6. Residue Calculus Let Z0be an Isolated Singularity of F(Z)

6. Residue calculus

Let z0 be an isolated singularity of f(z), then there exists a certain deleted neighborhood Nε = z : 0 < z z < ε such that f is { | − 0| } analytic everywhere inside Nε. We deﬁne 1 Res(f,z0)= f(z) dz, 2πi IC where C is any simple closed contour around z0 and inside Nε.

1 Since f(z) admits a Laurent expansion inside Nε, where

∞ n ∞ n f(z)= an(z z ) + bn(z z ) , − 0 − 0 − nX=0 nX=1 then 1 b1 = f(z) dz = Res(f,z0). 2πi IC

Example

1 1 if k = 1 Res ,z0 = (z z )k ! ( 0 if k = 1 − 0 6 1 1 2 1 Res(e1/z, 0) = 1 since e1/z =1+ + + , z > 0 1! z 2!z2 ··· | |

1 1 Res , 1 = by the Cauchy integral formula. (z 1)(z 2) ! 1 2 − − −

2 Cauchy residue theorem

Let C be a simple closed contour inside which f(z) is analytic everywhere except at the isolated singularities z ,z , ,zn. 1 2 ···

f(z) dz = 2πi[Res (f,z1)+ + Res(f,zn)]. IC ··· This is a direct consequence of the Cauchy-Goursat Theorem.

3 Example

Evaluate the integral z + 1 dz z =1 z2 I| | using

(i) direct contour integration,

(ii) the calculus of residues, 1 (iii) the primitive function log z . − z

Solution

(i) On the unit circle, z = eiθ and dz = ieiθ dθ. We then have 2π 2π z + 1 iθ 2iθ iθ iθ dz = (e− +e− )ie dθ = i (1+e− ) dθ = 2πi. z =1 z2 0 0 I| | Z Z

4 (ii) The integrand (z + 1)/z2 has a double pole at z = 0. The Laurent expansion in a deleted neighborhood of z = 0 is simply 1 1 + , where the coeﬃcient of 1/z is seen to be 1. We have z z2 z + 1 Res 0 = 1 2 , , z and so z + 1 z + 1 dz = 2πiRes , 0 = 2πi. z =1 z2 z2 I| | (iii) When a closed contour moves around the origin (which is the branch point of the function log z) in the anticlockwise direction, the increase in the value of arg z equals 2π. Therefore, z + 1 = change in value of ln + arg 1 in 2 dz z i z z I z =1 z | | − | | traversing one complete loop around the origin = 2πi.

5 Computational formula

Let z0 be a pole of order k. In a deleted neighborhood of z0, b b f(z)= ∞ a (z z )n + 1 + + k , b = 0. n 0 k k − z z0 ··· (z z0) 6 nX=0 − − Consider g(z) = (z z )kf(z). − 0 the principal part of g(z) vanishes since

k 1 ∞ n+k g(z)= bk + bk 1(z z0)+ b1(z z0) − + an(z z0) . − − ··· − − nX=0 By diﬀerenting (k 1) times, we obtain − (k 1) g − (z0) if (k 1)( ) is analytic at (k 1)! g − z z0  − k 1 k d (z z0) f(z) b1 = Res(f,z0)=  lim − − if z is a removable  z z k 1 0  → 0dz − " (k 1)! # − (k 1) singularity of g − (z)    6 Simple pole

k =1: Res(f,z )= lim (z z )f(z). 0 z z 0 → 0 − p(z) Suppose f(z)= where p(z0) = 0 but q(z0) = 0, q (z0) = 0. q(z) 6 ′ 6 Res(f,z ) = lim (z z )f(z) 0 z z 0 → 0 − p(z0)+ p′(z0)(z z0)+ = lim (z z0) − ··· z z0 q (z0) 2 → − q (z )(z z )+ ′′ (z z ) + ′ 0 − 0 2! − 0 ··· p(z ) = 0 . q′(z0)

7 Example Find the residue of e1/z f(z)= 1 z − at all isolated singularities.

Solution (i) There is a simple pole at z = 1. Obviously

Res(f, 1) = lim (z 1)f(z)= e1/z = e. z 1 − − − → z=1

(ii) Since

1 1 e1/z =1+ + + z 2!z2 ··· has an essential singularity at z = 0, so does f(z). Consider e1/z 1 1 = (1+z +z2 + ) 1+ + + , for 0 < z < 1, 1 z ··· z 2!z2 ··· | | − the coeﬃcient of 1/z is seen to be 1 1 1+ + + = e 1 = Res(f, 0). 2! 3! ··· −

8 Example

Find the residue of z1/2 f(z)= z(z 2)2 − at all poles. Use the principal branch of the square root function z1/2.

