1. the Calculus of Residues 1.1. the Residue Theorem. Let Ω ⊂ C Be A

MATH 722, COMPLEX ANALYSIS, SPRING 2009 PART 3

1. The calculus of residues 1.1. The Residue Theorem.

Let Ω ⊂ C be a connected open set, and let Z ⊂ Ω be a discrete subset. Thus Z is either a ﬁnite set of points, or a countable set with no limit point in Ω. Write Z = {a1, a2,...}. Let f :Ω\Z → C be a holomorphic function. Then at each point aj ∈ Z, the function f has one of three possible kinds of behavior:

0 (i) f can have a removable singularity, so that in some small punctured disk D (, aj), we can P∞ n write f(z) = n=0 cn(z − aj) . In this case f extends to a holomorphic function on the whole disk D(, aj).

0 (ii) f can have a pole of order N ≥ 1, so that in some small punctured disk D (, aj), we can P∞ n write f(z) = n=−N cn(z − aj) , with c−N 6= 0.

0 (iii) f can have an essential singularity at aj, so that in some small punctured disk D (, aj), we P∞ n can write f(z) = n=−∞ cn(z − aj) , with infinitely many non-zero negative coefficients. Definition 1.1. If ρ < , the number 1 Z Rj = f(ζ) dζ 2πi |ζ−aj |=ρ is called the residue of f at the point aj.

Note that this number is independent of the radius ρ, and equals the coeﬃcient a−1 in the Laurent expansion of f about the point aj. The residue is sometimes denoted by “Resz=aj f(z)”. If the function f has a removable singularity at aj, then Resz=aj f(z) = 0.

Let Γ be a piecewise differentiable, simple closed curve in Ω. Then C\Γ consists of two connected components, one of which is bounded, and the other unbounded. We call the bounded component the interior of Γ, and write it Int(Γ). If the interior of Γ is also a subset of Ω, then the curve Γ does not ‘go around’ any hole in Ω. Let Z = {aj} be a discrete set of points in Ω, and suppose that Γ does not pass through any of these points. Then Z ∩ Int(Γ) is a finite set of points {a1, . . . , am}. We can find radii {ρ1, . . . , ρm} so that each disk D(aj, ρj) is contained in the interior of Γ, and so that the disks {D(aj, ρj)} are iθ mutually disjoint. Let γj be the closed curve in Ω parameteried by γj(t) = aj + ρje , 0 ≤ θ ≤ 2π, If f is holomorphic on Ω \ Z, it follows that

m Z X Z f(ζ) dζ = f(ζ) dζ. Γ j=1 γj From this, we easily obtain the following. 1 COMPLEX ANALYSIS 2

Theorem 1.2 (Residue Theorem). Let Ω ⊂ C be a connected open set, and suppose that f(z) is holomorphic for z ∈ Ω except for z belonging to a discrete set Z = {aj} ⊂ Ω. Let Γ be a simple closed curve in Ω with the property that Int(Γ) ⊂ Ω. If Γ does not pass through any of the points {aj}, then 1 Z X f(ζ) dζ = Resz=aj f(z), 2πi Γ aj ∈int(Γ) where the sum is over the ﬁnite number of points {a1, . . . , am} contained in the interior of Γ. 1.2. The argument priniple. Recall that if Ω ⊂ C is a connected open set, a function f is meromorphic on Ω if there exists a discrete set E = {aj} ⊂ Ω so that f is holomorphic on Ω \ E, and f has either a removable singularity or a pole at each point aj. If f is not identically zero and is meromorphic on Ω, then g(z) = f(z)−1 is also meromorphic. In fact, g(z) is holomorphic at every point where f(z) 6= 0, g has poles at the points where f has zeros, and g has removable singularities (with value 0) at the poles of f. The following is then clear. Proposition 1.3. The set of meromorphic functions M(Ω) on a connected open set is a ﬁeld. Moreover, if f ∈ M(Ω), then f 0 ∈ M(Ω). We also have the following easy calculation. Proposition 1.4. Suppose that f is meromorphic on Ω. Then if a ∈ Ω,  0 if f is holomorphic and non-zero at a, f 0  Res (z) = N if f has a zero of order N at a, z=a f −N if f has a pole of order N at a. Proof. If f is holomorphic and non-zero at z = a, then f −1(z) is also holomorphic near a, and so f 0/f is holomorphic near a. It follows that the residue is zero. If f has a zero of order N, then near a we can write f(z) = (z − a)N g(z) with g holomorphic near a and g(a) 6= 0. Then f 0(z) = N(z − a)N−1g(z) + (z − a)N g0(z), so near z = a we have f 0 N(z − a)N−1g(z) + (z − a)N g0(z) N g0(z) (z) = = + . f (z − a)N g(z) (z − a) g(z) Since g0/g is holomorphic near a, it follows that f 0/f has a simple pole at z = a and the residue is N. If f has a pole of order N, then near a we can write f(z) = (z − a)−N g(z) with g holomorphic near a and g(a) 6= 0. Then f 0(z) = −N(z − a)−N−1g(z) + (z − a)−N g0(z), so near z = a we have f 0 −N(z − a)−N−1g(z) + (z − a)−N g0(z) −N g0(z) (z) = = + . f (z − a)−N g(z) (z − a) g(z) Since g0/g is holomorphic near a, it follows that f 0/f has a simple pole at z = a and the residue is −N. If we combine this Proposition with the Residue Theorem, we obtain: Theorem 1.5 (The Argument Principle). Let Ω ⊂ C be a connected open set, and let f be meromorphic on Ω. Let Γ be a closed curve in Ω which does not pass through any zero or pole of f, and such that the interior of Γ is contained in Ω. Then 1 Z f 0(ζ) dζ = Z(Γ) − P (Γ) 2πi Γ f(ζ) where Z(Γ) is the number of zeros of f (counted with multiplicity) inside Γ, and P (Γ) is the number of poles of f (counted with multiplicity) inside Γ. COMPLEX ANALYSIS 3

