Residues of functions Definition If f(z) =
∞
X
k=−∞
ck(z−α)k is holomorphic in a deleted neighborhood of α, then the residue of f(z) at the point α, is defined as
Res (f;α) =c−1.
From Cauchy’s integral formula, we recall that if D(α, R) = {z | |z−α| ≤ R} ⊂ D∪ {α} and CR(α) = {z | |z−α|=R},then
ck = 1 2πi
Z
CR(α)
f(w)
(w−α)k+1 dw for each k = 0, ±1,±2, . . . . Hence we have
c−1 = 1 2πi
Z
CR(α)
f(w)dw.
We see from this that the “limsup stuff” shows that Laurent series are uniformly convergent on A(R1, R2) by comparison with geometric series.
On a ball of radius ε, we can compute the residue of f at α by using the formula:
Res (f;α) = c−1 = 1 2πi
Z
Cε(0)
f(w−α)dw= 1 2πi
Z
Cε(α)
f(w)dw.
In general, one can compute residues as follows.
(a) Compute the integral Z
Cε(α)
f(w)dw.
(b) Compute the Laurent series forf centered at α.
(c) If f has a simple pole at α, then
c−1 = lim
z→α(z−α)f(z).
(d) Lemma Iff(z) = A(z)
B(z) for A a nonzero function, and B having a simple zero, then Res (f;α) = A(α)
B0(α). Examples
(1) Compute Res 1
1 +z2;i
by using the Laurent expansion ati.
(2) Compute Res 1
1 +z2;i
by using thati is a simple pole of 1 1 +z2. (3) Show that Res (cscz; 0) = 1
cos 0 = 1.
(4) Show that Res 1
z4−1;i
= 1 4i3 = i
4. (5) Show that Res
1 z3; 0
= 0.
(6) Show that Res (sin 1
z−1; 1) = 1.
(7) Show that Z ∞
−∞
1
1 +x2 dx = π by using that Z
CR
1
1 +z2 dz = π for each R > 1, and
R→∞lim Z
ΓR
1
1 +z2 dz = 0, where
(a) CR = ΓR∪[−R, R] (a closed contour),
(b) ΓR={|z|=R, Imz ≥0}={Reit |0≤t≤π}, (c) [−R, R] is the line segment from z=−R to z =R.
Winding numbers
These intuitive give the number of times a path loops around a point, and are also dealt with in topology.
Definition Letγ : [0,1]→Cbe a closed contour curve with γ(t)6=α for allt ∈[0,1].Then the winding number of γ about α, denoted n(γ, α), is defined by
n(γ, α) := 1 2πi
Z
γ
1
z−αdz = 1 2πi
Z 1
0
γ0(t) γ(t)−αdt.
Theorem n(γ, α)∈Z.
Proof For each s∈[0,1], letF : [0,1]→C be defined by F(s) =
Z s
0
γ0(t) γ(t)−αdt.
Since F0(s) = γ0(s)
γ(s)−α, the function G(s) := (γ(s)−α)e−F(s) satisfies that G0(s) = γ0(s)e−F(s)−γ0(s)e−F(s) = 0 ∀s∈[0,1].
SoG(s) is a constant and hence G(1) =G(0). Since F(0) = 0 andγ(1) =γ(0)6=α, (γ(1)−α)e−F(1) =G(1) =G(0) =γ(0)−α =⇒ F(1) = 2kπi for somek ∈Z. Hence n(γ, α) = F(1)
2πi ∈Z.
Corollary (version of Jordan curve theorem) Let γ : [0,1] → C be a closed curve. Then C\γ =C\ {γ(t)|t ∈[0,1]} is disconnected.
ProofSince the functionn(γ, α) :C\γ →Zis a continuous function and{n(γ, α)|α∈C\γ} ⊂Z is a disconected subset containing at least 2 integers, C\γ is disconnected.
Definition γ is called a regular closed curve if γ is a simple closed curve with n(γ, α) = 0 or 1 for all α /∈ γ. In that case, we will call {α | n(γ, α) = 1} the inside of γ. The set of points {α|n(γ, α) = 0} is called theoutside of γ.
The Jordan curve theorem shows that any closed curve has what we call aninteriorandexterior.
For interior regionB, we would haven = 16= 0.For the exterior regionB, we would haven = 0.
