Chapter 12: The residue theorem Flashcards
singularities and integral ∫_γ f(z)dz
Let D be a simply connected region containing a simple positively oriented contour γ.
Suppose f is analytic on D except for finitely many singularities β1, . . . , βn, none of which lie on γ.
Then
∫_γ f(z)dz
= 2πi × (sum of residues of f at the singularities of f inside γ).
EXAMPLE:
Evaluate
∫_γ dz/(z^2(z − 3))
where c : z = 71e^it (0 ≤ t ≤ 2π).
let f(z) = 1/(z^2(z-3))
Then f is analytic in C \ {0, 3}. Now z^2(z − 3) has a zero of order 2 at the origin
and a simple zero at the point z = 3.
Hence f has a double pole at the origin and a simple pole at z = 3. Both these
singularities lie inside the contour c, which is a simple contour described in the
positive direction. Now
Res{f;0} = (1/1!) lim as z tends to 0 of ((d/dz)(z^2 x [1/(z^2(z-3)])
=((-1)/(z-3)^2)_{z=0} = -1/9
Res{f;3}=([(1/z^2)]/[(d/dz)(z-3)])_{z=3} = 1/9
By Cauchy’s Residue Theorem
∫_c dz/(z^2(z − 3)) = 2πi(Res{f; 0} + Res{f; 3}) = 0.
EXAMPLE:
Evaluate
∫_γ dz/(z^4 +1)
where c denotes the semi-circular contour consisting of the
straight line from −2 to 2 along the real axis, followed by the semicircle z = 2e^it
(0 ≤ t ≤ π) of radius 2 in the upper half plane from 2 back to −2.
let g(z)= 1/(z^4 +1)
Then g is analytic in C except for the four points at which z^4 + 1 = 0. Now z^4 + 1 has simple zeros at the points exp (±πi/4), exp (±3πi/4) and, hence, g has simple poles, at exp (±πi/4), exp (±3πi/4) .
diagram contour semicircle with 2 points inside 2 outside
The simple poles at exp ( πi/4), exp ( 3πi/4) lie inside the contour whereas the other two lie in the lower half plane, so are outside the contour. Moreover c is a simple contour described in the positive direction. Now
Res{g; exp(πi/4)} = (1/[(d/dz)(z^4+1)]){z=exp(πi/4)}=(1/4z^3){z=exp(πi/4)}=1/(4e^{3πi/4})= -(e^{πi/4})/4
Res{g; exp(3πi/4)} = (1/[(d/dz)(z^4+1)]){z=exp(3πi/4)}=(1/4z^3){z=exp(3πi/4)}=1/(4e^{9πi/4})= (e^{-πi/4})/4
Thus the sum of the residues is
Res{g ; exp ( πi/4)} + Res{g ; exp ( 3πi/4)} =1/4(e^− πi/4 − e^πi/4 ) = (−1/4).2isin π/4 =−i/2√2
By Cauchy’s Residue Theorem
∫_c dz/(z^4 + 1) = 2πi ×−i/(2√2) = π/(√2)
EXAMPLE: Let γ be the square contour with vertices −3, −3i, 3, 3i described in the anticlockwise
direction. Evaluate
∫_γ z^3 cos(1/z) dz
By Cauchy’s Residue Theorem
∫_γ h(z) dz = ∫_γ z^3 cos(1/z) dz =2πi/24 = πi/12
Let h(z) = z^3cos (1/z), k(z) = cos (1/z).
Then h and k are both analytic in C \ {0} and have isolated essential singularities at the origin.
diagram inside contour
origin lies inside the contour γ, which is a simple contour described in the positive direction. Now. for z ≠ 0,
z^3cos(1/z) = z^3(1 −1/(z^2 2!) + 1/(z^4 4!) − · · · ) = z^3 − z/2! + 1/z 4! − · · · ,
cos(1/z) = (1 − 1/(z^2 2!) + 1/(z^4 4) − · · · )
Thus Res{h; 0} =1/4! = 1/24 and Res{k; 0} = 0
By Cauchy’s Residue Theorem ∫_γ h(z) dz = ∫_γ z^3cos(1/z) dz = 2πi/24 = πi/12
EXAMPLE: Let γ be the square contour with vertices −3, −3i, 3, 3i described in the anticlockwise
direction. Evaluate
∫_γ cos(1/z) dz
By Cauchy’s Residue Theorem
∫_γ k(z) dz = ∫_γ cos(1/z) dz = 0.
Let h(z) = z^3cos (1/z), k(z) = cos (1/z).
