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{SECT 0 {EXCHG {PARA 258 "" 0 "" {TEXT 257 41 "COMPLEX ANALYSIS: Maple
 Worksheets,  2001" }{TEXT 289 144 "\n(c) John H. Mathews          Rus
sell W. Howell\nmathews@fullerton.edu     howell@westmont.edu\n\nCompl
imentary software to accompany the textbook:" }}{PARA 258 "" 0 "" 
{TEXT 257 82 "COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed,
 2001, ISBN: 0-7637-1425-9" }{TEXT 288 197 "\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 0 "" 0 "" {TEXT 
279 1 "\n" }{TEXT 256 26 "CHAPTER 8   RESIDUE THEORY" }{TEXT 277 2 "\n
\n" }{TEXT 256 36 "Section 8.3  Trigonometric Integrals" }{TEXT 278 2 
"\n\n" }{TEXT -1 250 "   Amazingly, we can evaluate certain definite r
eal integrals with the aid of the residue theorem.  One way to do this
 by interpreting the definite integral as the parametric form of an in
tegral of an analytic function along a simple closed contour. " }}
{PARA 0 "" 0 "" {TEXT 291 74 "   The method in this section is used to
 evaluate integrals of the form   " }{XPPEDIT 18 0 "Int(F(cos(theta), \+
sin(theta)), theta=0..2*pi)" "6#-%$IntG6$-%\"FG6$-%$cosG6#%&thetaG-%$s
inG6#F,/F,;\"\"!*&\"\"#\"\"\"%#piGF5" }{TEXT 281 30 "    \nby using th
e contour  C: " }{XPPEDIT 18 0 "abs(z) = 1" "6#/-%$absG6#%\"zG\"\"\"" 
}{TEXT 282 33 " ,  and the change of variable \n " }}{PARA 0 "" 0 "" 
{XPPEDIT 18 0 "cos(theta) = (z+1/z)/2" "6#/-%$cosG6#%&thetaG*&,&%\"zG
\"\"\"*&F+F+F*!\"\"F+F+\"\"#F-" }{TEXT 283 5 " ,   " }{XPPEDIT 18 0 "s
in(theta) = (z - 1/z)/(2*i)" "6#/-%$sinG6#%&thetaG*&,&%\"zG\"\"\"*&F+F
+F*!\"\"F-F+*&\"\"#F+%\"iGF+F-" }{TEXT 284 5 " ,   " }{XPPEDIT 18 0 "d
*theta = d*z/(i*z)" "6#/*&%\"dG\"\"\"%&thetaGF&*(F%F&%\"zGF&*&%\"iGF&F
)F&!\"\"" }{TEXT 285 5 " ,   " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 292 26 "and the complex function  " }{XPPEDIT 
18 0 "f(z)" "6#-%\"fG6#%\"zG" }{TEXT 293 5 "  =  " }{XPPEDIT 18 0 "1/(
i*z);" "6#*&\"\"\"F$*&%\"iGF$%\"zGF$!\"\"" }{TEXT 294 1 " " }{XPPEDIT 
18 0 "F((z+1/z)/2,(z-1/z)/(2*i));" "6#-%\"FG6$*&,&%\"zG\"\"\"*&F)F)F(!
\"\"F)F)\"\"#F+*&,&F(F)*&F)F)F(F+F+F)*&F,F)%\"iGF)F+" }{TEXT 287 3 " ,
\n" }}{PARA 0 "" 0 "" {TEXT 295 49 "and evaluating the resulting conto
ur integral    " }{XPPEDIT 18 0 "Int(f(z), z=C..` `)" "6#-%$IntG6$-%\"
fG6#%\"zG/F);%\"CG%\"~G" }{TEXT 286 5 " .   " }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 290 93 "Load Maple's  \+
\"residue\" procedure.\nMake sure this is done only ONCE during a Mapl
e  session.\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "readlib(residue):
" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 260 143 "These proble
ms were also solved using Maple's table of integrals.\nThe method of s
olution is to make a solution and obtain a contour integral.\n\n" }
{TEXT 257 52 "We need to use the following substitution procedure." }
{TEXT 280 1 "\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 323 "Zsub := proc(F1
)\n  local f0;\n  f0 := F1;\n  f0 := subs(\{cos(t)=(z+1/z)/2,\n       \+
       sin(t)=-I*(z-1/z)/2\},f0);\n  f0 := subs(\{cos(2*t)=(z^2+1/z^2)
/2,\n              sin(2*t)=-I*(z^2-1/z^2)/2\},f0);\n  f0 := subs(\{co
s(3*t)=(z^3+1/z^3)/2,\n              sin(3*t)=-I*(z^3-1/z^3)/2\},f0);
\n  f0 := f0/(I*z);\n  f0 := normal(f0);\nend:" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT 261 1 "\n" }{TEXT 256 23 "Example 8.10, Page 318." }{TEXT 
276 67 "  Use substitution and an equivalent contour integral to evalu
ate:\n" }{XPPEDIT 18 0 "int(1/(1 + 3*cos(theta)^2), theta=0..2*pi)" "6
#-%$intG6$*&\"\"\"F',&F'F'*&\"\"$F'*$-%$cosG6#%&thetaG\"\"#F'F'!\"\"/F
/;\"\"!*&F0F'%#piGF'" }{TEXT 271 4 " . \n" }{TEXT -1 0 "" }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 110 "f:='f': F:='F': t:='t': z:='z':\nF := 1/(1 +
 3*cos(t)^2):\nf1 := Zsub(F):\n`Given  F(t) ` = F;\n`Use  f(z) ` = f1;
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 27 "Find the singularities of  \+
" }{XPPEDIT 18 0 "f(z) = - 4*i*z/(3*z^4 + 10*z^2 + 3)" "6#/-%\"fG6#%\"
zG,$**\"\"%\"\"\"%\"iGF+F'F+,(*&\"\"$F+*$F'F*F+F+*&\"#5F+*$F'\"\"#F+F+
F/F+!\"\"F5" }{TEXT 263 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 217 
"Zn := sort([solve(denom(f1)=0, z)]):\n`For  f(z) ` = f1;\n`The singul
arities are:`;\nz1 := subs(z=Zn[1],z): z[1] = z1;\nz2 := subs(z=Zn[2],
z): z[2] = z2;\nz3 := subs(z=Zn[3],z): z[3] = z3;\nz4 := subs(z=Zn[4],
z): z[4] = z4;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 264 50 "Find out which singularities lie within \+
a circle  " }{XPPEDIT 18 0 "abs(z) < 1" "6#2-%$absG6#%\"zG\"\"\"" }
{TEXT 272 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 251 "print(abs(z[1
]),`< 1   `, abs(z1)<1, evalb(evalf(abs(z1))<1));\nprint(abs(z[2]),`< \+
1   `, abs(z2)<1, evalb(evalf(abs(z2))<1));\nprint(abs(z[3]),`< 1   `,
 abs(z3)<1, evalb(evalf(abs(z3))<1));\nprint(abs(z[4]),`< 1   `, abs(z
4)<1, evalb(evalf(abs(z4))<1));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 265 
75 "Remark.  Sometimes Maple will form the list of values in a differe
nt order." }}{PARA 0 "" 0 "" {TEXT 274 79 "It is always necessary to v
isually inspect the above results before proceeding." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 275 25 "Compute the residues a
t  " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT 266 7 "  and  " }
{XPPEDIT 18 0 "z[3]" "6#&%\"zG6#\"\"$" }{TEXT 273 3 " .\n" }}{PARA 0 "
> " 0 "" {MPLTEXT 1 0 93 "r2 := residue(f1, z=z2): `Res[f`,z2,`] ` = r
2;\nr3 := residue(f1, z=z3): `Res[f`,z3,`] ` = r3;" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT 267 63 "The value of the integral is computed by the res
idue calculus:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "f:='f':\n`f(t)`
 = F;\nval := 2*Pi*I*(r3 + r2):\nprint(int(f(t),t=0..2*pi) = val);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
268 110 "Computer algebra systems such as Maple are capable of finding
 some of these difficult integrals.\nThe functon  " }{XPPEDIT 18 0 "F(
t) = 1/(1 - 3*cos(t)^2)" "6#/-%\"FG6#%\"tG*&\"\"\"F),&F)F)*&\"\"$F)*$-
%$cosG6#F'\"\"#F)!\"\"F2" }{TEXT 269 40 "  can be integrated directly \+
to obtain:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "`f(t) ` = F; \nprin
t(int(f1(t),t) = int(F,t));\nprint(int(f1(t),t=0..2*pi) = int(F,t=0..2
*Pi));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 270 19 "End of Section 8.3." }
}}}{MARK "0 0 0" 29 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 
1 2 33 1 1 }