Solution

The point z = 0 is not a simple pole since z1/2 has a branch point at this value of z and this in turn causes f(z) to have a branch point there. A branch point is not an isolated singularity.

However, f(z) has a pole of order 2 at z = 2. Note that

d z1/2 z1/2 1 Res(f, 2) = lim = lim = , z 2 dz z ! z 2 − 2z2 ! −4√2 → → where the principal branch of 21/2 has been chosen (which is √2).

9 Example

Evaluate Res(g(z)f ′(z)/f(z),α) if α is a pole of order n of f(z), g(z) is analytic at α and g(α) = 0. 6 Solution

Since α is a pole of order n of f(z), there exists a deleted neighborhood z : 0 < z α < ε such that f(z) admits the Laurent { | − | } expansion:

bn b b f(z)= + n 1 + + 1 + ∞ a (z α)n, b = 0. n − n 1 n n (z α) (z α) − ··· (z α) − 6 − − − nX=0 Within the annulus of convergence, we can perform termwise dif- ferentiation of the above series

10 nbn (n 1)bn b1 ∞ n 1 f ′(z)= − − + nan(z α) − . (z α)n+1 − (z α)n −···− (z α)2 − − − − nX=0 Provided that g(α) = 0, it is seen that 6 (z α) nbn (n 1)bn b1 + na (z α)n 1 (z −α)n+1 (z− α)n (z α)2 n∞=0 n − = lim ( ) − − − − −···− − − g z b z α bn + n 1 + + b1 + P a (z α)n → (z α)n (z α−)n 1 z α n∞=0 n − − − ··· − − = ng(α) = 0, P − 6 so that α is a simple pole of g(z)f ′(z)/f(z). Furthermore, f Res g ′,α = ng(α). f ! − Remark

When g(α) = 0,α becomes a removable singularity of gf ′/f.

11 Example

Suppose an even function f(z) has a pole of order n at α. Within the deleted neighborhood z : 0 < z α < ε , f(z) admits the Laurent { | − | } expansion

bn b1 ∞ n f(z)= + + + an(z α) , bn = 0. (z α)n ··· (z α) − 6 − − nX=0 Since f(z) is even, f(z)= f( z) so that − bn b1 ∞ n f(z)= f( z)= + + + an( z α) , − ( z α)n ··· ( z α) − − − − − − nX=0 which is valid within the deleted neighborhood z : 0 < z + α < ε . { | | } Hence, α is a pole of order n of f( z). Note that − − Res(f(z),α)= b and Res(f(z), α)= b 1 − − 1 so that Res(f(z),α) = Res(f(z), α). For an even function, if − − z = 0 happens to be a pole, then Res(f, 0) = 0.

12 Example tan z tan z π tan z π dz = 2πi Res , + Res , z =2 z z 2 z −2 I| | π since the singularity at z = 0 is removable. Observe that is a 2 π simple pole and cos z = sin z , we have − − 2 π tan z π z 2 tan z Res , = lim − z 2 z π z →2 z π sin z = lim − 2 π π 3 z 2 π (z 2) → z z 2 + −6 + "− − ··· # 1 2 = = . π −π −2

13 tan z π 2 As tan z/z is even, we deduce that Res , = using the z −2 π result from the previous example. We then have tan z dz = 0. z =2 z I| | Remark

π 2 π Let p(z) = sin z/z,q(z) = cos z, and observe that p = , q = 2 π 2 π 0 and q′ = 1 = 0, then 2 − 6 tan z π π π 2 Res , = p q′ = − . z 2 2 , 2 π

14 Example

Evaluate z2 2 2 2 dz. IC (z + π ) sin z Solution

z z z z lim = lim lim = 0 z 0 sin z (z2 + π2)2 z 0 sin z z 0 (z2 + π2)2! → → → so that z = 0 is a removable singularity.