Cauchy’s Residue Theorem Suppose f is analytic in a simply connected region D except for isolated singularities at α1, α2, . . . , αm. Let γ be a closed curve not intersecting any of the singularities.
Then
Z
γ
f = 2πi
m
X
k=1
n(γ, αk) Res (f, αk).
Proof If we subtract the principal parts P1
1 z−α1
,· · · , Pm
1 z−αm
fromf, the difference
g(z) = f(z)−
m
X
k=1
Pk 1
z−αk
(with the appropriate definitions at α1, . . . , αm) is an analytic function in D. Hence, by the Closed Curve Theorem
(∗) Z
γ
g = 0 =⇒ Z
γ
f =
m
X
k=1
Z
γ
Pk 1
z−αk
dz.
Furthermore, for any k and for any n 6= 1, since d
dz
(z−αk)1−n
1−n = 1
(z−αk)n =⇒ Z
γ
1
(z−αk)n =
(0 if n 6= 1, 2πi if n = 1, and if
Pk 1
z−αk
= C−1,k
z−αk
+ C−2,k
(z−αk)2 +· · · , then, by (∗),
Z
γ
f =
m
X
k=1
Z
γ
Pk 1
z−αk
dz =
m
X
k=1
Z
γ
C−1,k
z−αk
dz = 2πi
m
X
k=1
n(γ, αk) Res (f, αk).
Corollary If f is as above, and if γ is a regular closed curve in the domain of analyticity of f,
then Z
γ
f = 2πiX
k
Res (f, αk).
where the sum is taken over all the singularities of f inside γ.
Applications of Cauchy’s Residue Theorem
TheoremSupposeγ is a regular closed curve. Iff is meromorphic inside and onγ and contains no zeroes or poles on γ, and if
Z= number of zeroes off inside γ (a zero of order k being counted k times), P= number of poles off inside γ (again with multiplicity),
then 1
2πi Z
γ
f0
f =Z−P.
Proof Note thatf0/f is analytic except at the zeroes or poles of f.
If f has a zero of order k at z =a, that is, if
f(z) = (z−a)kg(z) with g(z)6= 0, then
f0(z) = (z−a)k−1
kg(z) + (z−a)g0(z) has a zero of order k−1 at z,and
f0(z)
f(z) = k
z−a + g0(z) g(z).
Hence, at each zero of f of orderk, f0/f has a simple pole with residue k.
Similarly, if f has a pole of order k at z =a,then
f(z) = (z−a)−kg(z) with g(a)6= 0,
and f0(z)
f(z) = −k
z−a + g0(z) g(z),
so that at each pole of f, f0/f has a simple pole with residue −k. By the preceding Corollary, we obtain
1 2πi
Z
γ
f0
f =X Res
f0 f
=Z−P.
Corollary (Argument Principle) If f is analytic inside and on a regular closed curve γ and contains no zeroes on γ, then
Z(f) = the number of zeroes of f inside γ = 1 2πi
Z
γ
f0 f = 1
2π∆Argf(z).
Remarks
(a) The above is known as the “Argument Principle” because ifγ is given byz(t),for 0≤t ≤1,
then 1
2πi Z
γ
f0
f = logf(z(1))−logf(z(0))
2πi = 1
2π∆Argf(z) (1)
as z travels around γ from the starting point z(0) to the terminal point z(1) = z(0), and where
∆Argf(z) = lim
t→1−Argf(z(t))− lim
t→0+Argf(z(t)).
To prove (1),we split γ into a finite number of simple arcs γ1 :z(t), 0≤t≤t1 γ2 :z(t), t1 ≤t ≤t2
· · ·
γn:z(t), tn−1 ≤t ≤tn= 1.
Since an analytic branch of logf can be defined in a simply connected domain containing γ1,
Z
γ1
f0
f = logf(z(t1))−logf(z(0)).
Similarly
Z
γk
f0
f = logf(z(tk))−logf(z(tk−1)), for k = 2, 3, . . . , n.
We note that
Z
γ
= Z
γ1
+ Z
γ2
+· · ·+ Z
γn
,
and the first equality in (1) follows. Note, also, that since z(0) =z(1) and since logw = log|w|+iArgw,
logf(z(1))−logf(z(0)) = i
Argf(z(1))−Argf(z(0)) , and the second equality follows.