d f 0 If we could deﬁne a branch of the logarithm on the range of f, then dz (log(f)) = f . Thus R f 0(ζ) Γ f(ζ) dζ is the change in log(f) as we go around the curve Γ. However, the real part of the logarithm does not change as we go around a closed curve, but the imaginary part gives the total 1 R f 0(ζ) change in the angle. Thus 2πi Γ f(ζ) dζ gives the total number of times that the curve t → f(Γ(t)) winds around the origin. 3/4/09 As an application, we have

Theorem 1.6 (Rouché’sTheorem). Let Ω ⊂ C be a connected open set, and let f and g be holomorphic on Ω. Let Γ ⊂ Ω be a closed curve such that the interior of Γ is contained in Ω. If |f(z) − g(z)| < |f(z)| for all z ∈ Γ, then f and g have the same number of zeros on Int(Γ). Proof. It follows from the assumption that neither f(z) nor g(z) can equal zero for z ∈ Γ. In particular, f is not identically zero, and so the function F (z) = g(z)/f(z) is meromorphic on Ω. Moreover, |F (z) − 1| < 1 for z ∈ Γ. Thus the image under F of the curve Γ is a compact subset of the open disk D(1, 1) centered at 1 with radius 1. Let log(w) be a branch of the logarithm defined on D(1, 1). Then log(F (z)) is defined and holomorphic in a connected open neighborhood U of Γ. Moreover, for z ∈ U, we have d F 0(z) log(F (z) = . dz F (z) Thus Z F 0(ζ) Z d dζ = log(F (ζ) dζ = 0. Γ F (ζ) Γ dζ But then it follows from Theorem 1.5 that the number of zeros of F inside Γ equals the number of poles of F inside Γ. Since the zeros of F are just the zeros of g, and the poles of F are just the zeros of f, this gives the desired equality, and completes the proof.

n n n−1 Example: Let f(z) = z and let g(z) = z + a1z + ··· + an−1z + an, so that g is a monic polynomial. If |z| = R > 1 we have n−1 n−1 |f(z) − g(z)| = |a1z + ··· + an−1z + an| ≤ |a1|R + ··· + |an−1|R + |an| n−1 n−1 ≤ max{|aj|} (R + ··· + R + 1) ≤ n max{|aj|} R n max{|a |} n max{|a |} = j Rn = j |f(z)| R R

Thus if R > n max{|aj|}, the functions f and g have the same number of zeros in the disk of radius R. But since f clearly has n zeros, it follows that the monic polynomial g(z) also has n zeros. This is a (slightly sharper) version of the fundamental theorem of algbera.

Exercise 1. Show that the equation z7 − 2z5 + 6z3 − z + 1 = 0 has exactly three solutions for |z| < 1.