Then h and k are both analytic in C \ {0} and have isolated essential singularities at the origin.
diagram inside contour
origin lies inside the contour γ, which is a simple contour described in the positive direction. Now. for z ≠ 0,
z^3cos(1/z) = z^3(1 −1/(z^2 2!) + 1/(z^4 4!) − · · · ) = z^3 − z/2! + 1/z 4! − · · · ,
cos(1/z) = (1 − 1/(z^2 2!) + 1/(z^4 4) − · · · )
Thus Res{h; 0} =1/4! = 1/24 and Res{k; 0} = 0
By Cauchy’s Residue Theorem ∫_γ k(z) dz = ∫_γ cos(1/z) dz = 0.
lemma 12.5
straight line segment on imaginary axis and ∫_L1 φ(z) e^{iλz} dz
Suppose that the function φ is analytic on the region D = {z ∈ C : |z| > r} for some r > 0. Let L_1 be the straight line segment from iR to iS on the imaginary axis, where R, S > r. Let λ be a positive real number and let t ∈ R. If |φ(it)| → 0 as t → ∞,
then
I_R =
∫_L1 φ(z) e^{iλz} dz → 0 as R, S → ∞.
∫_[-∞,∞] (xsinπx)/(x^2+2x+5) dx
let p(z)=z, q(z) = z^2 + 2z + 5, φ(z) = p(z)/q(z)
f(z) = φ(z)e^{iπz} = ze^{iπz}/(z^2 + 1)
Then
(i) q(z) = z^2 + 2z + 5 = (z + 1)^2 + 4 and so the polynomial q is non-zero on the real axis,
(ii) the degree of the polynomial q is bigger thn the degree of the polynomial p,
(iii) the function f is analytic in C except for simple poles at −1 ± 2i. The only
singularity of f in the upper half -plane H = {z ∈ C : Im z > 0} is the simple pole
at −1 + 2i.
Now Res{f; −1 + 2i} =
(ze^{iπz}/((d/dz) (z^2 + 2z + 5))){z=−1+2i} =
((ze^iπz)/(2z + 2)){z=−1+2i}
=(−1 + 2i)e^{iπ(−1+2i)}/4i
= (−1 + 2i)(e^−iπ e^−2π)/4i
= −((−1 + 2i)e^−2π)/4i
By theorem 12.1
∫_[-∞,∞] (x)/(x^2+2x+5)e^iπx dx =
∫_[-∞,∞]
φ(x)e^iπx dx = 2πi Res{f; −1 + 2i}
= 2πi [−((−1 + 2i)e^−2π)/4i]
=π/2e^−2π(1 − 2i)
Equating imaginary parts gives
∫_[-∞,∞] (xsinπx)/(x^2+2x+5) dx
=−π e^{−2π}
∫_[-∞,∞] (cosπx)/(x^2+2x+5) dx
Let
p(z) = 1, q(z) = (z^2 + 1)2
φ(z) = p(z)/q(z)
f(z) = φ(z)e^iπz =e^{iπz}/[(z^2 + 1)^2]
Then
(i) the polynomial q is non-zero on the real axis,
(ii) the degree of the polynomial q > the degree of the polynomial p,
(iii) the function f is analytic in C except for double poles at ± i. The only
singularity of f in the upper half -plane H = {z ∈ C : Im z > 0} is the double pole at i.
Now
Res{f;i} =(1/1!) limz→i d/dz[(z−i)^2f(z)]= limz→i d/dz[(z − i)^2 e^{iπz}/[(z − i)^2(z + i)^2]
= limz→i d/dz [e^iπz/(z + i)^2]
= limz→i[ [iπe^{iπz}/(z + i)^2]−[2 e^{iπz}/(z + i)^3]]
=
(iπe^−π)/4i^2− 2e^−π/8i^3
= − i e^{−π} (π + 1)/4
By Theorem 12.1,
∫[-∞,∞] [1/(1 + x^2)^2]e^iπx dx =
∫[-∞,∞] φ(x)e^{iπx} dx = 2πi Res{f;i} = 2πi × [−i e^{−π}(π + 1)/4]=(π/2)e^{−π}(π + 1)
and equating real parts gives
∫_[-∞,∞] cos πx /(1 + x^2)^2 dx =(π/2) e^−π (π + 1).
Using the substitution t = −x we see that
∫[0,∞] [cos πx/(1 + x^2)^2]dx =
∫[0,∞] [cos(−πt)/(1 + t^2)^2]dt =
∫_[0,∞] [cos(πt)/(1 + t^2)^2]dt
Thus
∫[0,∞] [cos πx/(1 + x^2)^2]dx =
(1/2)∫[-∞,∞] [cos πx/(1 + x^2)^2] dx = (π/4)e^{−π}(π + 1)
Let α > 0. ∫_[-∞,∞] evaluate (cosαx)/(1+x^2) dx
By using a suitable value for α and a standard
result, deduce that
∫_[0,∞] (cos^2 x)/(1 + x^2) dx =
(π/4e^2)(1 + e^2).
3. Let
p(z) = 1, q(z) = z
2 + 1, φ(z) = p(z)
q(z)
, f(z) = φ(z)e
iαz =
e
iαz
z
2 + 1
.
Then
(i) the polynomial q is non-zero on the real axis,
(ii) the degree of the polynomial q > the degree of the polynomial p,
(iii) the function f is analytic in C except for simple poles at ± i. The only
singularity of f in the upper half -plane H = {z ∈ C : Im z > 0} is the simple pole
at i.