(b) We may also view 1 2πi
Z
γ
f0
f = 1 2πi
Z
f(γ)
dz
z as the winding number of the curve f(γ(z)) around z = 0. Thus, if f is analytic inside and on γ, the number of zeroes of f inside γ is equal to the number of times that the curve f(γ) winds around the origin, i.e.
Z(f) = 1 2πi
Z
γ
f0
f =n(f(γ); 0).
By considering f(z)−a, it follows that the number of times thatf =a inside γ equals the number of times that f(γ) winds around the complex number a.
Rouch´e’s Theorem Suppose that f and g are analytic inside and on a regular closed curve γ and that
|f(z)|>|g(z)| for all z ∈γ.
Then
Z(f +g) =Z(f) inside γ.
Proof Note first that if f(z) =A(z)B(z), then f0
f = A0 A + B0
B =⇒ Z
γ
f0 f =
Z
γ
A0 A +
Z
γ
B0 B. Thus, if we write
f +g =f
1 + g f
=⇒ (f+g)0 f+g = f0
f +(1 + fg)0 1 + fg , then
Z(f+g) = 1 2πi
Z
γ
(f+g)0 f +g = 1
2πi Z
γ
f0 f + 1
2πi Z
γ
(1 + fg)0
1 + fg =Z(f) + 1 2πi
Z
γ
(1 + fg)0 1 + fg . Since |f(z)| > |g(z)| =⇒ 1 >
g(z) f(z)
=
1 + g(z) f(z)−1
for all z ∈ γ, the closed contour γ∗ defined by
γ∗ = (1 +g/f)(γ) ={1 + g(z)
f(z) |z ∈γ} ⊂ {w∈C| |w−1|<1}
remains within a disk of radius 1 around z = 1, hence the winding number of γ∗ around 0 is equal to 0,i.e. 0 =n(γ∗; 0) = 1
2πi Z
γ∗
dw w = 1
2πi Z
γ
(1 + gf)0 1 + fg .
Remark Iff and g are analytic inside and on a regular closed curve γ and that |f(z)| ≥ |g(z)|
and f(z) +g(z)6= 0 for each z ∈γ, then
Z(f +g) =Z(f) inside γ.
In the proof, note that
Z(f +g)f+g6=0 on= γ 1 2πi
Z
γ
(f+g)0
f+g =Z(f) + 1 2πi
Z
γ
(1 + gf)0 1 + fg ,
and
1≥
g(z) f(z)
=
1 + g(z) f(z)−1
f+g6=0 onγ
=⇒ 1>
1 + g(z) f(z) −1
.
Thus we have 0 =n(γ∗; 0) = 1 2πi
Z
γ
(1 + gf)0
1 + gf and Z(f +g) =Z(f) inside γ.
ExampleShow that 2z10±4z2+ 1 has exaactly two zeros in|z|<1.
Since| ±4z2|>|2z10+ 1|on|z|= 1,and since 4z2 has exactly one zero of multiplicity 2 at z = 0 in|z|<1, 2z10±4z2+ 1 has exactly two zeroes in |z|<1.
Generalized Cauchy’s Theorem Suppose that f is analytic in a simply connected region D and thatγ is a regular closed curve contained inD.Then for eachz insideγ andk = 0,1, 2, . . . ,
(∗) f(k)(z) = k!
2πi Z
γ
f(w)
(w−z)k+1dw.
Proof For each z inside γ, since f is analytic in D, there is a neighborhood D(z, r) inside γ, such that
f(w) = f(z) +f0(z)(w−z) +· · ·+f(k)(z)
k! (w−z)k+· · · ∀w∈D(z, r)¯
=⇒ f(w)
(w−z)k+1 = f(z)
(w−z)k+1 + f0(z)
(w−z)k +· · ·+f(k)(z) k!
1
(w−z) +· · · ∀w∈D(z, r)¯
=⇒ Res
f(w) (w−z)k+1;z
= f(k)(z)
k! = the coefficient for 1
w−z by the definition.
Since f(w)
(w−z)k+1 has no other singularities in D,so by the Cauchy’s Residue Theorem, we have 1
2πi Z
γ
f(w)
(w−z)k+1 dw= Res
f(w) (w−z)k+1;z
= f(k)(z) k!
Theorem Suppose a sequence of functionsfn, analytic in a region D, converges to f uniformly on compacta of D. Then f is analytic, fn0 → f0 in D and the convergence of fn0 is also uniform on compacta ofD.
