 1)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.2.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ StyleBox["Method (i).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " Expand the integrand f[t] into its real and imaginary parts, then \ integrate u[t] and v[t]." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[t, u, v, w];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, Int, ImInt, ImVal, ReInt, ReVal, Val];\)\ \), "\[IndentingNewLine]", \(\(w = \[ExponentialE]\^\(t\ + \ \[ImaginaryI]\ t\);\)\ \), "\n", \(\(f[t_]\ = \ ComplexExpand[w];\)\ \), "\n", \(\(u[t_]\ = \ ComplexExpand[Re[f[t]]];\)\ \), "\n", \(\(v[t_]\ = \ ComplexExpand[Im[f[t]]];\)\ \), "\[IndentingNewLine]", \(\(U[ t_]\ = \ \[Integral]u[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(V[t_]\ = \ \[Integral]v[t] \[DifferentialD]t;\)\ \), "\n", \(\(Int\ = \ U[t]\ + \ \[ImaginaryI]\ V[t];\)\ \), "\[IndentingNewLine]", \(\(ReVal\ = \ \ \[Integral]\_0\%\(\[Pi]/2\)u[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(ImVal\ = \ \ \[Integral]\_0\%\(\[Pi]/2\)v[ t] \[DifferentialD]t;\)\ \), "\n", \(\(Val\ = \ ReVal\ + \ \[ImaginaryI]\ ImVal;\)\ \), "\n", \(\(Print["\", f[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", u[t]];\)\ \), "\n", \(\(Print["\", v[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", U[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", V[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\\"", U[\[Pi]\/2], "\<, \>", \ "\", U[0]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", V[\[Pi]\/2], "\<, \>", "\", V[0]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\<\[Integral]u[t]\[DifferentialD]t + \[ImaginaryI] \ \[Integral]v[t]\[DifferentialD]t = \>", Int];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%\(\[Pi]/2\)\)\>\"", w, \*"\"\<\[DifferentialD]t = \!\(\[Integral]\_0\%\(\[Pi]/2\)\)u[t]\ \[DifferentialD]t + \[ImaginaryI] \!\(\[Integral]\_0\%\(\[Pi]/2\)\)v[t]\ \[DifferentialD]t\>\""];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%\(\[Pi]/2\)\)\>\"", w, "\<\[DifferentialD]t = (U[1]-U[0]) + \[ImaginaryI](V[1]-V[0])\>"];\ \)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%\(\[Pi]/2\)\)\>\"", w, "\<\[DifferentialD]t = \>", ReVal, "\< + \[ImaginaryI](\>", ImVal, "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%\(\[Pi]/2\)\)\>\"", w, "\<\[DifferentialD]t = \>", Val];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n\tUsing Definition (6-1) it is easy to show that the integral of their \ sum is the sum of their integrals, that is \n\n\t\t", Cell[BoxData[ \(\[Integral]\_a\%b\((f[t] + g[t])\) \[DifferentialD]t = \[Integral]\_a\%b f[ t] \[DifferentialD]t + \[Integral]\_a\%b g[ t] \[DifferentialD]t\)]], ". " }], "Text"], Cell[TextData[{ "\tIf we divide the interval ", Cell[BoxData[ \(a \[LessEqual] t \[LessEqual] b\)]], " into ", Cell[BoxData[ \(a \[LessEqual] t \[LessEqual] c\)]], " and ", Cell[BoxData[ \(c \[LessEqual] t \[LessEqual] b\)]], " and integrate f(t) over these subintervals using definition (6-1) we get \ \n\n\t\t\t", Cell[BoxData[ \(\[Integral]\_a\%b f[ t] \[DifferentialD]t = \[Integral]\_a\%c f[ t] \[DifferentialD]t + \[Integral]\_c\%b f[ t] \[DifferentialD]t\)]], ". " }], "Text"], Cell[TextData[{ "\tSimilarly, if ", Cell[BoxData[ \(\[Alpha] + \[ImaginaryI]\ \[Beta]\)]], " denotes a complex constant, then\n\n\t\t\t", Cell[BoxData[ \(\[Integral]\_a\%b\((\[Alpha] + \[ImaginaryI]\ \[Beta])\) f[t] \[DifferentialD]t = \((\[Alpha] + \[ImaginaryI]\ \[Beta])\) \ \(\[Integral]\_a\%b f[t] \[DifferentialD]t\)\)]], ". " }], "Text"], Cell[TextData[{ "\tIf the limits of integration are reversed, then\n\n\t\t\t", Cell[BoxData[ \(\[Integral]\_a\%b f[ t] \[DifferentialD]t = \(-\(\[Integral]\_b\%a f[ t] \[DifferentialD]t\)\)\)]], ". " }], "Text"], Cell[TextData[{ "\tThe integral of the product fg becomes\n\n", Cell[BoxData[ \(\(\(\[Integral]\_a\%b f[t] g[t] \[DifferentialD]t\)\(=\)\(\ \)\)\)]], Cell[BoxData[ \(\[Integral]\_a\%b\((u[t] + \[ImaginaryI]\ v[t])\) \((p[ t] + \[ImaginaryI]\ q[t])\) \[DifferentialD]t\)]], " \n\n", Cell[BoxData[ \(\(\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)\(\(=\)\(\ \)\)\)\)]], Cell[BoxData[ \(\(\(\[Integral]\_a\%b\((u[t]\ p[t] - v[t]\ q[t])\) \[DifferentialD]t\)\(\ \)\(+\)\)\)]], " ", Cell[BoxData[ \(\[ImaginaryI]\ \(\[Integral]\_a\%b\((u[t]\ q[t] + v[t]\ p[t])\) \[DifferentialD]t\)\)]], ". " }], "Text"], Cell[TextData[{ "\n", StyleBox["Example 6.3, Page 203.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Let us verify property (6-5): \n\n", Cell[BoxData[ \(\(\(\ \)\(\[Integral]\((c + \[ImaginaryI]\ d)\) \((u[ t] + \[ImaginaryI]\ v[t])\) \[DifferentialD]t = c\ \(\[Integral]u[t] \[DifferentialD]t\) - d\ \(\[Integral]v[ t] \[DifferentialD]t\) + \[ImaginaryI]\ d\ \(\[Integral]u[ t] \[DifferentialD]t\) + \[ImaginaryI]\ c\ \(\[Integral]v[ t] \[DifferentialD]t\)\)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.3.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[c, d, int1, int2, t, u, v];\)\ \), "\n", \(\(int1\ = \[Integral]\((c\ + \ \[ImaginaryI]\ d)\) \((u[ t]\ + \ \[ImaginaryI]\ v[ t])\) \[DifferentialD]t;\)\ \), "\n", \(\(int2\ = \ MapAll[Distribute, int1];\)\ \), "\n", \(\(Print[int1, "\< = \>"];\)\ \), "\n", \(\(Print[int2];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n\tIt is worthwhile to point out the similarity between equation (6-2) \ and its counterpart in calculus. Suppose that U and V are differentiable on \ ", Cell[BoxData[ \(a < t < b\)]], " and ", Cell[BoxData[ \(F \((t)\) = U \((t)\) + \[ImaginaryI]\ V \((t)\)\)]], ". Since ", Cell[BoxData[ \(F' \((t)\) = \(U' \((t)\) + \[ImaginaryI]\ V' \((t)\) = \(u \((t)\) + \ \[ImaginaryI]\ v \((t)\) = f \((t)\)\)\)\)]], ", equation (6-2) takes on the familiar form \n\n\t\t\t", Cell[BoxData[ \(\[Integral]\_a\%b f[ t] \[DifferentialD]t = \(\(F \((t)\)\)\( | \_\(t = a\)\%\(t = b\)\)\)\)]], Cell[BoxData[ \(\(\(=\)\(F \((b)\) - F \((a)\)\)\)\)]], "." }], "Text"], Cell[TextData[{ "\n", StyleBox["Example 6.4, Page 204.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use the above equation (6-8) to show that ", Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(\[ExponentialE]\^\(t + \[ImaginaryI]\ \ t\)\) \[DifferentialD]t = \(1\/2\) \((\[ExponentialE]\^\(\[Pi]/2\) - 1)\) + \(\(\(\[ImaginaryI]\)\(\ \)\)\/2\) \ \((\[ExponentialE]\^\(\[Pi]/2\) + 1)\)\)]], ". 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