(iv) α > 0.
Now
Res{f;i} =
e
iαz
d
dz (z
2 + 1)!
z=i
=
e
−α
2i
.
By Theorem 12.1
Z ∞
−∞
1
1 + x
2
e
iαx dx =
Z ∞
−∞
φ(x)e
iαx dx = 2πi Res{f;i} ) = 2πi ×
e
−α
2i
= πe−
and equating real parts gives Z ∞ −∞ cos αx 1 + x 2 dx = πe−α . Using the substitution t = −x we see that Z 0 −∞ cos αx 1 + x 2 dx = Z ∞ 0 cos(−αt) 1 + t 2 dt = Z ∞ 0 cos αt 1 + t 2 dt. Thus Z ∞ 0 cos αx 1 + x 2 dx = 1 2 Z ∞ −∞ cos αx 1 + x 2 dx = πe−α 2 . (1) We notice that Z ∞ 0 cos2 x 1 + x 2 dx = 1 2 Z ∞ 0 1 + cos 2x 1 + x 2 dx. (2) Putting α = 2 in equation (1) gives, Z ∞ 0 cos 2x 1 + x 2 dx = πe−2 2 = π 2e 2 (3) and Z ∞ 0 1 1 + x 2 dx =
tan−1 x ∞ 0 = π 2 . (4) From equations (1), (2), (3), (4) it follows that Z ∞ 0 cos2 x 1 + x 2 dx = π 4
1 e 2 \+ 1 = π(1 + e 2 ) 4e 2
lemma 12.4
∫_ΓR φ(z) e^iλz dz
Suppose that the function φ is analytic on the region D = {z ∈ C : |z| > r}
for some r > 0. Let ΓR be given by
z = Reit (θ1 ≤ t ≤ θ2),
where 0 ≤ θ1 < θ2 ≤ π and R > r. Let M(R) be the maximum value of |φ(z)| on ΓR and
let λ be a positive real number. If M(R) → 0 as R → ∞, then
IR =
∫_ΓR φ(z) e^iλz dz → 0 as R → ∞.
corollary 12.3
positive real number
For any positive real number R,
−R sin θ ≤ −2Rθ/π
(0 < θ ≤π/2)
lemma 12.2
θ and sinθ relation
For 0 < θ ≤π/2
2/π≤ (sin θ)/θ
Example. Show that
∫_[- ∞,∞] (cos x)/(x^2 + 1) dx = π/e
Deduce that
∫_[0,∞] (cos x)/(x^2 + 1) dx =π/2e
`
Let
p(z) = 1, q(z) = z^2 + 1,
φ(z) = p(z)/q(z)
f(z) = φ(z)e^iz =[1/(z^2 + 1)]e^iz
Then the function f is analytic in C except for simple poles at ± i.
using contour semicircle from -R to -to R to -R
consisting of the straight line segment L from −R to R followed by the semi-circle ΓR
given by z = Reit(0 ≤ t ≤ π), where R > 1. Now f is analytic in C except for simple
poles at ± i, neither of which lie on γ. By the Cauchy’s Residue Theorem
∫_γ f(z) dz = 2πiRes{f;i}
since the pole at i is inside γ and the pole at −i is outside. Now
Res{f;i} =[(e^iz)/(d/dz (z^2 + 1))]_z=i = e^−1/2i = 1/2ie
∫_L f(z) dz + ∫_ΓR f(z) dz =
∫_γ f(z) dz =2πi/2ie = π/e
i.e. ∫_[-R,R] f(x) dx + ∫_ΓR f(z) dz =π/e
We now show that ∫_ΓR f(z) dz → 0 as R → ∞.
…. by mod less than or equal to
equating real and imaginary parts substitution
….
complex integration helps us evaluate certain real integrals. Integrals of the form ∫_[- ∞,∞] φ(x) cos λx dx , ∫_[- ∞,∞] φ(x) sin λx dx , ∫_[- ∞,∞] φ(x) cos λx dx , ∫_[0,∞]φ(x) sin λx dx ,
where φ is a rational function and λ is a positive real number
Integrals of this form are evaluated by integrating
f(z) = φ(z) e^{iλz}
around a suitable contour
thm 12.1
useful*
no singularities on real axis
Suppose that
φ(z) = p(z)/q(z)
and f(z) = φ(z) e^{iλz}
where λ ∈ R, and p and q are polynomials with no common factor other than 1. Suppose
also that the polynomial q is non-zero on the real axis (i.e. none of the singularities
of the rational function φ lie on the real axis.)
If the degree of the polynomial q is greater that the degree of the polynomial p and λ > 0,
then
∫_[- ∞,∞] φ(x)e^{iλx}.dx
=2πiΣ from r=1 to k of [Res{f;z_r}]
where z1, z2, · · · zk are the zeros of the polynomial q in the upper half-plane
H = {z ∈ C : Im z > 0} (i.e. z1, z2, · · · zk are the singularities of f in the upper half plane
H.)
