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Mathews Rus sell W. Howell\nmathews@fullerton.edu howell@westmont.edu\n\nCompl imentary software to accompany the textbook:" }}{PARA 257 "" 0 "" {TEXT 257 82 "COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9" }{TEXT 304 197 "\nJones and Bartlett Publi shers, Inc., 40 Tall Pine Drive, Sudbury, MA 01776\nTel e. (800) 832-0034; FAX: (508) 443-8000, E-mail: mkt@jbpu b.com, http://www.jbpub.com/" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 303 1 "\n" }{TEXT 256 26 "CHAPTER 8 RESIDUE THEORY" }{TEXT 296 2 "\n \n" }{TEXT 256 53 "Section 8.4 Improper Integrals of Rational Functio ns" }{TEXT 297 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 120 " An importan t application of the theory of residues is the evaluation of certain t ypes of improper integrals. Let " }{XPPEDIT 18 0 "f" "6#%\"fG" } {TEXT -1 49 " be a continuous function of the real variable " } {XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 18 " on the interval " } {XPPEDIT 18 0 "0 <= x;" "6#1\"\"!%\"xG" }{XPPEDIT 18 0 "` < `*infinity ;" "6#*&%$~<~G\"\"\"%)infinityGF%" }{TEXT -1 52 ". Recall from calcul us that the improper integral " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 8 " over " }{XPPEDIT 18 0 "`[0`*`,`*infinity*`)`;" "6#**%#[0G\"\" \"%\",GF%%)infinityGF%%\")GF%" }{TEXT -1 14 " is defined by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 " " } {XPPEDIT 18 0 "Int(f(x),x = 0 .. infinity) = Limit(int(f(x),x = 0 .. b ),b = infinity);" "6#/-%$IntG6$-%\"fG6#%\"xG/F*;\"\"!%)infinityG-%&Lim itG6$-%$intG6$-F(6#F*/F*;F-%\"bG/F9F." }{TEXT -1 37 ", provided that the limit exists. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 4 "If " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 27 " is d efined for all real " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 24 ", the n the integral of " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 7 " over \+ " }{XPPEDIT 18 0 "`(-`*infinity*`,`*infinity*`)`;" "6#*,%#(-G\"\"\"%)i nfinityGF%%\",GF%F&F%%\")GF%" }{TEXT -1 16 " is defined by " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 " " } {XPPEDIT 18 0 "Int(f(x),x = -infinity .. infinity);" "6#-%$IntG6$-%\"f G6#%\"xG/F);,$%)infinityG!\"\"F-" }{TEXT -1 5 " = " }{XPPEDIT 18 0 " Limit(int(f(x),x = a .. 0),a = -infinity);" "6#-%&LimitG6$-%$intG6$-% \"fG6#%\"xG/F,;%\"aG\"\"!/F/,$%)infinityG!\"\"" }{TEXT -1 5 " + " } {XPPEDIT 18 0 "(Limit(int(f(x),x = 0 .. b),b = infinity))" "6#-%&Limit G6$-%$intG6$-%\"fG6#%\"xG/F,;\"\"!%\"bG/F0%)infinityG" }{TEXT -1 37 " \+ , provided that the limit exists. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 307 93 "Load Maple's \"residu e\" procedure.\nMake sure this is done only ONCE during a Maple sessi on.\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "readlib(residue):" }{TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 256 23 "Example 8.13, Page 3 23." }{TEXT 298 13 " Evaluate " }{XPPEDIT 18 0 "limit(int(x, x=-R.. R), R=infinity)" "6#-%&limitG6$-%$intG6$%\"xG/F);,$%\"RG!