15 It is easily seen that z = iπ is a pole of order 2. d Res(f,iπ) = lim [(z iπ)2f(z)] z iπ dz − → d z2 = lim z iπ dz "(z + iπ)2 sin z# → 2z(z + iπ) sin z z2[(z + iπ) cos z + 2 sin z] = lim − z iπ (z + iπ)3 sin2 z → 2 sinh π + ( π cosh π sinh π) 1 cosh π = − − = + . 4π sinh2 π −4π sinh π 4π sinh2 π − Recall that sin iπ = i sinh π and cos iπ = cosh π. Hence, z2 = 2 Res( ) 2 2 2 dz πi f,iπ IC (z + π ) sin z i 1 cosh π = + 2 . 2 −sinh π sinh π

16 Theorem

If a function f is analytic everywhere in the ﬁnite plane except for a ﬁnite number of singularities interior to a positively oriented simple closed contour C, then 1 1 ( ) = 2 Res 0 f z dz πi 2f , . IC z z

17 We construct a circle z = R which is large enough so that C is | | 1 interior to it. If C denotes a positively oriented circle z = R , 0 | | 0 where R0 > R1, then

∞ n f(z)= cnz , R1 < z < , (A) n= | | ∞ X−∞ where 1 f(z) c = dz n = 0, 1, 2, . n n+1 2πi IC0 z ± ± ··· In particular,

2πic 1 = f(z) dz. − IC0 How to ﬁnd c 1? First, we replace z by 1/z in Eq. (A) such that − the domain of validity is a deleted neighborhood of z = 0.

18 Now

1 1 cn c 1 f = ∞ = ∞ n 2, 0 < z < , 2 n+2 −n z z n= z n= z | | R1 X−∞ X−∞ so that 1 1 = Res 0 c 1 2f , . − z z

Remark

By convention, we may define the residue at infinity by 1 1 1 Res( )= ( ) = Res 0 f, f z dz 2f , , ∞ −2πi IC − z z where all singularities in the finite plane are included inside C. With the choice of the negative sign, we have

Res(f,z )+Res(f, ) = 0. i ∞ Xall

19 Example Evaluate 5z 2 − dz. z =2 z(z 1) I| | − Solution

5z 2 Write f(z)= − . For 0 < z < 1, z(z 1) | | − 5z 2 5z 2 1 2 − = − − = 5 ( 1 z z2 ) z(z 1) z 1 z − z − − − −··· − − so that Res(f, 0) = 2.

20 For 0 < z 1 < 1, | − | 5z 2 5(z 1) + 3 1 − = − z(z 1) z 1 1 + (z 1) − − 3 − = 5+ [1 (z 1) + (z 1)2 (z 1)3 + ] z 1 − − − − − ··· − so that Res(f, 1) = 3.

Hence, 5z 2 − dz = 2πi [Res(f, 0) + Res(f, 1)] = 10πi. z =2 z(z 1) I| | −

21 On the other hand, consider 1 1 5 2z 5 2z 1 = = 2f − − z z z(1 z) z 1 z 5 − − = 2 (1 + z + z2 + ) z − ··· 5 = +3+3z, 0 < z < 1, z | | so that 5z 2 − dz = 2πiRes(f, ) I z =2 z(z 1) − ∞ | | − 1 1 = 2 Res 0 = 10 πi 2f , πi. z z

22 Evaluation of integrals using residue methods

A wide variety of real deﬁnite integrals can be evaluated eﬀectively by the calculus of residues.

Integrals of trigonometric functions over [0, 2π]

We consider a real integral involving trigonometric functions of the form 2π R(cos θ, sin θ) dθ, Z0 where R(x,y) is a rational function deﬁned inside the unit circle z = 1,z = x+iy. The real integral can be converted into a contour | | integral around the unit circle by the following substitutions:

23 z = eiθ, dz = ieiθ dθ = iz dθ, eiθ + e iθ 1 1 cos θ = − = z + , 2 2 z eiθ e iθ 1 1 sin θ = − − = z . 2i 2i − z

The above integral can then be transformed into 2π R(cos θ, sin θ) dθ Z0 1 z + z 1 z z 1 = R − , − − dz z =1iz 2 2i ! I| | 1 z + z 1 z z 1 = 2πi sum of residues of R − , − − inside z = 1 . " iz 2 2i ! | | #

24 Example

2π cos 2θ Compute I = dθ. Z0 2 + cos θ Solution

1 2 1 2π cos 2θ 2 z + 2 dz dθ = i z 0 2 + cos θ − z =1 2+ 1 + 1 z Z I| | 2 z z z4 + 1 = i dz. − z =1 z2(z2 + 4z + 1) I| |

The integrand has a pole of order two at z = 0. Also, the roots of z2 + 4z + 1 = 0, namely, z = 2 √3 and z = 2+ √3, are simple 1 − − 2 − poles of the integrand.