Proof For each z0 ∈ D and for some r > 0, let D(z0;r) = {z | |z − z0| ≤ r} ⊂ D and C = {z | |z −z0| = r}. Then for each z ∈ D(z0;r/2) since |w−z| ≥ r/2, {fn(w)/(w−z)} is analytic and converges uniformly to f(w)/(w−z) on C,we have
f(z) = lim
n→∞fn(z) = lim
n→∞
1 2πi
Z
C
fn(w)
w−z dw= 1 2πi
Z
C n→∞lim
fn(w)
w−z dw= 1 2πi
Z
C
f(w) w−z dw, and for eachz ∈D(z0;r/2) and|h|< r/4,since|w−z−h| ≥r/4, 1
w−z−h converges to 1 w−z uniformly onC,
f0(z) = lim
h→0
f(z+h)−f(z)
h = lim
h→0
1 h · 1
2πi Z
C
f(w)
w−z−h − f(w) w−z
dw = 1 2πi
Z
C
f(w) (w−z)2 dw,
i.e. f is analytic in D(z0;r/2) and hence analytic in D.Also since fn0(z)−f0(z) = 1
2πi Z
C
fn(w)−f(w)
(w−z)2 dw ∀z ∈D(z0;r/2)
by the Generalized Cauchy Integral Formula,|w−z| ≥r2/4 for eachz ∈D(z0;r/2) andw∈C, fn(w)/(w−z)2 converges tof(w)/(w−z)2 uniformly onC,the sequence{fn0}converges to{f0} uniformly onD(z0;r/2) and hence uniformly on compacta ofD.
Remarks
• For each ε >0 and for each z ∈D(z0;r/2) since |w−z| ≥r/2, {fn(w)/(w−z)} and {fn} converges to f on C,there exists k ∈N such that if n≥k, then
|fn(w)−f(w)|< ε r
2 ∀w∈C =⇒ |fn(w)−f(w)|
|w−z| < ε r 2 · 1
r/2 =ε ∀w∈C.
This implies that{fn(w)/(w−z)} converges uniformly to f(w)/(w−z) on C.
• For eachε >0 and for eachz ∈D(z0;r/2),since |w−z−h| ≥r/2,chooseδ = min{r 4,ε r2
8 } such that if|h|< δ, then
1
w−z−h − 1 w−z
= |h|
|(w−z−h)(w−z)| < δ· 4 r · 2
r ≤ ε r2 8 · 8
r2 =ε ∀w∈C.
This implies that|w−z−h| ≥r/4, 1
w−z−h converges to 1
w−z uniformly onC.
• For each ε >0 and for each z ∈D(z0;r/2) since |w−z| ≥r/2, {fn(w)/(w−z)} and {fn} converges to f on C,there exists k ∈N such that if n≥k, then
|fn(w)−f(w)|< ε r2
4 ∀w∈C =⇒ |fn(w)−f(w)|
|w−z| < ε r2 4 · 1
r2/4 =ε ∀w∈C.
This implies thatfn(w)/(w−z)2 converges to f(w)/(w−z)2 uniformly onC.
Hurwitz’s Theorem Let {fn} be a sequence of non-vanshing analytic functions in a region D and suppose fn→f uniformly on compacta ofD. Then either
f ≡0 inD or f(z)6= 0 for all z∈D.
Proof Suppose f(z0) = 0 for some z0 ∈D and suppose that f 6≡0, then there is some circle C centered at z0 and such that f(z)6= 0 for all z 6=z0 on and inside C; hence
fn0 fn → f0
f uniformly onC =⇒ lim
n→∞
1 2πi
Z
C
fn0 fn = 1
2πi Z
C n→∞lim
fn0 fn = 1
2πi Z
C
f0 f. However this is a contradiction since fn(z)6= 0 for allz ∈D and for alln ∈N,
1 2πi
Z
C
fn0
fn =Z(fn) = 0 while 1 2πi
Z
C
f0
f =Z(f)≥1 =⇒ f ≡0 in D.
[Note that it is possible to have f ≡ 0 despite the fact that fn(z) 6= 0 for all n. Consider, for example, fn(z) = (1/n)ez.]
Example Let fn(z) =
n
X
k=0
(−1)kz2k+1
(2k+ 1)! and f(z) = sinz. Since f(π) = 0, there exists n0 such that fn has a zero in |z−π|<1 for alln > n0.
Corollary Suppose that fn is a sequence of analytic functions in a region D, that fn → f uniformly on compacta in D, and that fn6=a. Then either
f ≡a inD or f(z)6=a for all z ∈D.
Proof Consider gn(z) =fn(z)−a, etc.
Theorem Suppose that fn is a sequence of analytic functions, and that fn → f uniformly on compacta in a regionD. Iffn is 1−1 inD for alln, then eitherf is constant or f is 1−1 inD.
Proof If f is not 1−1, there exist z1 6= z2 ∈ D such that f(z1) = f(z2) = a. Since D is a region and z1 6= z2 ∈ D, there exist disjoint closed disks D1, D2 ⊂ D surrounding z1 and z2, respecitvely.
Suppose that f 6≡a in D, then f 6≡a in D1∪D2.
If there exists a subsequence fnk → f such that fnk(z)6= a for all z ∈D1 then f(z) 6=a for all z ∈D1 by Hurwitz’s Theorem which contradicts to our hypothesis f(z1) = a.Hence there exists N ∈N such that
if n ≥N, then there exists zn ∈D1 such that fn(zn) = a.
Since fn is 1−1 in D, D1∩D2 =∅ and f 6≡a in D2,
fn(z)6=a ∀ z ∈D2, n ≥N =⇒ f(z)6=a ∀ z ∈D2 by Hurwitz’s Theorem.
This contradicts to our hypothesis f(z2) = a. Hence f(z)≡a for all z ∈D.
Evaluation of Integrals and Sums
Recall that if f is analytic in a simply connected domain D except for isolated singularities at {αk}mk=1,and if γ is a closed curve not intersecting any of the singularities, then
Z
γ
f = 2πi
m
X
k=1
n(γ, αk) Res (f;αk),
Type (1) If P, Q are polynomials, Q(x) 6= 0 for x ∈ R, and degQ ≥ degP + 2, then by the Residue Theorem
R→∞lim Z
CR
P(z)
Q(z)dz = 2πi
m
X
k=1
Res
P(z) Q(z);αk
,
where {αk}mk=1 are singularities of P(z)
Q(z) in {z |Imz >0}, CR = ΓR∪[−R, R], and
ΓR iR
−R R
CR
• ΓR={|z|=R, Imz ≥0}={Reit |0≤t≤π},
• [−R, R] is the line segment from z =−R to z =R.
This implies that
Z ∞
−∞
P(x)
Q(x)dx= 2πi
m
X
k=1
Res
P(z) Q(z);αk
.
ExampleSince αk =ei(π/4+(k−1)π/2)
is a simple pole of 1
z4+ 1 inside CR,so Res
1 z4+ 1;αk
= 1
4α3k =−αk
4 for k = 1,2, and
Z ∞
−∞
1
x4+ 1 dx= 2πi
2
X
k=1
Res 1
z4+ 1;αk
= π√ 2 2 . ExampleEvaluate
Z ∞
−∞
1 x6+ 1dx.
Type (2) If P, Q are polynomials, Q(x) 6= 0 for real x (except perhaps at a zero of cosx or sinx), anddegQ >degP,then by applying the Residue Theorem to P(z)
Q(z)eiz aroundCR R→∞lim
Z
CR
P(z)
Q(z)eizdz = 2πi
m
X
k=1
Res
P(z) Q(z)eiz;αk
,
where {αk}mk=1 are singularities of P(z)
Q(z) in {z |Imz≥0}, CR = ΓR∪[−R, R] as above.
iR
A A
B B
ih θ
−R R
CR
LetK =R·max
|z|=R
P(z) Q(z)
, ΓR =A∪B, where
• A={|z|=R, Imz ≤h} with h≤
√3
2 R =⇒ Z
A
|dz|= 2R θ≤2Rtanθ≤4h,
• B ={|z|=R, Imz ≥h} =⇒ R
B|eiz| |dz| ≤e−hπR.
Since Z
A
P(z) Q(z)eizdz
≤ K
R ·4h for h≤
√3
2 R and Z
B
P(z) Q(z)eizdz
≤ K
R ·e−h·π R, so by setting h=√
R and lettingR → ∞,we have Z ∞
−∞
P(x)
Q(x)eixdx= 2πi
m
X
k=1
Res
P(z) Q(z)eiz;αk
.
ExampleSince
eix =
∞
X
k=0
(ix)k k! =
∞
X
k=0
(ix)2k (2k)! +
∞
X
k=0
(ix)2k+1
(2k+ 1)! = cosx+isinx, and since lim
x→0
sinx
x = 1, we have sinx x = Im
eix−1 x
and hence Z ∞
−∞
sinx
x dx= Im Z ∞
−∞
eix−1 x dx.
Since eiz−1
z is holomorphic for all z 6= 0 and has a removable singularity at z = 0, so by Cauchy’s Theorem,
0 = Z
CR
eiz−1 z dz =
Z R
−R
eix−1 x dx+
Z
ΓR
eiz−1 z dz.
Thus
R→∞lim Z R
−R
eix−1
x dx= lim
R→∞
Z
ΓR
1−eiz
z dz = lim
R→∞
Z
ΓR
1
z dz− lim
R→∞
Z
A∪B
eiz
z dz =πi, where A={|z|=R, Imz ≤√
R}and B ={|z|=R, Imz ≥√
R},and Z ∞
−∞
sinx
x dx=π.
The Fourier Transform Let f :R→R orf :R→C. The Fourier transform off is given by fˆ(y) =
Z ∞
−∞
f(x)e−2πixydx.
The inversion formula is
f(x) = Z ∞
−∞
f(y)eˆ 2πiyxdy.
ExampleLet f(x) = 1
1 +x2.Then f(y) =ˆ
Z ∞
−∞
1
1 +x2e−2πixydx.
Fory ≤0, since
R→∞lim Z
CR
1
1 +z2 e−2πiyzdz = 2πiRes 1
1 +z2e−2πiyz;i
= 2πi· e2πy
2i =πe2πy.
and for y ≥ 0, by using the same contour reflected over the real axis, i.e. CR = ΓR∪[−R, R], with ΓR={|z|=R, Imz ≤0}={Re−it|0≤t≤π},we have
R→∞lim Z
CR
1
1 +z2e−2πiyzdz = 2πi n(CR,−i) Res 1
1 +z2 e−2πiyz;−i
=−2πi· e−2πy
−2i =πe−2πy. Hence
fˆ(y) = Z ∞
−∞
1
1 +x2e−2πixydx=πe−2π|y| ∀y∈R.
Type (3A) Let P, Q be polynomials, Q(x) 6= 0 for x ≥ 0, with degQ ≥ degP + 2, and let CR,ε=I1∪ΓR∪I2∪Cε, where
• I1 is the line segment from z =iε to z =√
R2−ε2+iε,
• ΓR is the circular arc of radius R fromz =√
R2−ε2+iε to z=√
R2−ε2−iε,
• I2 is the line segment from z =√
R2−ε2−iε to z=−iε,
• Cε is the circular arc of radius ε from z=−iε to z =iε.
iR
ΓR
Cε
iε
−iε
I1
I2
−R
CR,ε
Note that the inside ofCR,ε is a simply connected domain not containing 0 and hence logz may be defined there as an analytic function. (For simplicity, we choose 0<Argz <2π.)
By applying the Residue Theorem to P(z)
Q(z) logz around CR,ε R→∞lim lim
ε→0
Z
CR,ε
P(z)
Q(z) logz dz = 2πi
m
X
k=1
Res
P(z)
Q(z) logz;αk
,
where {αk}mk=1 are singularities of P(z) Q(z).
Since P(z)
Q(z) is continuous at 0,|logz|<log|z|+ 2π for z∈Cε,and since |P(z)|
|Q(z)| ≤ B
|z|2 for some constant B and for z ∈ΓR,
R→∞lim lim
ε→0
Z
ΓR∪Cε
P(z)
Q(z) logz dz = 0,
R→∞lim lim
ε→0
Z
I1
P(z)
Q(z) logz dz = Z ∞
0
P(x)
Q(x) logx dx,
R→∞lim lim
ε→0
Z
I2
P(z)
Q(z) logz dz =− Z ∞
0
P(x)
Q(x) logx+ 2πi dx, we have
Z ∞
0
P(x)
Q(x)dx=−
m
X
k=1
Res
P(z)
Q(z) logz;αk
,
where {αk}mk=1 are singularities of P(z) Q(z). ExampleSince αk =ei(π/3+2(k−1)π/3)
is a simple pole of logz z3+ 1, so Res
logz z3+ 1;αk
= i[π/3 + 2(k−1)π/3]
3α2k =−i[π/3 + 2(k−1)π/3]αk
3 for k= 1,2,3, and
Z ∞
0
1
x3+ 1dx=−
3
X
k=1
Res
logz z3+ 1;αk
=
3
X
k=1
i[π/3 + 2(k−1)π/3]αk
3 =−2π√
3 9 . Type (3B) If P, Q are polynomials, Q(x) 6= 0 for x ≥ a, and degQ ≥ degP + 2, then can compute
Z ∞
a
P(x)
Q(x)dx, by considering Z
CR,ε
P(z)
Q(z) log(z−a)dz =⇒ Z ∞
a
P(x)
Q(x)dx=−
m
X
k=1
Res
P(z)
Q(z) log(z−a);αk
,
where CR,ε is a contour as follows and{αk}mk=1 are singularities of P(z) Q(z).
ΓR
−R 0 a R
CR,ε
Type (3C)If Qis a polynomial, Q(x)6= 0 forx≥0,with degQ≥1,and if 0< α <1,then we can compute
Z ∞
0
xα−1
Q(x)dx, by considering Z
CR,ε
zα−1 Q(z)dz, where CR,ε is the “keyhole” contour as in (3A). Since
zα−1 =e(α−1) logx =xα−1 along I1, zα−1 =e(α−1) logx+2πi
=xα−1e2πi(α−1) along I2, we have
1−e2πi(α−1) Z ∞
0
xα−1
Q(x)dx= 2πi
m
X
k=1
Res
zα−1 Q(z);αk
, where {αk}mk=1 are zeros of Q(z).
ExampleSince z =−1 is the only simple pole of z−1+1/2 1 +z ,so Res
z−1+1/2 1 +z ;−1
=−i, and
1−e2πi(−1+1/2) Z ∞
0
x−1+1/2
1 +x dx= 2πiRes
z−1+1/2 1 +z ;−1
= 2π =⇒ Z ∞
0
√ 1
x(1 +x)dx=π.
Type (4) If R(z) is a rational function, then we can compute Z 2π
0
R(cosθ,sinθ)dθ by setting z =eiθ, and observing that
dθ = dz iz, cosθ = eiθ+e−iθ
2 = 1
2
z+ 1 z
, sinθ = eiθ−e−iθ
2 = 1
2i
z− 1 z
.
Example Z 2π
0
dθ
2 + cosθ = 2 i
Z
|z|=1
dz
z2+ 4z+ 1 = 4πRes
1
z2+ 4z+ 1;√ 3−2
= 2 3π√
3.
Type (5) Assume that f satisfies that lim
z→∞zf(z) = 0. Note that cotπz = cosπz
sinπz has a simple pole at each integer z =n, and
Res πcosπz sinπz ;n
= πcosnπ πcosnπ = 1.
By applying the Residue Theorem to the integral Z
CN
f(z)·π cotπz dz,
whereCN is a simple closed contour enclosing the integersn= 0, ±1, ±2, . . . ,±N and the poles of f which we assume to be finite in number, we have
Z
CN
π f(z) cotπz dz = 2πi
" N X
n=−N, n6=αk
Res (πf(z) cotπz;n) +
m
X
k=1
Res (πf(z) cotπz;αk)
#
= 2πi
" N X
n=−N, n6=αk
f(n) +
m
X
k=1
Res (πf(z) cotπz;αk)
# , where{αk}mk=1 are poles of f.
Note that
cotπz= cosπz
sinπz =ieiπz+e−iπz
eiπz−e−iπz =iei2πz+ 1
ei2πz−1 =iei2πx−2πy + 1 ei2πx−2πy−1,
CN
−(N+12)
CN
i(N+12)
−i(N+12)
N N+ 1
and
• if z =± N + 12
+iy∈CN,then
|cotπz|=
e±πi−2πy+ 1 e±πi−2πy −1
=
1−e−2πy 1 +e−2πy
< 1 +e−2πy 1 +e−2πy = 1,
• if z =x± N + 12
i∈CN,then
|cotπz| ≤
1 +e∓π(2N+1) 1−e∓π(2N+1)
=
1 +e−π(2N+1) 1−e−π(2N+1)
≤
1 +e−π 1−e−π
<2.
This implies that there exists a constant A such that
Z
CN
π f(z) cotπz dz
≤Amax
z∈CN|zf(z)|.
Furthermore, if we assume that|f(z)| ≤ B
|z|2, so that
z→∞lim zf(z) = 0,
then we have 0 = lim
N→∞
Z
CN
π f(z) cotπz dz= 2πi
" ∞ X
n=−∞, n6=αk
f(n) +
m
X
k=1
Res (πf(z) cotπz;αk)
# ,
and hence
(∗)
∞
X
n=−∞, n6=αk
f(n) =−
m
X
k=1
Res (πf(z) cotπz;αk),
where {αk}mk=1 are poles of f.
ExampleSince
Res
πcotπz z2 ;n
= 1
n2 ∀n∈Z, n 6= 0, by setting f(z) = 1
z2 in (∗), we have
∞
X
1
1 n2 = 1
2
∞
X
n=−∞, n6=0
1
n2 =−1 2Res
πcotπz z2 ; 0
.
Since cot(−z) = cotz which implies that c2k = 0 for all k in the Laurent expansion of cotz at z = 0, andc−1 = lim
z→0zcotz, c1 = lim
z→0
cotz− c−1
z 1
z, c3 = lim
z→0
cotz− c−1
z −c1z 1
z3, . . . , etc., that is
cotz = 1 z − z
3− z3
45+· · · =⇒ πcotπz z2 = 1
z3 − π2
3z − π4z
45 +· · · , we have
∞
X
1
1
n2 =−1 2Res
πcotπz z2 ; 0
= π2 6 . ExampleSince
Res π sinπz;n
= 1
cosπn = (−1)n ∀n∈Z, and since
csc2πz= 1 + cot2πz =⇒ cscπz is bounded on CN, Thus we may conclude that
N→∞lim Z
CN
π f(z) cscπz dz= 0, and by the Residue Theorem, that
∞
X
n=−∞, n6=αk
(−1)nf(n) = −
m
X
k=1
Res (πf(z) cscπz;αk), where {αk}mk=1 are poles of f.
So, by setting f(z) = 1
z2,we have
∞
X
1
(−1)n−1
n2 =−1 2
∞
X
n=−∞, n6=0
(−1)n
n2 =−1
2Res πcscπz z2 ; 0
= π2 12
since
πcscπz z2 = 1
z3 + π2
6z + 7π4z
360 +· · ·. ExampleFor each n ≥0,since
2n n
= 1 2πi
Z
C
(1 +z)2n
zn+1 dz =⇒ 2n
n 1
5n = 1 2πi
Z
C
(1 +z)2n (5z)n
dz z , where C is any simple contour surrounding the origin, and since |1 +z|2
5|z| ≤ 4
5 on|z|= 1,
∞
X
n=0
(1 +z)2n (5z)n+1 = 1
5z
∞
X
n=0
(1 +z)2 5z
n
= 1
5z · 1
1− (1 +z)2 5z
= 1
3z−1−z2 uniformly on|z|= 1, we have
∞
X
n=0
2n n
1 5n = 5
2πi Z
|z|=1
dz
3z−1−z2dz = 5 Res 1
3z−1−z2;3−√ 5 2
!
=√ 5.
ExampleFor each k ≥1,since
(−1)k n
k 2n
k
= 1 2πi
Z
C
(1 +z)n
1−1 z
2n
z dz = 1
2πi Z
C
(z+ 1) (z−1)2n
z2n+1 dz, where C is any simple contour surrounding the origin, and since
(z+ 1) (z−1)2 ≤ 16
9
√
3 on |z|= 1,
by using the Lagrange multiplier method to maximizef(a, b) =a2bsubject tog(a, b) = a2+b2 = 4,wherea=|z−1|2 andb =|z+ 1|,(by solving (2ab, a2) =∇f =λ∇g =λ(2a,2b),we getb=λ, a2 = 2λ2,and by substituting intog(a, b) = a2+b2 = 4,we getλ = 2
√3 and f(√
2λ, λ) = 16 9
√ 3.)
|z−1|
|z+ 1|
−1 1
we have
n
X
k=1
(−1)k n
k 2n
k
≤ 16
9
√ 3
n
.
