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Unformatted text preview: $9 Ave/7% W} 356% SECTION 69
1. (a) Write
1 Hz) .
csc =————=——, whet =1 d = ,
z Sing gm e [2(2) an q(z) smz
Since P(0)=1¢0, 4(0)mSiﬂO=0, and q’(0)=cosO=1;¢{}, z = 0 must. be a simple poke of cscz, with residue 3:92.. .._. ._1_ :1
q’( ) 1
(b) From Exercise 2, Sec“ 61, we know that
,. 3. 3. I “a; a
esezz+3ﬁz~i~[(3f)2 5!]: +~~ (O <Ezi< 77:).
Since the coefﬁcient of —1 here is I, it foﬁows that z m 0 is a simpie pole of cscz, the‘
z residue being 1. 110
2. {0) Write zusinhz = p(z) , where m m ' h z 2 ' .
zzsmhz (1m 10(2) 2: 5m 2 and q(z) z smhz it Since
p(7ri) = m" ¢ 0, q(m’) : 0, and q' (7131‘) m 752 :5 0, it follows that Ra$z;emhz a Mm?
Fm z smhz q'(m) 7: 11: Mi
""5" (b) Write t
6:513: : 5.5% Where 17(3) 2 “13(3) and (1(2) = sinhz, It is easy to see that expert) m pew) m , muff—TWCX ii?!" and R68 — ~' 1“.
2"” sth q (m) p( ) z=m' sinhz Q'Cwm') MM m )
Evidently, then,
RES eXPCZI) + Res. CXpC’it) : m2 exp(zm)+exp(—mt) : _ZCOS 7”.
PM 8111112: WW smhz 2
3. ((1) Write
f (E) “"' "gag, Where ME) 3 z and q(z) : 0032:,
9(2)
Observe that
It
‘1(”2“"+mr)=0 (n m0,i1,ﬂ,...)n Also, for the stated values of n, 75 717 p[£+””) = "3+ "7‘ i 0 and 61("2" + 127:) m msin(—i— +1175) 2: (“*1)"+1 :5 0" 2 112
4. Let C be the positively oriented circle tel: 2, shown just below. (a) To evaluate the integral L: tanz dz , we write the integrand as tanz = Pig—l, where 10(2): sinz and q(z) = (3052, W)
and recall that the zeros of cosz are 2 2 25+ mt (n = 0,i1,i2,..,). Only two of those zeros, namely 2 3int! 2, are interior to C, and they are the isolated singularities of
tanz interior to C. Observe that Res tanzm pot/2) mwl and Res marw:m1n
“"12 Mir/2) We” q’(~ Int/2) Hence
Jctanzdz m Zm'(—l  1) = —4m'. (b) The problem here is to evaluate the integral , dz . To do this, we write the
C smh 22
integrand as
_ I =EQQ, where p(z)=1 and q(z)ﬂsinh22:.
sthZ q(z) Now sinh 2:: = 0 when 22 2 nm’ (n 2 0,i1,ﬂ,r.n), or when
2:331— (n=0,i1,i2,...). Three of these zeros of sinh 22, namely 0 Midi—1g, are inside C and are the isolated singularities of the integrand that need to be considered here" It is straightforward to
show that 1:&_11 Res , .—  ,
==0 sthz q’(0) ZooshO 2 ’ 113
I “(xi/22L 1 1 1 qwm/z)"m=mm?
and
Resw:mmimi_=__im=mi
mellsinhZZ q’(~wm'/2) 2cosh(m€) Zcos(7r) 2'
Thus 1 . Within (3”, the function —2—,—— has isolated singularities at
z 8111:: 2:0 and z=imr (n31,2,...,N). To find the residue at z 2 0, we recall the Laurent series for 0802: that was found in Exercise
2, Sec. 61, and write =_1?+_1_._1.+[m1m_,1_]z+... (O<Izi<7r).
6 z . 114 . This tells us that 2 1 , has a pole of order 3 at z = 0 and that
z 3111: Res 21_ ==° z smz 6 As for the points z =1tmt (n = 1,2, ,N), write 1 = p(z)
zzsﬂnz q(z)’ where p(z) =1 and q(z) = 22 sinz.
