A.6 The Poisson Integral Formula
A.6.1 Formula for the Half Plane
In the case of the right half plane, we are faced with integration over a contour that becomes arbitrarily long. To deal with this, we will consider a class of functions with restricted behavior at infinity, and then use the bounding technique of (A.26) to estimate the integral over the infinite con- tour.
In particular, given a functionf, define m(R) =sup
θ
|f(Rejθ)|, θ∈[−π/2, π/2]. (A.40) Thenf(s)is said to be of classRif
Rlim
m(R)
R =0. (A.41)
For example, the functions considered in Examples A.4.1 and A.4.3 are in this class. More generally, iffis analytic and of bounded magnitude in the CRHP, thenfis of classR.
The Poisson integral formula for the half plane is given in the following result.
Theorem A.6.1 (Poisson Integral Formula for the Half Plane). Letfbe analytic in the CRHP and suppose thatfis of classR. Lets0 = σ0+jω0
be a point in the complex plane withσ0> 0. Then f(s0) = 1
π Z
−
f(jω) σ0
σ20+ (ω0−ω)2 dω. (A.42) Proof. Letfbe as in the statement of the theorem and lets0 =σ0+jω0be any point such thatσ0 > 0. Consider the clockwise oriented semicircular contour Cshown in Figure A.11, whereR is large enough so thats0 is interior to the contour.13 Thus Cconsists of the segment s = jω,ω ∈ [−R, R], together with the arcCRgiven bys=Rejθ,θ∈[−π/2, π/2].
Sincefis analytic on and insideC, then Cauchy’s integral formula (A.36) gives
f(s0) = − 1 2πj
I
C
f(s) s−s0
ds.
13Recall that the interior is the domainboundedby the curve, and it is defined independent of the orientation.
A.6 The Poisson Integral Formula 299
¯
FIGURE A.11. Contour for the Poisson integral formula
Now consider the point−s0, which is outsideC. Thus, the Cauchy integral theorem gives
0= 1 2πj
I
C
f(s) s+s0ds.
Adding the above two equations, we obtain f(s0) = − 1
2πj I
C
f(s) s0+s0
(s−s0)(s+s0)ds, (A.43) which can be decomposed into the sum of two integrals as
f(s0) = − 1 π
ZR
−R
f(jω) σ0
(jω−s0)(jω+s0)dω
− 1 πj
Z
CR
f(s) σ0
(s−s0)(s+s0)ds.
(A.44)
AsR→ ∞, the first integral in (A.44) becomes 1
π Z
−
f(jω) σ0
σ20+ (ω0−ω)2 dω. (A.45) Comparing to (A.42), it thus remains to show that the second integral in (A.44) vanishes asR→ ∞. Using (A.40) and the fact that, forRsufficiently large, the denominator in the second integral has magnitudeR2, we have
1 πj
Z
CR
f(s) σ0
(s−s0)(s+s0)ds ≤ 1
π
m(R)σ0π R R2 ,
which tends to zero as R → ∞ since fis of class Rand hence satisfies
(A.41). The result then follows.
Iffis analytic in the ORHP and on the imaginary axis except for singu- larities of a particular type, then the Poisson integral formula is still valid, as shown next.
300 Appendix A. Review of Complex Variable Theory Lemma A.6.2. Letfbe analytic in the CRHP, except for singular pointssk
on the imaginary axis that satisfy
slim sk
Res≥0
(s−sk)f(s) =0. (A.46) Suppose further that f is of class R. Then the Poisson integral formula (A.42) holds at each complex points0 =σ0+jω0withσ0> 0.
Proof. Consider the contour shown in Figure A.12, i.e., a semicircle of ra- diusRencircling the points0 and such that the portion of curve on the imaginary axis has semicircular indentations of radiusδinto the ORHP at each singularityskoff.
FIGURE A.12. Contour for with singularities on the -axis.
We next proceed as in the proof of Theorem A.6.1, but in this case the integral (A.43) will be decomposed into the sum of the integral over the semicircular contourCR, the integrals over the small semicircular contours of radiiδ, and the integrals over the remaining portions of the imaginary axis. This last sum of integrals over the portions of imaginary axis between the small semicircles will tend to (A.45) asR→ ∞andδ→0. The integral overCRvanishes as in the proof of Theorem A.6.1. It thus remains to show that the integrals over each semicircle of radiusδalso vanish asδ→0.
Consider then one of the semicirclesCδ in Figure A.12, centered atsk, say. On this contour,s=sk+δejθ,θ∈[−π/2, π/2]. Then
Z
Cδ
f(s)σ0
(s−s0)(s+s0)ds=j Zπ/2
−π/2
f(sk+δejθ)δejθσ0
(sk+δejθ−s0)(sk+δejθ+s0)dθ.
Note that (A.46) implies that limδ 0f(sk+δejθ)δejθ=0. Hence the inte- grand on the RHS above vanishes and the result follows.
Other forms of the Poisson formula are obtained by separating into real and imaginary parts. For example, iff(s) = u(σ, ω) +jv(σ, ω), then the
A.6 The Poisson Integral Formula 301 formula (A.42) gives two real equations of the same structure, namely
u(σ0, ω0) = 1 π
Z
−
u(0, ω) σ0
σ20+ (ω0−ω)2 dω, v(σ0, ω0) = 1
π Z
−
v(0, ω) σ0
σ20+ (ω0−ω)2 dω.
(A.47)
Sinceuandvare harmonic whenfis analytic, each of the formulae (A.47) is the Poisson integral formula for harmonic functions.
Note that the integrals in (A.47) are improper integrals of the form I=
Z
−
w(ω)dω.
Every such integral will be evaluated based on itsCauchy principal value, i.e.,
I= lim
R
ZR
−R
w(ω)dω.
Existence of the Cauchy principal value of an integral does not, in gen- eral, guarantee the existence of the two limits limR R0
−Rw(ω)dωand limR RR
0 w(ω)dω. However, ifw(ω)is even, i.e.,w(−ω) =w(ω), then existence of the Cauchy principal value implies existence of these two lim- its.
A useful result for a particular harmonic function is given next.14 Corollary A.6.3. Let f be analytic and nonzero in the CRHP except for possible zeros on the imaginary axis and/or zeros at infinity. Assume that logfis in classR. Then, at each complex points0=σ0+jω0,σ0> 0,
log|f(s0)|= 1 π
Z
−
log|f(jω)| σ0
σ20+ (ω0−ω)2 dω. (A.48) Proof. Iffis as in the statement of the theorem, then logfis analytic in the CRHP except for singularities at the imaginary zeros offand/or at zeros offat infinity. Iffhas imaginary zerossk, it is not difficult to prove that logfsatisfies (A.46), and hence Lemma A.6.2 shows that these singulari- ties do not affect the Poisson integral. Iffhas zeros at infinity, then logf has a singularity at infinity, but the contour of integration in Figure A.12 has an indentation around the point at infinity (i.e., the large semicircle into the ORHP). The fact that logfis in classRshows that the integral on this semicircle vanishes as the radius tends to infinity. Then (A.42) holds for logf(s)and (A.47) holds for log|f(s)|=Re logf(s).
14This result, to the best of our knowledge, first appeared in Freudenberg and Looze (1985).
302 Appendix A. Review of Complex Variable Theory Note that zeros offat infinity can be treated as zeros offon the imagi- nary axis, since both types of zeros are in fact singularities of logfon the contour of integration that encircles the ORHP. The procedure that we take to deal with these singularities consists of two steps: first, indentations around these singularities have to be made on the contour of integration;
second, precautions should be taken to show that the integrals on those indentations converge (go to zero in our case) as the indentations vanish, i.e., condition (A.46) is assumed for imaginary zeros, and the property of logfbeing in classRis assumed for zeros at infinity.