As a further application, we can give another derivation of the results on the local behavior of a holomorphic mapping. Lemma 1.7. Suppose that f is holomorphic in the disk D(0,R), that f has a zero of order n at z = 0, and that f has no other zeros in this disk. Then there exists > 0 so that if |w| < , the equation f(z) = w has exactly n solutions in the disk D(0,R). Also COMPLEX ANALYSIS 4

Proof. Choose any 0 < r < R. Then by compactness, there exists > 0 so that |f(reiθ)| > for 0 ≤ θ ≤ 2π. Thus if |w| < , the function g(z) = f(z) − w has no zeros on the circle |ζ| = r. It 1 R g0(ζ) follows from the argument principle that 2πi |ζ|=r g(ζ) dζ is the number of zeros of g(z) inside the disk of radius r. Of course, this is just the number of solutions of f(z) = w inside this disk. But now 1 Z g0(ζ) 1 Z f 0(ζ) dζ = dζ 2πi |ζ|=r g(ζ) 2πi |ζ|=r f(ζ) − w is integer-valued, is continuous in w, and takes the value n when w = 0. It follows that the integral is independent of w for small |w|, and this completes the proof. Exercise 2. Suppose that f is holomorphic in the disk D(0,R), that f has a zero of order n at z = 0, and that f has no other zeros in this disk. Put 1 Z ζmf 0(ζ) gm(w) = dζ. 2πi |ζ|=r f(ζ) − w th Show that gm is holomorphic in w for |w| suﬃciently small, and equals the sum of the m powers of the n roots of the equation f(z) = w. Exercise 3. In particular, if f is as in Exercise 2, and if n = 1 so that f 0(0) 6= 0, show that 1 Z ζf 0(ζ) g1(w) = dζ 2πi |ζ|=r f(ζ) − w is a holomorphic function deﬁned for |w| small such that f(g(w)) = w.

2. Evaluation of integrals

2.1. Calculation of residues. 3/6/09 Suppose that f is holomorphic in D0(a, R) and has a pole of order N at a. Then for z ∈ D0(a, R) we can write f(z) = (z − a)−N g(z) where g is holomorphic on D(a, R). We want to calculate the residue of f at a. If we write N−1 N g(z) = c0 + c1(z − a) + ··· + cN−1(z − a) + cN (z − a) + ··· , then ∞ c0 c1 cN−1 X f(z) = + + ··· + + c (z − a)j, (z − a)N (z − a)N−1 (z − a) N+j j=0 and so the residue of f at a is the constant cN−1. But from the theory of Taylor expansions, (N−1) cN−1 = (N − 1)! g (a). Thus we see that (N−1) Resz=af(z) = (N − 1)! g (a). ez ez Example: Let f(z) = = . z3 − z4 z3(1 − z) (a) Compute the residue of f at z = 0 and z = 1. 1 Z ez (b) Evaluate dz. 2πi 1 z3 − z4 |z|= 2 1 Z ez (c) Evaluate 3 4 dz. 2πi |z|=3 z − z 1 Z ez (d) Evaluate dz. 2πi 7 z3 − z4 |z−4|= 2 COMPLEX ANALYSIS 5

Solutions: g(z) ez (a1) f has a pole of order 1 at z = 1, and we can write f(z) = where g(z) = . Thus z − 1 z3 Resz=−1f(z) = g(1) = e. h(z) ez (a2) f has a pole of order 3 at z = 0, and we can write f(z) = where h(z) = = (z−1)−1ez. z3 z − 1 00 Thus Resz=1f(z) = 2h (0). But h0(z) = −(z − 1)−2ez + (z − 1)−1ez h00(z) = 2(z − 1)−3ez − 2(z − 1)−2ez + (z − 1)−1ez.

00 Thus h (0) = −2 − 2 − 1 = −5, and so Resz=0f(z) = −10.

1 (b) The only pole inside |z| = 2 is at z = 0 so the sum of the residues inside is e. (c) Both poles are inside |z| = 3, so the sum of the residues is e − 10.

7 (d) The only pole inside |z − 4| = 2 is at z = 1, so the sum of the residues is −10.