25 z4 + 1 Write f(z)= . Note that z1 is inside but z2 is outside z2(z2 + 4z + 1) z = 1. | |

26 d z4 + 1 Res(f, 0) = lim z 0 dz z2 + 4z + 1 → 3z3(z2 + 4z + 1) (z4 + 1)(2z + 4) = lim − = 4 z 0 (z2 + 4z + 1)2 − →

z4 + 1 d Res(f, 2+ √3) = (z2 + 4z + 1) − z2 ,dz z= 2+√3 z= 2+√3 − − ( 2+ √3)4 + 1 1 7 = − = . ( 2+ √3)2 · 2( 2+ √3) + 4 √3 − − I = ( i)2πi Res(f, 0) +Res(f, 2+ √3) − − h 7 i = 2π 4+ . − √3!

27 Example

Evaluate the integral π 1 I = dθ, a>b> 0. Z0 a b cos θ Solution −

Since the integrand is symmetric about θ = π, we have 1 2π 1 2π eiθ I = dθ = dθ. 2 0 a b cos θ 0 2aeiθ b(e2iθ + 1) Z − Z − The real integral can be transformed into the contour integral 1 I = i dz. z =1bz2 2az + b I| | − The integrand has two simple poles, which are given by the zeros of the denominator.

28 Let α denote the pole that is inside the unit circle, then the other 1 pole will be α. The two poles are found to be

a a2 b2 1 a + a2 b2 α = − − and = − . q b α qb 1 Since a>b> 0, the two roots are distinct, and α is inside but α is outside the closed contour of integration. We then have 1 1 I = dz −ib z =1(z α) (z 1) I| | − − α 2πi 1 = Res ,α − ib (z α) (z 1)  − − α 2πi  π  = = . −ib α 1 2 2 − α a b q −

29 Integral of rational functions

∞ f(x) dx, Z−∞ where

1. f(z) is a rational function with no singularity on the real axis,

2. lim zf(z) = 0. z →∞

It can be shown that

∞ f(x) dx = 2πi [sum of residues at the poles of f in the upper Z half-plane].−∞

30 Integrate f(z) around a closed contour C that consists of the upper semi-circle C and the diameter from R to R. R −

By the Residue Theorem R f(z) dz = f(x) dx + f(z) dz C R C I Z− Z R = 2πi [sum of residues at the poles of f inside C].

31 As R , all the poles of f in the upper half-plane will be enclosed →∞ inside C. To establish the claim, it suﬃces to show that as R , →∞ lim f(z) dz = 0. R C →∞ I R

The modulus of the above integral is estimated by the modulus inequality as follows:

π f(z) dz f(Reiθ) R dθ

ZCR ≤ Z0 | |

π max f(Reiθ) R dθ ≤ 0 θ π | | 0 ≤ ≤ Z = max zf(z) π, z C | | ∈ R which goes to zero as R , since lim zf(z) = 0. z →∞ →∞

32 Example

Evaluate the real integral x4 ∞ dx 1+ x6 Z−∞ by the residue method.

Solution

z4 √3+ i The complex function f(z) = has simple poles at i, 1+ z6 2 √3+ i and − in the upper half-plane, and it has no singularity on 2 the real axis. The integrand observes the property lim zf(z)=0. z We obtain →∞ √3+ i √3+ i ∞ f(x) dx = 2πi Res(f,i)+ Res f, + Res f, − . " 2 ! 2 !# Z−∞

33 The residue value at the simple poles are found to be

1 i Res(f,i)= = ,

6z −6 z=i

√3+ i 1 √3 i Res f, = = − ,

2 ! 6z √3+i 12 z= 2 and √3+ i 1 √3+ i Res f, − = = ,

2 ! 6z √3+i − 12 z=− 2 so that x4 i √3 i √3+ i 2π ∞ dx = 2πi + − = . 1+ x6 −6 12 − 12 ! 3 Z−∞

34 Integrals involving multi-valued functions

Consider a real integral involving a fractional power function f(x) ∞ dx, 0 <α< 1, α Z0 x

1. f(z) is a rational function with no singularity on the positive real axis, including the origin.

2. lim f(z) = 0. z →∞

f(z) We integrate φ(z)= along the closed contour as shown. zα

35 The closed contour C consists of an inﬁnitely large circle and an inﬁnitesimal circle joined by line segments along the positive x-axis.

36 (i) line segment from ε to R along the upper side of the positive real axis: z = x, ε x R; ≤ ≤ iθ (ii) the outer large circle CR : z = Re , 0 <θ< 2π;

(iii) line segment from R to ε along the lower side of the positive real axis z = xe2πi, ε x R; ≤ ≤

(iv) the inner inﬁnitesimal circle Cε in the clockwise direction

z = εeiθ, 0 <θ< 2π.

