31 Residue Integration Method

(1)

SEOUL NATIONAL UNIVERSITY

School of Mechanical & Aerospace Engineering

400.002

Eng Math II

31 Residue Integration Method

31.1 Residue Theorem Theorem 1 [Residue theorem]

Let f(z) be analytic inside a simple closed path C and on C, except for finitely many singular points z₁, z₂,· · · , z_kinsideC. Then the integral off(z) taken counterclockwise around C equals 2πi times the sum of the residues of f(z) at z₁,· · · , z_k.

6C uz₁

uz₂ · · · uz_k '

&

$

H %

Cf(z)dz= 2πiP_k

j=1Res_z=z_jf(z)

Proof. I

C

f(z)dz+ I

C1

f(z)dz+ I

C2

f(z)dz+· · ·+ I

Ck

f(z)dz = 0 C : ccw, C₁, C₂,· · ·, C_k: cw

I

C

f(z)dz= I

C1

f(z)dz+ I

C2

f(z)dz+· · ·+ I

Ck

f(z)dz C, C₁, C₂,· · · , C_k: ccw

I

Cj

f(z)dz = 2πiRes_z=z_jf(z), j= 1,· · ·, k Example 1. Integration by the residue theorem.

I

C

4−3z

z²−zdz (C : ccw) Solution.

Res_z=0 4−3z z(z−1) =

·4−3z z−1

¸

z=0

=−4, Res_z=1 4−3z z(z−1) =

·4−3z z

¸

z=1

= 1 (a) 0 & 1 are inside C : 2πi(−4 + 1) =−6πi

(b) 0 is inside, 1 is outside : 2πi(−4) =−8πi (c) 1 is inside, 0 is outside : 2πi(1) = 2πi (d) 0 & 1 are outside : 2πi(0) = 0

(2)

Example 2. Poles and essential singularities.

C : 9x²+y² = 9 (ccw) I

c

µ ze^πz

z⁴−16 +ze^π/z

¶ dz Solution.

±2i insideC ; ±2 outsideC Res_z=2i ze^πz

z⁴−16 =

·ze^πz 4z³

¸

z=2i

=−1

16 ; Res_z=−2i ze^πz z⁴−16 =

·ze^πz 4z³

¸

z=−2i

=− 1 16 z·e^π/z =z

µ 1 +π

z + π²

2!z² + π³ 3!z³ +· · ·

¶

=z+π+π² 2 ·1

z +· · ·

∴ 2πi µ

−1 16 − 1

16 +π² 2

¶

=π µ

π²−1 4

¶

i= 30.221i . 31.2 Integrals of Rational Function of cosθ and sinθ

I = Z _2π

0

F(cosθ,sinθ)dθ =? F : real rational ftn Sete^iθ =z, thendz/dθ=ie^iθ ⇒ dθ=dz/iz, and

cosθ= 1

2(e^iθ+e^−iθ) = 1 2(z+1

z) sinθ= 1

2i(e^iθ−e^−iθ) = 1 2i(z−1

z)





 F Ãf(z)

∴ I = I

C

f(z)dz

iz C:|z|= 1. Example 3.

Show that Z _2π

0

√ dθ

2−cosθ = 2π Solution.

cosθ= 1

2(z+ 1/z), dθ=dz/iz I

c

dz/iz

√2−¹₂(z+ 1/z) = I

c

dz

−₂ⁱ(z²−2√

2z+ 1) =−2 i

I

c

dz (z−√

2−1)(z−√ 2 + 1) z₁ =√

2 + 1 : ( outside of C), z₂=√

2−1 (inside of C)

Res_z=z₂ 1

(z−√

2−1)(z−√

2 + 1) =

· 1

z−√ 2−1

¸

z=√ 2−1

=−1 2

∴ Z _2π

0

√ dθ

2−cosθ = 2πi(−2/i)(−1/2) = 2π .

(3)

31.3 Improper Integrals of Rational Functions.

When the interval of integration is not finite:

Z _∞

−∞

f(x)dx= lim

a→−∞

Z ₀

z

f(x)dx+ lim

b→∞

Z _b

0

f(x)dx If both limits exist, Z _∞

−∞

f(x)dx= lim

R→∞

Z _R

−R

f(x)dx

If f(x) is a real rational function whose denominator is different from zero for all realx and is of degree at least two units higherthan the degree of the numerator, then the both limits exist.

Consider I

C

f(z)dz around a pathC .

Choose R large enough so that C encloses all the upper-half plane (UHP) poles (finitely many), then I

C

f(z)dz = Z

S

f(z)dz+ Z _R

−R

f(x)dx= 2πiX

Resf(z)

=⇒ Z _R

−R

f(x)dx= 2πiX

Resf(z)− Z

S

f(z)dz . (1)

Setz=Re^iθ, then onS,R=const, 0≤θ≤π. By assumption,

|f(z)|< k

|z|² (|z|=R > R₀) for somek and R₀. ¯

¯¯

¯ Z

S

f(z)dz

¯¯

¯¯< k

R²πR= kπ

R (R > R₀) Take R−→ ∞ in (1). Then¯

¯R

Sf(z)dz¯

¯−→0.