\"\"F-/F-%)in finityG" }{TEXT 272 3 " . " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "f:=' f': F:='F': x:='x':\nf := x -> x:\ng := int(f(x),x):\n`F(x) ` = f(x); \+ \n`G(x) = `, int(F(x),x) = int(f(x),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 186 "limit(int(F(x),x=-R..R), R=infinity) = int(f(x),x=-i nfinity..infinity);\nint(F(x),x) = int(f(x),x);\nlim := limit(int(f(x) ,x=-R..R),R=infinity):\nlimit(int(F(x),x=-R..R), R=infinity) = lim;" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 39 "Definition 8.2: Cauchy \+ principal value" }{TEXT 324 3 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 5 "Let " }{XPPEDIT 18 0 "f" "6#%\"fG" } {TEXT -1 47 " be a continuous real valued function for all " } {XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 325 22 "Cauchy principal value" }{TEXT -1 25 " (P.V .) of the integral " }{XPPEDIT 18 0 "Int(f(x),x = -infinity .. infini ty)" "6#-%$IntG6$-%\"fG6#%\"xG/F);,$%)infinityG!\"\"F-" }{TEXT -1 16 " is defined by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 " " }{XPPEDIT 18 0 "`P.V. `*Int(f(x),x = -infinity .. \+ infinity) = Limit(int(f(x),x = -R .. R),R = infinity);" "6#/*&%&P.V.~G \"\"\"-%$IntG6$-%\"fG6#%\"xG/F-;,$%)infinityG!\"\"F1F&-%&LimitG6$-%$in tG6$-F+6#F-/F-;,$%\"RGF2F>/F>F1" }{TEXT -1 5 " " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "provided that the limit exists. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 256 23 "Example 8.14, Page 323." }{TEXT 299 26 " E valuate the integral " }{XPPEDIT 18 0 "int(1/(1+x^2), x=-infinity..in finity)" "6#-%$intG6$*&\"\"\"F',&F'F'*$%\"xG\"\"#F'!\"\"/F*;,$%)infini tyGF,F0" }{TEXT 273 4 " . \n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "f:= 'f': F:='F': z:='z':\nf := z -> 1/(z^2 + 1):\n`f(z) ` = f(z);" }}} {EXCHG {PARA 257 "" 0 "" {TEXT 260 27 "Find the singularities of " } {XPPEDIT 18 0 "f(z) = 1/(z^2 + 1)" "6#/-%\"fG6#%\"zG*&\"\"\"F),&*$F'\" \"#F)F)F)!\"\"" }{TEXT 275 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 153 "Zn := sort([solve(denom(f(z))=0, z)]):\n`For f(z) ` = f(z);\n`Th e singularities are:`;\nz1 := subs(z=Zn[1],z): z[1] = z1;\nz2 := subs( z=Zn[2],z): z[2] = z2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 257 "" 0 "" {TEXT 261 42 "Which poles lie in the upper ha lf plane ?\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "print(`0 < `,Im(z[ 1]), ` `, Im(z1)>0, evalb(evalf(Im(z1))>0));\nprint(`0 < `,Im(z[2]), \+ ` `, Im(z2)>0, evalb(evalf(Im(z2))>0));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 280 75 "Remark. Sometimes Maple will form the list of values in a different order." }}{PARA 0 "" 0 "" {TEXT 281 79 "It is always nece ssary to visually inspect the above results before proceeding." }} {PARA 257 "" 0 "" {TEXT 262 25 "\nCompute the residue at " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT 274 3 " .