37 Establish: lim φ(z)=0 and lim φ(z) = 0. R C ε 0 C →∞ Z R → Z ε 2π φ(z) dz φ(Reiθ)Reiθ dθ 2π max zφ(z) C ≤ 0 | | ≤ z C | | Z R Z ∈ R 2π iθ φ(z) dz φ(εe ) ε dθ 2π max zφ(z) . C ≤ 0 | | ≤ z Cǫ | | Z ε Z ∈

It suﬃces to show that zφ(z) 0 as either z or z 0. → →∞ →

1. Since lim f(z) = 0 and f(z) is a rational function, z →∞ deg (denominator of f(z)) 1 + deg (numerator of f(z)). ≥ Further, 1 α< 1, zφ(z)= z1 αf(z) 0 as z . − − → →∞ 2. Since f(z) is continuous at z = 0 and f(z) has no singularity at the origin, zφ(z)= z1 αf(z) 0 f(0) = 0 as z 0. − ∼ · →

38 The argument of the principal branch of zα is chosen to be 0 θ < ≤ 2π, as dictated by the contour.

φ(z) dz = φ(z) dz + φ(z) dz IC ZCR ZCǫ Rf(x) ǫf(xe2πi) + dx + dx α α 2απi Zǫ x ZR x e = 2πi [sum of residues at all the isolated singularities of f enclosed inside the closed contour C].

By taking the limits ǫ 0 and R , the ﬁrst two integrals vanish. → →∞ The last integral can be expressed as f(x) f(x) ∞ dx = e 2απi ∞ dx. α 2απi − α −Z0 x e − Z0 x

Combining the results, f(x) 2πi ∞ dx = [sum of residues at all the isolated α 2απi Z0 x 1 e− singularities− of f in the ﬁnite complex plane].

39 Example

1 Evaluate ∞ dx, 0 <α< 1. α Z0 (1 + x)x Solution

1 f(z) = is multi-valued and has an isolated singularity at (1 + z)zα z = 1. By the Residue Theorem, − 1 dz α IC (1 + z)z R dx dz dz = (1 e 2απi) + + − α α α − Zǫ (1 + x)x ZCR (1 + z)z ZCǫ (1 + z)z 1 2πi = 2πi Res , 1 = . (1 + z)zα − ! eαπi

40 The moduli of the third and fourth integrals are bounded by

1 2πR dz R α 0 as R , α α − ZCR (1 + z)z ≤ (R 1)R ∼ → →∞ −

1 2πǫ dz ǫ1 α 0 as ǫ 0. α α − ZCǫ (1 + z)z ≤ (1 ǫ)ǫ ∼ → → −

On taking the limits R and ǫ 0, we obtain →∞ → 1 2πi (1 e 2απi) ∞ dx = ; − α απi − Z0 (1 + x)x e so 1 2πi π ∞ dx = = . 0 (1 + x)xα eαπi (1 e 2απi) sin απ Z − −

41 Example

Evaluate the real integral eαx ∞ dx, 0 <α< 1. 1+ ex Z−∞ Solution

The integrand function in its complex extension has inﬁnitely many poles in the complex plane, namely, at z = (2k + 1)πi, k is any integer. We choose the rectangular contour as shown

l : y = 0, R x R, 1 − ≤ ≤ l : x = R, 0 y 2π, 2 ≤ ≤ l : y = 2π, R x R, 3 − ≤ ≤ l : x = R, 0 y 2π. 4 − ≤ ≤

42 The chosen closed rectangular contour encloses only one simple pole at z = πi.

43 The only simple pole that is enclosed inside the closed contour C is z = πi. By the Residue Theorem, we have eαz R eαx 2π eα(R+iy) dz = dx + idy C 1+ ez R 1+ ex 0 1+ eR+iy I Z− Z R eα(x+2πi) 0 eα( R+iy) + − dx + − idy x+2πi R+iy ZR 1+ e Z2π 1+ e− eαz = 2πi Res ,πi 1+ ez ! eαz = 2πi = 2πieαπi. ez − z=πi

44 Consider the bounds on the moduli of the integrals as follows:

2π eα(R+iy) 2π eαR idy dy O(e (1 α)R), R+iy R − − Z0 1+ e ≤ Z0 e 1 ∼ −

0 eα( R+iy) 2π e αR − idy − dy O(e αR). R+iy R − Z2π 1+ e− ≤ Z0 1 e− ∼ −

(1 α)R αR As 0 <α< 1, both e− − and e− tend to zero as R tends to inﬁnity. Therefore, the second and the fourth integrals tend to zero as R . On taking the limit R , the sum of the ﬁrst and → ∞ → ∞ third integrals becomes eαx (1 e2απi) ∞ dx = 2πieαπi; − 1+ ex − Z−∞ so eαx 2πi π ∞ dx = = . 1+ ex eαπi e απi sin απ Z−∞ − −