Since
Mimic) m1 at 0, inmt) m 0, and q’(:i:mz:) m 1227:! cos mt m (——1)"nz:'r2 ad 0, it follows that l 1 (I)" ("1)" "MIaw il Res 2 _ ,
z=tmrz smz ("1) n 7: («1)” 1221sz So, by the residue theorem, d .1 N—N
JCN‘ z dz=2m[g+22(2 )2]. .22 sin 2 Rewriting this equation in the form ikﬂimﬁmﬁt &
":1 n2 12 41' szzsinz and recalling from Exetcise 7, Sec. 41, that the value of the integral here tends to zero as N
tends to infinity, we arrive at the desired summation formuia: 2 z n 12 ~ we 1:: 6. The path C here is the positively oriented boundary of the rectangle with vertices at the
points :2 and i2 + i. The problem is to evaluate the integral J dz
W6~m+i 115
The isolated singnlaxities of the integrand are the zeros of the polynomial 4(z) m (212 _1)2 + 3 Setting this polynomial equal to zero and solving for zz, we find that any zero 2: of q(z) has
the property 22 =1$wf§i . It is straightforward to find the two square roots of 1+\/§i and also the two square roots of 1— ﬁt”. These are the four zeros of q(z). Only two of those
zeros, «5H Zozﬁemi6= J5 and ~§o=~ﬁewiafﬁ=”\f3—+l .ﬁ’ lie inside C. They are shown in the figure below. To ﬁnd the residues at 2,, and we, , we write the integrand of the integral to be evaluated as ngg' Where M3313“ (1(Z)m(zl—1)1+3, This polynomial 9(a) is, of course, the same q(z) as above; hence q(zo) m 0. Note, too, that
p and q are analytic at an and that p(zo) e 0. Finally, it is straightforward to show that q’(z) = 42(22 "1) and hence that
q’(zo) 2 4.20053 — 1) m “Q's/6 +6J'i: a 0,
We may conclude, then, that 20 is a simple pole of the integrand, with residue 9(20) 2 1
5112,) 446 + (S‘s/5i" Similar results are to be found at the singular point «ﬂ, To be specific, it is easy to see that {IT—Eu) m ~q’(i'"o) = "—q’(zo) = 2J6— + éﬁi :5 0,
the residue of the integrand at “2}, being P(“fo) : 1 n
q’(~:z",) Mme/'2? 116’ 8. Finally, by the residue theorem, “MAE” EM[W£WW+W) bi
C(z2 — 132 +3 #245 +6f2‘i 246 + 6451' 2J5 ' We are given that f (z) w 1/ [q(z)}2, where q is analytic at 20, q(zo)=0, and q’(ze)¢0.
These conditions on q tell us that g has a zero of order mm} at 20" Hence q(z) : (z  29 )g(z), where g is a function that is analytic and nonzero at 30; and this enables
us to write ¢(z) __ 1
)2, where ¢(z)— (“any So f has a pole of order 2 at 2:0, and w 23130)
[3(30 )]3 . Efﬁe.) m ¢’(ZO) =
But, since (1(2) 3 (z  Zo)g(z), we know that q’(z) = (z  zg)g’(z) + 3(2) and q”(z) m (z — zo)g”(z) + 2g’(z) Then, by setting 2: m 20 in these last two equations, we ﬁnd that 4’(Ze)=s(20) and q”(zo)=23'(zg)w Consequently, our expression for the residue of f at 20 can be put in the desired form: ‘1”(30 )
R m * .
gm) [attain]3 (a) To find the residue of the function (:502 2: at z m 0, we write 1
WE)? ’ (:302 z = where (1(2) 7 Sine, Since :1 is entire, (1(0): 0, and q’(0) m 1 e 0, the result in Exercise 7 telis us that Res (:302 z = u q, (0)3
F“ [q (0)] 118 SECTION 72 , we integrate the function f(z) == z2+1 1. To evaluate the integral
g! 2:"1 +1 closed contour shown below, where R > 1. Weseethat
dx dz
+ =2 .3,
.‘LxNl ﬁzzH m
where
BzRe‘s 11 535.5 1 ="L”] mi”
3:: 3 +1 z=t (z—z)(z+l) Z+i an: 2i
Thus
R
dx d
I: 5M :2
_Rx +1 Caz +1 Now if z is apoint; on CR, tzl +1i21szzm11m R2 —1; around the simple 121
xzdx We use the simple closed contour {x2 + 1)(x2 + 4) 4. We wish to evaluate the integral J
0 shown below, where R > 2. We must find the residues of the function g x z t ' 
f ( ) (124.3002er 4) 3 Its mmple poles
z=i and 2:213 They are
z2 1
B =Res (z =~————————— my...
‘ f ) (42+:sz +4)L 6i
and
2
1
=Res (new mm‘
Bl :in f (I:2 +1)(z+2i)]z=2i 3i
Thus
R 2 2
x dx 3 dz
+ W: '
£sz + I)(x2 + 4) '[Ca (2:2 +1)(z2 +4) zmwl +32%
or
RH. xzdx ZZ “(xi —I~l)(.x2 + 4) 3 Ca (:2 §~1)(z2 + 4) ' If 2: is a point on CR, then ﬁzz +1E2Itz12~1 = R2 w: and I2”1 +4123izF«4I m R2 "4a Consequently,
2 3 if.
I 2 z a": s 2 “R ngugOastaoo'
cfz we +4)! (R ~1)(R~“4) 1...“: 1 .4 ’
R1 — R2 and we may conclude that (x2+1)(.x2+4) 3' or a?+l)(x2+4) I xzdx 717 xzdx 123
7, In order to show that V? xdx m n:
' (x2+1)(x2+2x+2)“ 5’ «no we introduce the function Z ﬂz) m (z2 +1)(z2 + 22: + 2) and the simple closed contour shown below. Observe that the singularities of f(z) are at i, 20 =1+i and their conjugates mi,
2‘0 x —1i in the lower half plane. Also, if R > J5, we see that R
j f(x)dx+jc f(z)dz:2m'(Bo +31),
3 R where
1 3
B mRes mﬂz =+—'
0 m) (21+1)(zfu)]z= 10 10'
and
z 1 I
BmR 2W "gmmm"
I if”) (z+i)(zz+2z+2):z=i 10 5‘
Evidently, then,
f W “22.4 mid—3M
"R(.x2+1)(.x2+2.x+2) 5 c«(z2+1)(zz+2z+2)'
Since
zdz _ zdz 11:}?2
I 2 2 "J. Mg 2 2"3'0
Cu(z +1)(Z +2z+2) “(Z +1)(zZO)(zhfo) (R “IXR"a/E) as R we no, this means that
R 111“ W = 
“"40: +1)(x +2x+2) 5
This is the desired result. xdx 7t 124 8. The problem here is to establish the integration formula J ? 2:31:— using the simple
0 x +1 3J5
closed contour shown below, where R > 1.
There is oniy one singularity of the function f(Z) = 3: 1, namely z{3 m em”, that is interior
2 to the closed centour when R > 1“ According to the residue theorem, I :13 +1 :13 4 dz =2m‘Res 1
Q: +1 Caz +1 C: ! z3+l 2% z3+l where the legs of the closed contour are as indicated in the figure. Since C1 has parametric
representation 2: = r (0 S r ..<_ R),
I dz 3:?" dr '
Cl 5+1 9 r3+1, and, since ~C2 can be represented by z : mm”3 (0 S r S. R), R than R
L dz m”chz3d:1=~JM——mmeizmaj d!" 223+: “(reWHMI" or344"
Furthermore,
1 1 1
Res 2 — 2
2:70 ZS +1 3612)!”
Consequently,
R
v dr Zm‘ dz
1__ 121143 3  _
( e r3 36121!” C 23+].
But
dz 1 211'}?
S —— no
Liza“ R3m1 3 ’ioaSR") This gives as the desired result, with the variable of integration r instead of .x: _. .— ra +1 3(ei2m'3 w(franc/:3 ( gsam) _ 3(ei21ﬂ3 wean/'3) : 35inwﬁl3) 3J3“ 7 dr 2m 2m“ 7: 2;:
G 129
This enables us to see that lq(z)l 2 (Rwlzol)4 when 2 is on CR. Thus 1 < mg
[91(2)]2 ‘— (RelZn!)a
for such points, and we arrive at the inequality
7:
I 1 2 dz 5 “Emma, =WJ:Z____S_,
Ca [61(2)] (Rlzoi) [ 4591)
R
which tells us that the value of this integral does, indeed, tend to 0 as R tends to co.
Consequently,
w "_  2
p_VNJ_?m£2mmT=_’E§,Im Wﬂ_
w[(x ma) +1} 4A 20 But the integrand here is even, and Im a+i(2a2+3) Mm“? wa+i(2a2+3) _JA+a_iJAma
go mum 1?? s/m " So, the desired result is °° dx
“(xi“afar = edgy [(2“2+3“A+“+“A‘“al'
9 where A = V02 +1. SECTION 74
cosx dx (x2 + azXJt:2 + £22)
“mg—w
(ZZ + alxz‘l +172)
inside the simple closed contour shown below, where R > a, The other singularities are, of 1. The problem here is to evaluate the integral I , where a > b > 0. To do this, we introduce the function f(z) = , whose singularities of and bi lie coarse, in the lower half plane. 130 According to the residue theorem, R .
fab: .
wmwm——~m+ “ = '
£sz +az)(x2+b2) iﬂae dz 2m(B‘+BZ)’ where B 31233 HI: :—_'£I:——'——* :—————e—:n—— 1 F“ UTE) ] (z+ai)(zz+b2)]mi 2a(b2va2)i
and
Ez _& B :ResL (z e“ :___——L—.——— x ‘3 _, 2 I?“ f ) 1 (£2 +a2)(z+bi):lzahi 2b(a2 That is, R . emdx *— ﬂ ewb emu . fz £(x2+a2)(x2+b2)ma2mbi a )wé‘lf(Z)e dz:
or R 1, . cosxdx 7!: [a e ")
“Mm"m ~ “Re f(z)e"dz
L(x2+a2)(x2+b) (12wa b a 6’;
Now,ifzis apoint on CR,
Ef(z)l$MR where MR 2 1 (R2 ma2)(R2 *b2)
and ieizim 8" 51. Hence 75R W122 maZXRZ ﬁbz) ~> 0 as R m} cm“ LR f(z)e‘zdzl S M Rer = Re ICE f(z)e"zdz S So it follows that (a>b>0). on 2. This problem is to evaluate the integral [63361: air, where £120" The function
0 x +
1 . . . .
f (z) = 22 +1 has the smgulantles it; and so we may integrate around the simpie closed contour shown below, where R > 1" 131 We start with
R fax
3 .
dx+ ““d 22 '
ixhrl iﬂzk 2: 15:13,
Where
in: em: emu
BxRes[f(z)e ]=——: m _.
== 2 I Fl 21
Hence
R efar ‘
LX2+1dx=ﬂe "6’;ka de,
or
fCOdedxm ﬂeaa _Re‘[f(z)eiazdz
_R at1 +1 CR '
Since
If(z)is MR where MR 2 21 ,
R 1
we know that
 v 717R
Re “dz S “(1 S ;
iftzk f(z)e 2. R2 _1
and so
I gﬂdx = 72'2"“.
m x +1
That is,
” cosax 7?:
dx = m "“ a 2 0 ‘
sz +1 2 e ( ) D . “ xsin2x . . .
To evaluate the mtegral I 75—h, we ﬁrst Introduce the functton
x Z Z z2+3 (zmzlxzm'z'ﬂ' f(Z) = where 2.1 m «Iii. The point 21 lies above the x axis, and 2'1 lies below it. If we write f(z)ei2z = 45(2) where ¢(z)=26xp(i2z)
zq 3mg ’ 132 we see that 2:1 is a simple pole of the function _f(z)e‘2‘ and that the corresponding residue is . ' F ‘
amenawzw' Now consider the simple closed contour shown in the ﬁgure below, where R > V? Integrating ﬁrs)?“ around the closed contour, we have
' R xerz: I
Thus
fﬁmxﬁmhemmwmjjnwhe
"R .x + 3 C3 Now, when 2 is apointon CR, I_f(z)l$ MR, where MR 2 RZR —>0 as R—éoo; and so, by Iimit (1), Sec. 74, ﬁre L f (z)e‘2‘dz = 0. Ram9 Consequently, since 5 lImJ'Cﬂ 10(2)?” dz Lﬂmmﬂ, we arrive at the result no on xsinx m _ xsinx __ at
I x2 +3 dx — ﬂexp(~—2\/§), or Ix: +3 dxnaexmZxﬁ). .4, 0 135 Now if z is a point on CR, then R Ef(z)l$ MR where ME =: (R2 “MR2 m9) as R ——> 00‘
So, in view of limit ( 1), Sec. 74, S 'ImJC f(z)ei‘dz ~—>0 as Rm) on; LR f (z)e"‘dz and this means that I xasinxdx m£(§__1) or]: .xssinxdx _i[2___1)
ennui—+9) Se e" ' 0(x2+1)(x2+9)_168 e2 a” in.” m sinxdx
x2 +4x +5 9. The Cauchy principal value of the integral I can be found with the aid of the 1
z2+4z+5 Using the quadratic formula to solve the equation zz+4z+5m0, we find that f has
1 (Z*ZI)(Z_EI) function f(z)~ and the simple closed contour shown below, where R>\/§. singularities at the points zI m ~2 +1? and fl = "2 i. Thus f (z) a , where zE is interior to the closed contour and E; is beiow the real axis. The residue theorem teHs us that R v
e“dx 
m4. '3 a '
if +4x+5 LRszk dz 27:13, where 81.: 651]
B 2 Res = ———;
W: (z ~zl)(z—2"1) (2:i %”5)
and so if sinxdr m Zm'eiz'
x2+4x+5 wR 136
01' sinx dx 7——— ~£sin2lm
x +4x+5 e r. LR f(z)e"‘dz. Now, if zis apointon CR, then ieizlze’y $1 and 1 lf(z)iS M"R where W = 1
M = W"
R (Ia—«Bf Hence 717R .<.. W R m,
WWW/5240215 ~> Ln f(z)ei‘dzl S Mam? = llm [CR f(z)ei‘dz and we may conclude that RV”! 2511M612:
x +4x+5 J'L‘ .
~sm2. e (x+1)cosx 2 dx, we shall use
x +4x+ 5 10. To ﬁnd the Cauchy principal value of the improper integral I 3+1 z+1 2:2 +4z+ 5 (z "31)(2 "‘21)
same simple closed contour as in Exercise 9. In this case, MZ—vl, and the the function f (z) = , where zi m —2 H, and f: R .
(x+1)e“dx E. ,
w+ =d m2 3,
_sz+4x+5 Jcnf(2)e Z m
where
BER“ (2+1)?z :(zi+1)eiz' m(—1+i)e"2‘
H (z"lezwEl) (2"?!) M '
Thus
R
(x+1)cosx _ _ _ 1.:
g x2 +4x+5 dx m Rgcsz) Laﬂzk ’
or
R
(x+1)cosx 7r . .
dezm sn2—c 2 w "d"
ix2+4x+5 (3(1 OS) J‘cnﬂz)‘; 2 Finally, we observe that if z is a point on C R, then R+1 R+l (RuizllXR—dfl!) = (12—4332 ""9 0 as R ""9 °°‘ If(z)[$ MR where MR= ...
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This note was uploaded on 12/27/2011 for the course MATH 5B taught by Professor Rickrugangye during the Fall '07 term at UCSB.
 Fall '07
 RickRugangYe

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