2.2. Evaluation of deﬁnite integrals.

“The technique can be learned at the hand of typical examples, but even complete mastery does not guarantee success.” (from the text by Ahlfors)

2.2.1. Integrals involving rational functions. Z +∞ dx Example 1: Evaluate 4 . −∞ 1 + x The meromorphic function f(z) = (1 + z4)−1 has simple poles at the fourth roots of −1. There are the points 1 1 1 1 a1 = √ (+1 + i), a2 = √ (−1 + i), a3 = √ (−1 − i), a4 = √ (+1 − i). 2 2 2 2 We have

4 −1 Resz=a1 (1 + z ) = (a1 − a2)(a1 − a3)(a1 − a4) = R1, 4 −1 Resz=a2 (1 + z ) = (a2 − a1)(a2 − a3)(a2 − a4) = R2.

Let R >> 1 and consider the closed curve in Γ ⊂ C consisting of the interval [−R, +R] on the real axis, together with the semi circle in the upper half plane with radius R. Thus Γ = γ1 + γ2 where

γ1(t) = t, −R ≤ t ≤ R, it γ2(t) = Re , 0 ≤ t ≤ π.

The residues of f inside this curve are at a1 and at a2. Thus for all R >> 1, we have Z dz Z dz Z dz 2πi[R1 + R2] = 4 = 4 + 4 . Γ 1 + z γ1 1 + z γ2 1 + z COMPLEX ANALYSIS 6

Z dz Z +R dt Z dz Z π iReit dt Z dz But = , and = . We will show lim = 0, 4 4 4 4 4it R→∞ 4 γ1 1 + z −R 1 + t γ2 1 + z 0 1 + R e γ2 1 + z Z +∞ dx so that 4 = 2πi[R1 + R2]. But −∞ 1 + x Z dz Z π R dt ≤ 4 4 it γ2 1 + z 0 |1 + R e | Z π R dt πR ≤ 4 = 4 , 0 R − 1 R − 1 and this clearly goes to zero as R tends to inﬁnity. More generally, let Q(z) be a polynomial of degree n ≥ 2 with no zeros on the real axis, and let P (z) P (z) be a polynomial of degree at most n − 2. If {R1,...,Rk} are the residues of R(z) = Q(z) in the upper half plane, the same technique shows that k Z +∞ P (x) X dx = 2πi R . Q(x) j −∞ j=1

Z +∞ x sin(x) Example 2: Evaluate 2 dx. Note that this improper integral does not converge 3/9/09 −∞ 1 + x absolutely; the integrand decays only like |x|−1 for |x| large. We deﬁne the integral as Z +∞ x sin(x) Z +N x sin(x) 2 dx = lim 2 dx. −∞ 1 + x M,M→+∞ −M 1 + x x sin(x) h xeix i Z +∞ xeix Note that 2 = =m 2 . Thus we can compute 2 dx, which is deﬁned in the 1 + x 1 + x −∞ 1 + x zeiz z eiz same way. Note that the only pole of = in the upper half-plane is at z = i, 1 + z2 (z + i)(z − i) 1 −1 and the residue at that point is 2 e .

Let Γ = ΓM,N,y be the closed rectangle with vertices at N, N + iy, −M + iy, and −M. If M, N, and y are all greater than 1, we have Z iz z e 1 −1 πi 2 dz = 2πi e = . Γ 1 + z 2 e We break up Γ into four pieces by setting

γ1(t) = t −M ≤ t ≤ N,

γ2,y(t) = N + it 0 ≤ t ≤ y,

γ3(t) = −t + iy −N ≤ t ≤ M,

γ4,y(t) = −M + i(y − t)t 0 ≤ t ≤ y. We have Z zeiz Z M (−t + iy)ei(−t+iy) Z N (t + iy)eit e−y 2 dz = − 2 i dt = − 2 2 dt. γ3 1 + z −N 1 + (−t + iy) −M 1 + t − 2ity − y 1 2 2 Let A = max{M,N}. We have 2|t|y ≤ 4 y + 4t , and so for t ∈ [−M,N], 3 3 |1 + t2 − 2ity − y2| ≥ y2 − 1 − t2 − 2|t|y ≥ y2 − 1 − 5t2 ≥ y2 − 1 − 5A2. 4 4 COMPLEX ANALYSIS 7

3 2 2 1 2 If we choose y so large that y > A and 4 y − 1 − 5A ≥ 2 y , we have Z zeiz Z N |t| + y 2y dz ≤ e−y dt ≤ e−y (M + N). 1 + z2 y2 − 1 − t2 − 2|t||y| 1 2 γ3 −M 2 y Thus for M and N ﬁxed, we have Z zeiz lim dz = 0. y→∞ 2 γ3 1 + z

Next, we have Z iz Z y iN −t Z y −t Z y −t ze (N + it) e e dt iN e iN t e 2 dz = 2 = Ne 2 dt + ie 2 dt. γ2,y 1 + z 0 1 + (N + it) 0 1 + (N + it) 0 1 + (N + it) Since |1 + (N + it)2| ≥ |N + it|2 − 1 ≥ N 2 − 1, we have Z zeiz N Z ∞ 1 Z ∞ dz ≤ e−t dt + t e−t dt → 0 2 2 2 γ2,y 1 + z N − 1 0 N − 1 0 Z zeiz as N → ∞, independently of y. The same is true for dz . 2 γ4,y 1 + z Thus if we ﬁrst choose M and N large, we can make the second and fourth integrals small, independently of y. With M and N then ﬁxed, we can choose y large to make the third integral small. Thus we have Z +∞ x sin(x) π 2 dx = −∞ 1 + x e

Z ∞ sin(x) Example 3: Evaluate dx. 0 x sin(x) −1 The function x has a continuous extension to x = 0, but decays only like |x| as |x| → +∞, and the integral is not absolutely convergent. Since the integrand is even, we write " # Z ∞ sin(x) Z sin(x) Z eix dx = lim dx = lim =m dx . 0 x →0+ ≤|x|≤R x →0+ ≤|x|≤R x R→+∞ R→+∞ We introduce four curves

γ1(t) = t ≤ t ≤ R, it γ2(t) = Re = R cos(t) + iR sin(t) 0 ≤ t ≤ π,

γ3(t) = −R + t 0 ≤ t ≤ R − , it γ4(t) = e π ≤ t ≤ 2π.

eiz and let Γ = γ1 + γ2 + γ3 + γ4. The meromorphic function z has one pole inside Γ, and the residue at that point is 1. Thus Z eiz dz = 2πi. Γ z Also Z eix Z eiz dx = , dz. ≤|x|≤R x γ3+γ1 z COMPLEX ANALYSIS 8

Now Z iz Z π Z π e exp i(R cos(t) + iR sin(t) it dz = iRe dt ≤ exp − R sin(t)] dt z Reit γ2 0 0 π −1/2 Z 2 −R Z π = exp − R sin(t)] dt + exp − R sin(t)] dt π −1/2 0 2 +R π −1/2 Z 2 +R + exp − R sin(t)] dt π −1/2 2 −R = I + II + III Since sin(t) ≥ 0 for 0 ≤ t ≤ π, the integrand is bounded by 1, and we can estimate the third − 1 integral by the length of the interval; i.e. III ≤ 2R 2 . The ﬁrst two integrals are equal, and here we can integrate by parts since we are staying away from the place where cos(t) = 0. Thus

π −R−1/2 π −R−1/2 Z 2 Z 2 d h i −1 I = exp − R sin(t)] dt = exp − R sin(t) dt 0 0 dt R cos(t) 1 π −R− 2 2 π −R−1/2 − exp − R sin(t) Z 2 sin(t) = − exp − R sin(t) dt. 2 R cos(t) 0 R cos (t) 0 Now for R large, it follows from Taylor approximations that

π − 1 − 1 −1 −2 sin − R 2 = cos R 2 = 1 − R + O(R ) 2 π − 1 − 1 − 1 − 3 cos − R 2 = − sin R 2 = −R 2 + O(R 2 ). 2 Thus 1 π −R− 2 √ − exp − R sin(t) 2 exp − R 1 ≈ √ − → 0 R cos(t) R R 0 as R → ∞. On the other hand,

π −R−1/2 π −R−1/2 Z 2 sin(t) 1 Z 2 sin(t) − exp − R sin(t) dt ≤ dt 2 2 0 R cos (t) R 0 cos (t) 1 π −R− 2 −1 2 = R cos(t) 0 1 1 = − 1 R π − 2 R cos( 2 − R ) 1 1 ≈ + √ → 0 R R as R → ∞. It follows that Z iz Z π e 1 dz ≤ exp − R sin(t) dt = O √ −→ 0 γ2 z 0 R as R → +∞. COMPLEX ANALYSIS 9

iz It remains to study the integral R e dz. We have γ4 z Z iz Z 2π i(cos(t)+i sin(t)) e e it dz = it ie dt γ4 z π e Z 2π = i exp [− sin(t) + i cos(t)] dt −→ πi π as → 0+. Putting all our estimates together, it follows that Z ∞ sin(x) π dx = . 0 x 2 √ Z ∞ 3 x Example 4: Evaluate 2 dx. 0 1 + x 5 − 3 In this case,√ the integrand behaves like x for x large, and so the integral converges. However, 3 z the function 1+z2 is not single-valued on the complex plane. We should write √ 3 z 1 = (1 + z2)−1 exp log(z) 1 + z2 3 Let us deﬁne a branch of log(z) deﬁne on C minus the ray making an angle − 2 with the positive x-axis, with > 0. We can make this branch equal to the usual function log(x) on the positive real −i 1 1 i 2π− axis. However, if z = re , we have log(z) = log(r) + i(2π − ), and so z 3 = |z| 3 e 3 . Now consider the following curves:

γ1(t) = t ≤ t ≤ R, it γ2(t) = Re 0 ≤ t ≤ 2π − , i(2π−) γ3(t) = (R − t)e 0 ≤ t ≤ R − , i(2π−t) γ4(t) = e ≤ t ≤ 2π. √ √ 3 z 3 z If Γ = γ1 + γ2 + γ3 + γ4, then the function 1+z2 = (z+i)(z−i) has two poles inside Γ at the points z = i and z = −i. With our choice of branch of the logarithm, we have √ √ 1 1 π h π i 3 i 3 i = exp log(i) = exp log(1) + i = exp i = + , 3 3 6 6 2 2 √ 1 1 π h π i 3 −i = exp log(−i) = exp log(1) + i = exp i = i, 3 3 2 2 and so √ √ √ "√ # Z 3 z h 3 i 3 −ii h h π i h π ii 3 i 2 dz = 2πi + = π exp i − exp i = π − . Γ 1 + z 2i −2i 6 2 2 2 As we did before, we can show that √ √ Z 3 z Z 3 z lim dz = lim dz = 0. R→∞ 2 →0 2 γ2 1 + z γ4 1 + z On the other hand, √ √ Z 3 z Z R 3 x 2 dz = 2 dx γ 1 + z 1 + x 1 √ √ 3 R 3 Z z 2π− Z x i 3 2 dz = −e 2 dx γ1 1 + z 1 + x COMPLEX ANALYSIS 10

Thus √ "√ # 2π 2π Z ∞ 3 x 3 i 1 − cos − i sin 2 dx = π − , 3 3 0 1 + x 2 2 or equivalently " √ # √ "√ # 3 i 3 Z ∞ 3 x 3 i − 2 dx = π − . 2 2 0 1 + x 2 2 Thus √ √ Z ∞ 3 x 3 − i π √ √ 2 dx = π = . 0 1 + x 3 − i 3 3

2 Example 4: Find the Fourier transform of the function f(x) = e−πx .

The Fourier transform of an integrable function f(x) is Z +∞ F[f](ξ) = fb(ξ) = e−2πix ξ f(x) dx. −∞

Z 2 Thus we must evaluate e2πixξe−πx dx. We have R Z 2 Z 2 e2πixξe−πx dx = e−π[x −2ix ξ] dx R R 2 Z 2 2 = e−πξ e−π[x −2ix ξ−ξ ] dx R 2 Z 2 = e−πξ e−π(x−iξ) dx. R Formally, we would like to make a change of variables, and replace x − iξ by t. If we could do this, we would get Z 2 2 Z 2 2 e2πixξe−πx dx = e−πξ e−π(t) dt = e−πξ R R since +∞ Z 2 e−πt dt = 1. −∞ We can justify this change of variables by appealing to Cauchy’s theorem, and changing the contour Z 2 in e−π(z−iξ) dz from the real axis to the horizontal line =m[z] = iξ. R Z +∞ 1 i[ 1 x3+sx] Example 5: Discuss the behavior of the Airy function A(s) = e 3 dx. 2π −∞ Theorem 2.1. Suppose that s > 0. Then as s → +∞, 1 − 1 2 3 h − 3 i Ai(s) = √ s 4 exp − s 2 1 + O(s 4 ) ; 2 π 3

1 − 1 2 3 π h − 3 i Ai(−s) = √ s 4 cos s 2 − 1 + O(s 4 ) . π 3 4 Remark: Note that for s > 0 we get exponential decay, while for s < 0 we get polynomial decay with oscillation.