\n" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 48 "r1 := residue(f(z), z=z1): `Res[f`,z1,`] ` = r1;" } }}{EXCHG {PARA 257 "" 0 "" {TEXT 263 63 "The value of the integral is \+ computed by the residue calculus:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "`F(x)` = f(x);\nval := 2*Pi*I*r1:\nprint(int(F(x),x=-infinity..inf inity) = val);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 264 56 "Or, we can evaluate the integral with \+ anti-derivatives:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 373 "F:='F': g:= 'g': G:='G':\ng := z -> subs(Z=z,int(f(Z),Z)):\n`F(x) ` = f(x); \n`G(x ) = `, int(F(x),x) = g(x);\n`G(-oo)` = simplify(g(-infinity)),` and \+ `,\n`G(oo)` = simplify(g(infinity));\n`G(oo) - G(-oo)` = g(infinity) \+ - g(-infinity);\n`G(oo) - G(-oo)` = simplify(g(infinity) - g(-infinity ));\nval := int(f(x),x=-infinity..infinity):\nprint(int(F(x),x=-infini ty..infinity) = val);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 257 "" 0 "" {TEXT 265 32 "The above answer is correct too ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 " If " }{XPPEDIT 18 0 " f(z) = P(z)/Q(z);" "6#/-%\"fG6#%\"zG*&-%\"PG6#F'\"\"\"-%\"QG6#F'!\"\" " }{TEXT -1 9 ", where " }{XPPEDIT 18 0 "P" "6#%\"PG" }{TEXT -1 5 " a nd " }{XPPEDIT 18 0 "Q" "6#%\"QG" }{TEXT -1 24 " are polynomials, then " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 14 " is called a " }{TEXT 326 17 "rational function" }{TEXT -1 197 ". You probably learned tech niques in calculus to integrate certain types of rational functions. W e now show how to use the residue theorem to obtain the Cauchy princip al value of the integral of " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 7 " over " }{XPPEDIT 18 0 "`(-`*infinity*`,`*infinity*`)`;" "6#*,%#(- G\"\"\"%)infinityGF%%\",GF%F&F%%\")GF%" }{TEXT -1 2 ". " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 11 "Theorem 8.3" }{TEXT 313 8 " Let " } {XPPEDIT 18 0 "f(z) = P(z)/Q(z)" "6#/-%\"fG6#%\"zG*&-%\"PG6#F'\"\"\"-% \"QG6#F'!\"\"" }{TEXT 314 9 " where " }{XPPEDIT 18 0 "P" "6#%\"PG" } {TEXT 315 7 " and " }{XPPEDIT 18 0 "Q" "6#%\"QG" }{TEXT 316 29 " ar e polynomials, of degree " }{XPPEDIT 18 0 "m" "6#%\"mG" }{TEXT 317 7 " and " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT 318 19 " , respectively. " }}{PARA 0 "" 0 "" {TEXT 327 4 "If " }{XPPEDIT 18 0 "Q(x) <> 0" "6 #0-%\"QG6#%\"xG\"\"!" }{TEXT 319 16 " for all real " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT 320 7 " and " }{XPPEDIT 18 0 "m+2 <= n" "6#1,&% \"mG\"\"\"\"\"#F&%\"nG" }{TEXT 321 8 " , then" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 322 6 " " }{XPPEDIT 18 0 "` P.V. `*Int(P(x)/Q(x),x = -infinity .. infinity) = 2*pi*i*sum(`Res[P/Q, `*z[j]*`]`,j = 1 .. k);" "6#/*&%&P.V.~G\"\"\"-%$IntG6$*&-%\"PG6#%\"xG F&-%\"QG6#F.!\"\"/F.;,$%)infinityGF2F6F&**\"\"#F&%#piGF&%\"iGF&-%$sumG 6$*(%*Res[P/Q,~GF&&%\"zG6#%\"jGF&%\"]GF&/FC;F&%\"kGF&" }{TEXT 308 4 " \+ ,\n" }}{PARA 0 "" 0 "" {TEXT 323 7 "where " }{XPPEDIT 18 0 "z[1]" "6 #&%\"zG6#\"\"\"" }{TEXT 309 3 " , " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6# \"\"#" }{TEXT 310 8 " , ..., " }{XPPEDIT 18 0 "z[k]" "6#&%\"zG6#%\"kG " }{TEXT 311 20 " are the poles of " }{XPPEDIT 18 0 "P/Q" "6#*&%\"PG \"\"\"%\"QG!