45 Example

Evaluate ∞ 1 3 dx. Z0 1+ x

Solution

Since the integrand is not an even function, it serves no purpose to extend the interval of integration to ( , ). Instead, we consider −∞ ∞ the branch cut integral Log z 3 dz, IC 1+ z where the branch cut is chosen to be along the positive real axis whereby 0 Arg z < 2π. Now ≤ Log z R ln x ǫ Log (xe2πi) dz = dx + dx 3 3 2πi 3 IC 1+ z Zǫ 1+ x ZR 1 + (xe ) Log z Log z + + 3 dz 3 dz ICR 1+ z ICǫ 1+ z

46 3 Log z = 2πi Res ,zj , 1+ z3 jX=1 3 where zj, j = 1, 2, 3 are the zeros of 1/(1 + z ). Note that Log z ǫ ln ǫ dz = O 0 as ǫ 0; C 1+ z3 1 −→ → I ǫ

Log z R ln R dz = O 0 as R . C 1+ z3 R3 −→ →∞ I R

Hence

Log z ln x 0 Log (xe2iπ) lim dz = ∞ dx + dx R C 1+ z3 0 1+ x3 1 + (xe2iπ)3 ǫ →∞0 I Z Z∞ → 1 = 2 ∞ πi 3 dx, − Z0 1+ x thus giving 3 ∞ 1 Log z dx = Res ,zj . 0 1+ x3 − 1+ z3 Z jX=1

47 The zeros of 1 + z3 are α = eiπ/3,β = eiπ and γ = e5πi/3. Sum of residues is given by Log z Log z Log z Res + Res + Res 3,α 3,β 3, γ 1+ z 1+ z 1+ z Log α Log β Log γ = + + (α β)(α γ) (β α)(β γ) (γ α)(γ β) −π − − 5π− − − 3(β γ)+ π(γ α)+ 3 (α β) 2π = i − − − = . − h (α β)(β γ)(γ α) i −3√3 − − − Hence, 1 2π ∞ dx = . 3 √ Z0 1+ x 3 3

48 Evaluation of Fourier integrals

A Fourier integral is of the form

∞ eimx f(x) dx, m> 0, Z−∞

1. lim f(z) = 0, z →∞ 2. f(z) has no singularity along the real axis.

Remarks

1. The assumption m > 0 is not strictly essential. The evaluation method works even when m is negative or pure imaginary.

2. When f(z) has singularities on the real axis, the Cauchy principal value of the integral is considered.

49 Jordan Lemma

We consider the modulus of the integral for λ> 0

π iθ f(z)eiλz dz f(Reiθ) eiλRe R dθ

ZCR ≤ Z0 | || |

π λR sin θ max f(z) R e− dθ ≤ z CR | | Z0 ∈ π 2 λR sin θ = 2R max f(z) e− dθ z CR | | Z0 ∈ π 2 λR2θ 2R max f(z) e− π dθ ≤ z C | | 0 ∈ R Z π λR = 2R max f(z) (1 e− ), z C | | 2Rλ − ∈ R which tends to 0 as R , given that f(z) 0 as R . →∞ → →∞

50 51 To evaluate the Fourier integral, we integrate eimzf(z) along the closed contour C that consists of the upper half-circle CR and the diameter from R to R along the real axis. We then have − R eimzf(z) dz = eimxf(x) dx + eimzf(z) dz. C R C I Z− Z R Taking the limit R , the integral over C vanishes by virtue of → ∞ R the Jordan Lemma.

Lastly, we apply the Residue Theorem to obtain

∞ eimxf(x) dx = 2πi [sum of residues at all the isolated Z −∞ singularities of f in the upper half-plane] since C encloses all the singularities of f in the upper half-plane as R . →∞

52 Example

Evaluate the Fourier integral sin 2x ∞ dx. x2 + x + 1 Z−∞ Solution

It is easy to check that f(z) = 1 has no singularity along z2+z+1 the real axis and lim 1 = 0. The integrand has two simple z z2+z+1 →∞2πi 2πi poles, namely, z = e 3 in the upper half-plane and e− 3 in the lower half-plane. By virtue of the Jordan Lemma, we have

53 sin 2x e2ix e2iz ∞ dx = Im ∞ dx = Im dz, x2 + x + 1 x2 + x + 1 C z2 + z + 1 Z−∞ Z−∞ I where C is the union of the inﬁnitely large upper semi-circle and its diameter along the real axis.

Note that

e2iz e2iz 2πi = 2 Res 3 2 dz πi 2 , e IC z + z + 1 z + z + 1 ! 2πi e2iz e2ie 3 = 2πi = 2πi . 2πi 2z + 1 2πi 2e 3 + 1 z=e 3

Hence,

2πi 2ie 3 sin 2x e 2 √3 ∞ dx = Im 2πi  = πe sin 1. 2 2πi √ − Z x + x + 1 2e 3 + 1 − 3 −∞      

54 Example

Show that 1 π ∞ sin x2 dx = ∞ cos x2 dx = . Z0 Z0 2r2 Solution

55 π 2 R 2 4 2 2iθ 0= eiz dz = eix dx + eiR e iReiθ dθ IC Z0 Z0 0 2 iπ/2 + eir e eiπ/4 dr. ZR

Rearranging R R π/4 2 2 iπ/4 r2 iR2 cos 2θ R2 sin 2θ iθ (cos x +i sin x ) dx = e e− dr e − iRe dθ. Z0 Z0 −Z0 Next, we take the limit R . We recall the well-known result →∞ π i ∞ r2 √π iπ/4 1 π i π e 4 e− dr = e = + . Z0 2 2r2 2r2 2φ π Also, we use the transformation 2θ = φ and observe sin φ , 0 φ , ≥ π ≤ ≤ 2 to obtain

56 π/4 π/4 iR2 cos 2θ R2 sin2 θ iθ R2 sin 2θ e − iRe dθ e− R dθ 0 ≤ 0 Z Z π/2 R R2 sin φ = e− dφ 2 Z0 π/2 R 2R2φ/π e− dφ ≤ 2 Z0 π R2 = (1 e− ) 0 as R . 4R − −→ →∞ We then obtain 1 π 1 π ∞(cos x2 + i sin x2) dx = + i Z0 2r2 2r2 so that 1 π ∞ cos x2 dx = ∞ sin x2 dx = . Z0 Z0 2r2

57 Example

ln(x2 + 1) Evaluate ∞ . 2 dx Z0 x + 1

Hint: Use Log(i x) + Log(i + x) = Log(i2 x2) = ln(x2 +1)+ πi. − − Solution

Log(z + i) Consider around as shown. 2 dz C IC z + 1

58 Log(z + i) The only pole of in the upper half plane is the simple z2 + 1 pole z = i. Consider Log(z + i) 2πiRes ,i z2 + 1 ! (z i)Log(z + i) π2 = 2πi lim − = πLog 2i = π ln2+ i. z i (z i)(z + i) 2 → − Log(z + i) (ln R)R = 0 as 2 dz O 2 R ZCR z + 1 R ! → →∞ R Log(i x) R Log(x + i) Log(z + i) π2 + + = ln2+ 2 − dx 2 dx 2 dz π i Z0 x + 1 Z0 x + 1 ZCR z + 1 2 and Log(i x) + Log(i + x) = ln(x2 +1)+ πi. −

ln(x2 + 1) πi π2 dx π From ∞ dx + ∞ dx = π ln2+ i and ∞ = 0 x2 + 1 0 x2 + 1 2 0 1+ x2 2 so thatZ Z Z ln(x2 + 1) ∞ = ln 2 2 dx π . Z0 x + 1

59 Cauchy principal value of an improper integral

Suppose a real function f(x) is continuous everywhere in the interval [a, b] except at a point x0 inside the interval. The integral of f(x) over the interval [a, b] is an improper integral, which may be deﬁned as b x ǫ b 0− 1 f(x) dx = lim f(x) dx + f(x) dx , ǫ1, ǫ2 > 0. a ǫ1,ǫ2 0 " a x +ǫ # Z → Z Z 0 2 In many cases, the above limit exists only when ǫ1 = ǫ2, and does not exist otherwise.

60 Example

Consider the following improper integral 2 1 dx, 1 x 1 Z− − show that the Cauchy principal value of the integral exists, then ﬁnd the principal value.

Solution 2 1 Principal value of dx exists if the following limit exists. 1 x 1 Z − 1 ǫ 1 2 1 lim − dx + dx ǫ 0+ " 1 x 1 1+ǫ x 1 # → Z− Z − 1 ǫ −2 − = lim ln x 1 + ln x 1 ǫ 0+  | − | | − |  1 1+ǫ → − = lim [(ln ǫ ln 2) + (ln 1 ln ǫ )] = ln 2. ǫ 0+ − − − → 2 1 Hence, the principal value of dx exists and its value is 1 x 1 ln 2. Z− − − 61 Lemma

If f has a simple pole at z = c and Tr is the circular arc deﬁned by iθ Tr : z = c + re (θ θ θ ), 1 ≤ ≤ 2 then

lim f(z) dz = i(θ2 θ1)Res (f, c). r 0+ Tr − → Z In particular, for the semi-circular arc Sr

lim f(z) dz = iπRes(f, c). r 0+ Sr → Z

62 Proof

Since f has a simple pole at c,

a 1 ∞ k f(z)= − + a (z c) , 0 < z c < R for some R. z c k − | − | − kX=0 g(z) | {z } 1 For 0

For 0

g(z) dz M arc length of Tr = Mr(θ2 θ1) T ≤ · − Z r and so

lim g(z) dz = 0. r 0+ Tr → Z Finally,

1 θ2 1 θ2 dz = ireiθ dθ = i dθ = i(θ θ ) iθ 2 1 Tr z c θ re θ − Z − Z 1 Z 1 so that

lim f(z) dz = a 1i(θ2 θ1) = Res(f, c)i(θ2 θ1). r 0+ Tr − − − → Z

64 Example

Compute the principal value of xe2ix ∞ dx. x2 1 Z−∞ Solution −

The improper integral has singularities at x = 1. The principal ± value of the integral is deﬁned to be 2 1 r1 1 r2 R xe ix lim − − + − + 2 dx. R R 1+r1 1+r2! x 1 r ,r→∞0+ Z− Z− Z − 1 2→

65 Let ze2iz = I1 2 dz Sr z 1 Z 1 − ze2iz = I2 2 dz Sr z 1 Z 2 − ze2iz IR = dz. C z2 1 Z R − ze2iz Now, f(z)= is analytic inside the above closed contour. z2 1 − By the Cauchy Integral Theorem

1 r 1 r R 2ix − − 1 − 2 xe + + dx + I1 + I2 + IR = 0. R 1+r 1+r ! x2 1 Z− Z− 1 Z 2 −

66 z By the Jordan Lemma, and since 0 as z , so z2 1 → →∞ − lim IR = 0. R →∞

Since z = 1 are simple poles of f, ± lim I1 = iπRes(f, 1) = iπ lim (z + 1)f(z) r 0+ − − − z 1 1→ →− = ( iπ)e 2i/2. − − iπe2i Similarly, lim I2 = iπRes(f, 1) = − . r 0+ − 2 2→

xe2ix iπe 2i iπe2i P V ∞ dx = − + = iπ cos 2. x2 1 2 2 Z−∞ −

67 Poisson integral formula 1 f(s) f(z)= ds. 2πi C s z I − Here, C is the circle with radius r0 centered at the origin. Write s = r eiφ and z = reiθ, r > r . We choose z such that z z = r2 0 0 1 | 1| | | 0 and both z1 and z lie on the same ray so that 2 2 r0 iθ r0 ss z1 = e = = . r z z

68 Since z1 lies outside C, we have 1 1 1 f(z) = f(s) ds 2πi C s z − s z ! I − − 1 1 2π s s = f(s) dφ. 2π 0 s z − s z ! Z − − 1 The integrand can be expressed as s 1 s z r2 r2 = + = 0 − s z − 1 s/z s z s z s z 2 − − − − | − | r2 r2 2π f(r eiθ) and so f(reiθ)= 0 − 0 dφ. 2π 0 s z 2 Z | − |

69 Now s z 2 = r2 2r r cos(φ θ)+ r2 > 0 (from the cosine rule). | − | 0 − 0 − Taking the real part of f, where f = u + iv, we obtain

2π 2 2 1 r0 r u(r, θ) = − u(r0, φ) dφ, r < r0. 2π 0 r2 2r r cos(φ θ)+ r2 Z 0 − 0 − P (r ,r,φ θ) 0 − Knowing u(r0, φ) on| the boundary,{z u(r, θ) is} uniquely determined.

The kernel function P (r , r, φ θ) is called the Poisson kernel. 0 − r2 r2 s z ( ) = 0 = Re + P r0, r, φ θ − 2 − s z s z s z | − | s z − − = Re + s z s z s −+ z − = Re which is harmonic for z < r0. s z | | −