∴ R_∞

−∞f(x)dx= 2πiP

U HPResf(z) . (summation over all UHP poles)

−R R

-x 6y

S (semicircle) o

R

-

C (contour) uz₁

uz₂ · · ·

uz_k

(4)

Example 4. An improper integral from 0 to∞ Z _∞

0

dx

1 +x⁴ = π 2√

2 Solution.

f(z) = 1/(1 +z⁴) ⇒z² =±i Four simple poles z₁=e^πi/4, z₂=e^3πi/4, z₃ =e^−3πi/4, z₄ =e^−πi/4

z₁ & z₂ lie in the upper half-plane Res_z=z₁f(z) =

· 1 (z+z⁴)⁰

¸

z=z1

=

· 1 4z³

¸

z=z1

= 1

4e^−3πi/4 =−1 4e^πi/4 Res_z=z₂f(z) =

· 1 (z+z⁴)⁰

¸

z=z2

=

· 1 4z³

¸

z=z2

= 1

4e^−9πi/4= 1 4e^−πi/4 (∵ e^πi =−1, e^−2πi= 1)

Z _∞

−∞

dx

1 +x⁴ =−2πi

4 (e^πi/4−e^−πi/4) =−2πi

4 2isinπ 4 = π

√2

∴ Z _∞

0

dz 1 +x⁴ = 1

2 Z _∞

−∞

dz

1 +x⁴ = π 2√

2 . 31.4 Fourier Integrals

f: rational ftn satisfying the degree assumption as before Z _∞

−∞

f(x)e^isxdx= 2πiX

Res[f(z)e^isz] (s >0)

⇒

½ R_∞

−∞f(x) cosxdx=−2πP

ImRes[f(z)e^isz] R_∞

−∞f(x) sinxdx= 2πP

ReRes[f(z)e^isz] Since s >0 and S lies in the upper half-plane y≥0,

Example 5.

Z _∞

−∞

cossx

k²+x²dx= π ke^−ks,

Z _∞

−∞

sinsx

k²+x²dx= 0. (s >0, k >0) Solution. z=ik in the upper half-plane.

Res_z=ik e^isz k²+z² =

·e^isz 2z

¸

z=ik

= e^−ks 2ik Z _∞

−∞

e^isx

k²+x²dx= 2πie^−ks 2ik = π

ke^−ks Z _∞

−∞

cossx

k²+x²dx= π ke^−ks,

Z _∞

−∞

sinsx

k²+x²dx= 0

(5)

31.5 Another Kind of Improper Integral.

A≤α≤B, lim

x→α|f(x)|=∞, Z _B

A

f(x)dx=?

Z _B

A

f(x)dx= lim

ε→0

Z _α−ε

A

f(x)dx+ lim

η→0

Z _B

α+η

f(x)dx Cauchy principal value of the integral:

pr.v.

Z _B

A

f(x)dx= lim

ε→0

·Z _α−ε

A

f(x)dx+ Z _B

α+ε

f(x)dx

¸

For example,

pr.v.

Z ₁

−1

dx x³ = lim

ε→0

·Z _−ε

−1

dx x³ +

Z ₁

ε

dx x³

¸

= 0 Theorem 1 [Simple poles on the real axis]

If f(z) has a simple pole at z=aon the real axis, then

r→0lim Z

c2

f(z)dz=πiRes_z=af(z) Proof.

f(z) = b₁

z−a+g(z), b₁ = Res_z=af(z) g(z) is analytic on the semicircle of integration.

C₂ :z=a+re^iθ 0≤θ≤π, Z

C2

f(z)dz = Z _π

0

b₁

re^iθire^iθdθ+ Z

C2

g(z)dz Z _π

0

b₁

re^iθire^iθdθ= Z _π

0

ib₁dθ=b₁πi Z

C2

g(z)dz ≤M πr asr→0, M πr→0.

Q. principal value of the integral of a rational functionf(z) from−∞ to∞? AsR−→ ∞,R

Sf(z)−→0.

Asr−→0,R

C2f(z)−→ −πiRes_z=af(z).

−R R

-x 6y

¾»

u a -C₂

S (semicircle) o

R

- -

C (contour) uz₁

uz₂ · · ·

uz_k

(6)

∴ pr.v.

Z _∞

−∞

f(x)dx= 2πi X

U HP

Resf(z) +πiX

real

Resf(z)

(the first sum over all the UHP poles, the second sum over all poles on the real axis) Example 6. Poles on the real axis

pr.v Z _∞

−∞

dx

(x²−3x+ 2)(x²+ 1) =?

Solution. Simple polesz= 1, z= 2 (real),z=i(in the UHP) Res_z=1f(z) =

· 1

(z−2)(z²+ 1)

¸

z=1

=−1 2 Res_z=2f(z) =

· 1

(z−1)(z²+ 1)

¸

z=2

= 1 5 Res_z=if(z) =

· 1

(z²−3z+ 2)(z+i)

¸

z=i

= 1

6 + 2i = 3−i 20

∴ pr.v.

Z _∞

−∞

dx

(x²−3x+ 2)(x²) = 2πi µ3−i

20

¶ +πi

µ

−1 2+ 1

5

¶

= π 10 .