\"\"" }{TEXT 312 37 " that lie in the upper half plane. \+ " }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 279 1 "\n" }{TEXT 256 23 "Example 8.15, Page 3 25." }{TEXT 300 53 " Use the residue calculus to evaluate the integr al\n" }{XPPEDIT 18 0 "int(1/((1+x^2)*(4+x^2)), x=-infinity..infinity) \+ = pi/6" "6#/-%$intG6$*&\"\"\"F(*&,&F(F(*$%\"xG\"\"#F(F(,&\"\"%F(*$F,F- F(F(!\"\"/F,;,$%)infinityGF1F5*&%#piGF(\"\"'F1" }{TEXT 276 3 " . " }} {PARA 257 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "f :='f': F:='F': z:='z':\nf := z -> 1/((z^2 + 1)*(z^2 + 4)):\n`f(z) ` = \+ f(z);" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 266 27 "Find the singularitie s of " }{XPPEDIT 18 0 "f(z) = 1/((z^2 + 1)*(z^2 + 4))" "6#/-%\"fG6#% \"zG*&\"\"\"F)*&,&*$F'\"\"#F)F)F)F),&*$F'F-F)\"\"%F)F)!\"\"" }{TEXT 301 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 221 "Zn := sort([solve(d enom(f(z))=0, z)]):\n`For f(z) ` = f(z);\n`The singularities are:`;\n z1 := subs(z=Zn[1],z): z[1] = z1;\nz2 := subs(z=Zn[2],z): z[2] = z2;\n z3 := subs(z=Zn[3],z): z[3] = z3;\nz4 := subs(z=Zn[4],z): z[4] = z4;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 267 41 "Which poles lie in the upper half plane?\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 247 "print(`0 <`,Im(z[1]),` `, Im(z1)>0, evalb(e valf(Im(z1))>0));\nprint(`0 <`,Im(z[2]),` `, Im(z2)>0, evalb(evalf(Im (z2))>0));\nprint(`0 <`,Im(z[3]),` `, Im(z3)>0, evalb(evalf(Im(z3))>0 ));\nprint(`0 <`,Im(z[4]),` `, Im(z4)>0, evalb(evalf(Im(z4))>0));" }} }{EXCHG {PARA 0 "" 0 "" {TEXT 282 75 "Remark. Sometimes Maple will fo rm the list of values in a different order." }}{PARA 0 "" 0 "" {TEXT 283 79 "It is always necessary to visually inspect the above results b efore proceeding." }}{PARA 257 "" 0 "" {TEXT 268 26 "\nCompute the res idues at " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT 277 7 " a nd " }{XPPEDIT 18 0 "z[3]" "6#&%\"zG6#\"\"$" }{TEXT 278 3 " .\n" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "r1 := residue(f(z), z=z1): `Res[f`, z1,`] ` = r1;\nr3 := residue(f(z), z=z3): `Res[f`,z3,`] ` = r3;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 269 63 "The value of the integral is computed by the residue cal culus:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "`F(x)` = f(x);\nval := \+ 2*Pi*I*(r1 + r3):\nprint(int(F(x),x=-infinity..infinity) = val);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 270 56 "Or, we can evaluate the integral with anti-derivatives: \n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 349 "g := z -> subs(Z=z,int(f(Z), Z)):\n`F(x) ` = f(x); \n`G(x) =`, int(F(x),x) = g(x);\n`G(-oo)` = simp lify(g(-infinity)),` and `,\n`G(oo)` = simplify(g(infinity));\n`G(oo ) - G(-oo)` = g(infinity) - g(-infinity);\n`G(oo) - G(-oo)` = simplify (g(infinity) - g(-infinity));\nval2 := int(f(x),x=-infinity..infinity) :\nprint(int(F(x),x=-infinity..infinity) = val2);" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 271 32 "The above answer is correct too." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 284 1 " \n" }{TEXT 256 23 "Example 8.16, Page 325." }{TEXT 302 54 " Use the \+ residue calculus to evaluate the integral:\n" }{XPPEDIT 18 0 "int(1/(x ^2 + 4)^3, x=-infinity..infinity) = 3*pi/256" "6#/-%$intG6$*&\"\"\"F(* $,&*$%\"xG\"\"#F(\"\"%F(\"\"$!\"\"/F,;,$%)infinityGF0F4*(F/F(%#piGF(\" $c#F0" }{TEXT 293 4 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "f:='f ': F:='F': z:='z':\nf := z -> 1/(z^2 + 4)^3:\n`f(z) ` = f(z);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 285 27 "Find the singularities of " } {XPPEDIT 18 0 "f(z) = 1/(x^2 + 4)^3" "6#/-%\"fG6#%\"zG*&\"\"\"F)*$,&*$ %\"xG\"\"#F)\"\"%F)\"\"$!\"\"" }{TEXT 286 3 " .\n" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 289 "Zn := sort([solve(denom(f(z))=0, z)]):\n`For f(z) ` = f(z);\n`The singularities are:`;\nz1 := subs(z=Zn[1],z): z[1] = z 1;\nz2 := subs(z=Zn[2],z): z[2] = z2;\nz3 := subs(z=Zn[3],z): z[3] = z 3;\nz4 := subs(z=Zn[4],z): z[4] = z4;\nz5 := subs(z=Zn[5],z): z[5] = z 5;\nz6 := subs(z=Zn[6],z): z[6] = z6;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 287 41 "Which poles lie \+ in the upper half plane?\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 365 "prin t(`0 <`,Im(z[1]),` `,Im(z1)>0, evalb(evalf(Im(z1))>0));\nprint(`0 <`, Im(z[2]),` `,Im(z2)>0, evalb(evalf(Im(z2))>0));\nprint(`0 <`,Im(z[3]) ,` `,Im(z3)>0, evalb(evalf(Im(z3))>0));\nprint(`0 <`,Im(z[4]),` `,Im (z4)>0, evalb(evalf(Im(z4))>0));\nprint(`0 <`,Im(z[5]),` `,Im(z5)>0, \+ evalb(evalf(Im(z5))>0));\nprint(`0 <`,Im(z[6]),` `,Im(z6)>0, evalb(ev alf(Im(z6))>0));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 294 75 "Remark. Sometimes Maple will form the l ist of values in a different order." }}{PARA 0 "" 0 "" {TEXT 295 79 "I t is always necessary to visually inspect the above results before pro ceeding." }}{PARA 0 "" 0 "" {TEXT 288 25 "\nCompute the residue at " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT 289 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "r1 := residue(f(z), z=z1): `Res[f`,z1,`] \+ ` = r1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 290 63 "The value of the inte gral is computed by the residue calculus:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "`F(x)` = f(x);\nval := 2*Pi*I*r1:\nprint(int(F(x),x=- infinity..infinity) = val);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT 291 56 "Or, we can evaluate the integra l with anti-derivatives:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 407 "G:=' G':\ng := z -> subs(Z=z,int(f(Z),Z)):\n`F(x) ` = f(x); \n`G(x) = `, in t(F(x),x) = g(x);\nprint(`Now do the above computation with limits!`); \ng1 := limit(g(z),z=-infinity):\ng2 := limit(g(z),z= infinity):\nLimi t(G(x), x=-oo) = g1;\nLimit(G(x), x=+oo) = g2;\n`G(-oo) = ` = g1;\n`G( oo) = ` = g2;\n`G(oo) - G(-oo) = ` = g2 - g1;\nval2 := int(f(x),x=-in finity..infinity):\nprint(int(F(x),x=-infinity..infinity) = val2);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT 292 32 "The above answer is correct too. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 306 19 "End of Section 8.4." }{TEXT -1 0 "" }}}}{MARK "0 0 0" 10 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }