Jump to ContentJump to Main Navigation
Foundation of Statistical Energy Analysis in Vibroacoustics$

A. Le Bot

Print publication date: 2015

Print ISBN-13: 9780198729235

Published to Oxford Scholarship Online: August 2015

DOI: 10.1093/acprof:oso/9780198729235.001.0001

Show Summary Details
Page of

PRINTED FROM OXFORD SCHOLARSHIP ONLINE (www.oxfordscholarship.com). (c) Copyright Oxford University Press, 2017. All Rights Reserved. Under the terms of the licence agreement, an individual user may print out a PDF of a single chapter of a monograph in OSO for personal use (for details see http://www.oxfordscholarship.com/page/privacy-policy). Subscriber: null; date: 25 February 2017

(p.291) Appendix C Useful Integrals

(p.291) Appendix C Useful Integrals

Source:
Foundation of Statistical Energy Analysis in Vibroacoustics
Publisher:
Oxford University Press

The frequency response function of a simple resonator is

(C.1)
Hi(ω)=1miωi21+2ıζiωωiω2ωi2

where ζi>0 is the damping ratio, ωi>0 the natural frequency, and mi>0 the modal mass. This is a complex-valued function defined on R.

The following integral is given in (Newland 1975, Appendix 1) and (Gradshteyn et al. 2000, Section 3.112, eqn (5)).

(C.2)
b0+ıωb1ω2b2ıω3b3a0+ıωa1ω2a2ıω3a3+ω4a42dω=πa0b32(a1a2a0a3)+a0a1a4(b222b1b3)+a0a3a4(b122b0b2)+a4b02(a2a3a1a4)a0a4(a1a2a3a0a32a12a4)

The above integral holds if all roots of the denominator lie in the upper half-plane Im(z)>0 (stable system).

Calculation of ωpHinHjmdω

Equation (C.1) shows that Hi is a rational function with two poles, ωiα1 and ωiα2,

(C.3)
{αk=ıζi±1ζi2when 0ζi<1αk=ı(ζi±ζi21)when 1ζi

where k=1,2 depending on the plus or minus sign. It is clear that Im(α1,2)>0 and therefore the two poles of Hi are located in the half-plane Im(z)>0.

(p.292) Let p, n, and m be three non-negative integers. The function ωωpHin(ω)Hjm(ω) is rational and therefore is holomorphic on Cz1,z2,z3,z4 where zi, i=1,,4 are the four poles of Hi and Hj. They are all located in the half-plane Im(z)>0.

Since Hi has degree 2, the rational function ωpHinHjm has degree p2(n+m). The condition lim|ω|ωp+1Hin(ω)Hjm(ω)=0 is fulfilled under the assumption 2(n+m)p2. Using Jordan’s lemma,

(C.4)
ωpHin(ω)Hjm(ω)dω=0

Calculation of I(ζ)=dz(1z2)2+4ζ2z2

Let Q(z) be the integrand of I. This is a rational function Q(z)=1/P(z2) where P is the polynomial,

(C.5)
P(z)=(1z)2+4ζ2z=z22(12ζ2)z+1

The poles of Q are found by solving the biquadratic equation P(z2)=0. The discriminant of P is

(C.6)
Δ=4ζ2(ζ21)

We consider successively the three cases ζ<1, ζ>1, and ζ=1.

When ζ<1, P has two roots,

(C.7)
zi=(12ζ2)±2ıζ1ζ2=(1ζ2±ıζ)2

where i=1,2. The poles of Q are the square roots of zi. These are α0, α¯0, α0, and α¯0 where

(C.8)
α0=1ζ2+ıζ

Factorizing Q gives

(C.9)
Q(z)=1(zα0)(z+α0)(zα¯0)(z+α¯0)

In the complex plane, the two poles located in the half-plane Im(z)>0 are α0 and α¯0. The closed path γ‎ is chosen as shown in Fig. C.1. Then,

(C.10)
I=2ıπ[Res(Q,α0)+Res(Q,α¯0)]
Appendix C Useful Integrals

Figure C.1 Closed path of integration for I.

Since all poles are simple, their residues are calculated by

(C.11)
Res(Q,αi)=limzαi(zαi)Q(z)

where αi=α0, α0. One obtains

Res(Q,α0)=12α0×2ıIm(α0)×2Re(α0)=18ıζ1ζ21ζ2+ıζ

(p.293) and

Res(Q,α¯0)=12Re(α0)×2ıIm(α0)×2α¯0=18ıζ1ζ2[1ζ2ıζ]

By summing the two above expressions,

(C.12)
I=2ıπ8ıζ1ζ211ζ2+ıζ+11ζ2ıζ

And finally,

(C.13)
I=π2ζ

When ζ>1, Δ>0 and the two roots of P are

(C.14)
zi=(12ζ2)±2ζζ21=ζ±ζ212

with i=1,2. The four poles of Q are then α1, α1, α2, and α2 where

(C.15)
α1=ıζ+ζ21
(C.16)
α2=ıζζ21

and Q becomes

(C.17)
Q(z)=1(zα1)(z+α1)(zα2)(z+α2)

(p.294) By choosing the same integration path (see Fig. C.1), the two residues to be computed are

Res(Q,α1)=12α1(α12α22)=18ıζ1ζ2ζ+ζ21

and

Res(Q,α2)=12α2(α22α12)=18ıζ1ζ2ζζ21

Hence,

(C.18)
I=2ıπ8ıζ1ζ21ζζ211ζ+ζ21

And finally,

(C.19)
I=π2ζ

The last case ζ=1 is straightforward. Since Q(z)=1/(1+z2)2, the change of variable z=tanu, dz=(1+tan2u)du gives

(C.20)
I=π/2π/2(1+tan2u)(1+tan2u)2du=π/2π/2cos2udu=π2

The same result is therefore obtained for all cases,

(C.21)
dz(1z2)2+4ζ2z2=π2ζ

for all ζ>0. This integral is also given in (Gradshteyn et al. 2000, Section 3.112, eqn (3)).

Application to |Hi|2dω

Substituting the expression of Hi(ω) given by eqn (C.1),

(C.22)
|Hi|2(ω)dω=1mi2ωi4dω1+2ıζiωωiω2ωi22

By the change of variable z=ω/ωi,

(C.23)
|Hi|2(ω)dω=1mi2ωi3dz(1z2)2+4ζ2z2=I(ζ)m2ωi3

(p.295) And finally,

(C.24)
|Hi|2(ω)dω=π2ζimi2ωi3

Application to ω2|Hi|2dω

Again with eqn (C.1) and the change of variable z=ω/ωi, one obtains

(C.25)
ω2|H|2(ω)dω=2m2ω00z2dz(1z2)2+4ζ2z2

The factor 2 stems from the fact that the integrand is odd. Factorizing z4 and performing the change of variable u=1/z gives

(C.26)
ω2|Hi|2(ω)dω=2m2ωi0dz/z2(1/z21)2+4ζ2/z2
(C.27)
=2mi2ωi0du(u21)2+4ζ2u2
(C.28)
=1mi2ωiI(ζ)

And finally,

(C.29)
ω2|Hi|2(ω)dω=π2ζimi2ωi

Application to ıωHidω

Multiplying the numerator and denominator of ıωHi(ω) by Hi gives

(C.30)
ıωHi(ω)dω=ıωmiωi212ıζiωωiω2ωi2|Hi|2(ω)dω

where the limits of integration are arbitrary and finite. By the change of variable z=ω/ωi,

(C.31)
ıωHi(ω)dω=2ζimiz21z22+4ζ2z2dz+ımiz(1z2)1z22+4ζ2z2dz

The first integral of the right-hand side is convergent when the limits become infinite and its value is I(ζ). Therefore,

(C.32)
Re[ıωHi(ω)]dω=πmi

(p.296) But the second integral is clearly divergent at infinite. However, the integrand is an odd function, so

(C.33)
XXIm[ıωHi(ω)]dω=0for\, allX

We shall write

(C.34)
ıωHi(ω)dω=πmi

which must be understood in the sense of principal values.

Calculation of Jn(r,ζ1,ζ2)=ız2n+1dz(1+2ıζ2rzr2z2)((1z2)2+4ζ12z2)

C.6

The integer n has values 0 or 1. Let Q(z) be the integrand of Jn. The function zQ(z) is rational and may be written as

(C.35)
Q(z)=ız2n+1Pζ2(rz)Pζ1(z)Pζ1(z)

where

(C.36)
Pζi(z)=1+2ıζizz2

Since Pζi may be factorized as

(C.37)
Pζi(z)=(zαi)(zβi)

where

(C.38)
{αi=ıζi+1ζi2, βi=ıζi1ζi2  if|ζi|<1αi=ı(ζi+ζi21),βi=ı(ζiζi21)  otherwise

we get

(C.39)
Q(z)=ız2n+1(rzα2)(rzβ2)(zα1)(zβ1)(zα¯1)(zβ¯1)

The six poles of Q are therefore α1, β1, α2, β2, α¯1, and β¯1. The poles are plotted in Fig. C.2. In all cases, only two poles are in the half-plane Im(z)<0, while the four others are in Im(z)>0. The better choice of closed path to minimize the number of residues to be calculated is to turn clockwise in the lower half-plane as shown in Fig. C.2. Application of Jordan’s lemma then leads to

(C.40)
Jn=2ıπ[Res(Q,α¯1)+Res(Q,β¯1)]

(p.297) where

(C.41)
Res(Q,α¯1)=ıα¯12n+1(rα¯1α2)(rα¯1β2)(α¯1α1)(α¯1β1)(α¯1β¯1)

and

(C.42)
Res(Q,β¯1)=ıβ¯12n+1(rβ¯1α2)(rβ¯1β2)(β¯1α1)(β¯1β1)(β¯1α¯1)
Appendix C Useful Integrals

Figure C.2 Closed path of integration for Jn.

From now on, we consider the case 0<ζ1<1 and 0ζ2<1. Then, from eqn (C.38), β1=α¯1 and β2=α¯2. And

(C.43)
Res(Q,α¯1)=ıα¯12n+1(rα¯1α2) (rα¯1+α¯2) (α¯1α1) (α¯1+α¯1) (α¯1+α1)

and

(C.44)
Res(Q,α1)=ıα12n+1(rα1α2)(rα1+α¯2)(α1α1)(α1+α¯1)(α1α¯1)

By adding Res(Q,α1) and Res(Q,α1) and applying (C.40),

(C.45)
Jn=[ıα¯12n(rα¯1α2) (rα¯1+α¯2)+ıα12n(rα1+α2) (rα1α¯2)]×2ıπ8ıIm(α1)Re(α1)

(p.298) As, Re(α1)=1ζ12 and Im(α1)=ζ1

(C.46)
Jn=π4ζ11ζ12×2Im[α¯12n(rα1+α2)(rα1α¯2)]|rα¯1α2|2|rα1+α2|2

Since

(C.47)
rα1+α2=r1ζ12+1ζ22+ı(rζ1+ζ2)
(C.48)
rα1α¯2=r1ζ121ζ22+ı(rζ1+ζ2)

it yields

(C.49)
Im[(rα1+α2)(rα1α¯2)]=2r(rζ1+ζ2)1ζ12
(C.50)
Im[α¯12(rα1+α2)(rα1α¯2)]=2(ζ1+rζ2)1ζ12

Finally,

(C.51)
J0=πζ1×r(rζ1+ζ2)(r1ζ12+1ζ22)2+(rζ1+ζ2)2(r1ζ121ζ22)2+(rζ1+ζ2)2

and

(C.52)
J1=πζ1×ζ1+rζ2(r1ζ12+1ζ22)2+(rζ1+ζ2)2(r1ζ121ζ22)2+(rζ1+ζ2)2

Application to ıωHj|Hi|2dω

(C.53)
ıωHj|Hi|2dω=1mjωj2mi2ωi4ıωdω1+2ıζjωωjω2ωj21+2ıζiωωiω2ωi22

By introducing the ratio r=ωi/ωj and performing the change of variable z=ω/ωi, it yields

(C.54)
ıωHj|Hi|2dω=1mjωj2mi2ωi2ızdz1+2ıζjrzr2z21+2ıζizz22

We recognize the expression of J0(r,ζi,ζj). Therefore by substituting eqn (C.51), it leads to

(C.55)
ıωHj|Hi|2dω=1mjωj2mi2ωi2×πζi   ×r(rζi+ζj)(r1ζi2+1ζj2)2+(rζi+ζj)2(r1ζi21ζj2)2+(rζi+ζj)2

(p.299) After simplification,

(C.56)
ıωHj(ω)|Hi|2(ω)dω=πmjmi2ωiζi×ωiζi+ωjζjωi1ζi2+ωj1ζj22+ωiζi+ωjζj2ωi1ζi2ωj1ζj22+ωiζi+ωjζj2

This result is valid when ζi,ζj<1.

Application to ıω3Hj|Hi|2dω

The calculation is similar. With r=ωi/ωj, we get

(C.57)
ıω3Hj|Hi|2dω=1mjωj2mi2ız3dz1+2ıζjrzr2z21+2ıζizz22

where the integral of the right-hand side is J1(r,ζi,ζj). By substituting eqn (C.52),

(C.58)
ıω3Hj|Hi|2dω=1mjωj2mi2×πζi×ζi+rζj(r1ζi2+1ζj2)2+(rζi+ζj)2(r1ζi21ζj2)2+(rζi+ζj)2

After simplification,

(C.59)
ıω3Hj(ω)|Hi|2(ω)dω=πωjmjmi2ζi×ωiζj+ωjζi(ωi1ζi2+ωj1ζj2)2+(ωiζi+ωjζj)2(ωi1ζi2ωj1ζj2)2+(ωiζi+ωjζj)2

This result is valid when ζi,ζj<1.

Application to ıω3Hj|Hi|2dω (other method)

Again, by setting r=ωi/ωj and performing the change of variable z=ω/ωi,

(C.60)
ιω3Hj|Hj|2dω=1mjωj2mi2ιz3dz(1+2ιζjrzr2z2)|1+2ιζizz2|2
(C.61)
=2mjωj2mi2Reιz3dz(1+2ιζjrzr2z2)|1+2ιζizz2|2

(p.300) where the last equality stems from Q(z)=Q¯(z) where Q is the integrand. By the change of variable u=1/z,

(C.62)
ıω3Hj|Hi|2dω=2mjωj2mi2Re0ı×1/u3×du/u21+2ıζjrur2u21+2ıζi1u1u22
(C.63)
=2mjωj2mi2r2Re0ıuduu2r2+2ıζjur11+2ıζiuu22
(C.64)
=1mjωj2mi2r2ıudu12ıζjuru2r21+2ıζiuu22

The integral J0(1/r,ζi,ζj) may be recognized,

(C.65)
ıω3Hj|Hi|2=1mjωi2mi2J¯0(1r,ζi,ζj)

Expanding J¯0(1/r,ζi,ζj) in the above equality reestablishes eqn (C.59).

Calculation of Kn=z2ndz((1ω12z2)2+4ζ12ω12z2)((1ω22z2)2+4ζ22ω22z2)

The integer n has values 0,1, or 2. Let Q(z) be the integrand. This is a rational function Q(z)=|zn/P(z)|2 where P is the polynomial,

(C.66)
P(z)=(1+2ıζ1ω1zω12z2)(1+2ıζ2ω2zω22z2)

Factorizing P gives

(C.67)
P(z)=(ω1zα1)(ω1zβ1)(ω2zα2)(ω2zβ2)

where

(C.68)
{αi=ıζi+1ζi2, βi=ıζi1ζi2  if|ζi|<1αi=ı(ζi+ζi21),βi=ı(ζiζi21)  otherwise

Thus, all zeros of P are localized in the upper complex half-plane Re(z)>0. By developing the polynomial P, we get

P(z)=1+2ız(ζ1ω1+ζ2ω2)z2(ω12+4ζ1ζ2ω1ω2+ω22)2ız3ω1ω2(ζ1ω2+ζ2ω1)+z4ω14ω24

(p.301) We can therefore apply eqn (C.2). The coefficients ai and bi take the values

(C.69)
a0=1b0=1b0=0b0=0a1=2(ζ1ω1+ζ2ω2)b1=0b1=1b1=0a2=ω12+4ζ1ζ2ω1ω2+ω22b2=0b2=0b2=1a3=2ω1ω2(ζ1ω2+ζ2ω1)b3=0b3=0b3=0a4=ω12ω22

for where the three columns of bi apply respectively to n=0, n=1, and n=2. Now, substituting these values successively into eqn (C.2) gives

(C.70)
K0=πa2a3a1a4a0(a1a2a3a0a32a12a4)
(C.71)
K1=πa0a3a0(a1a2a3a0a32a12a4)
(C.72)
K2=πa0a1a0(a1a2a3a0a32a12a4)

Now the denominator is

(C.73)
a0(a1a2a3a0a32a12a4)=4ω1ω2D

where

(C.74)
D=(ζ1ω1+ζ2ω2)(ω12+4ζ1ζ2ω1ω2+ω22)(ζ1ω2+ζ2ω1)ω1ω2(ζ1ω2+ζ2ω1)2+(ζ1ω1+ζ2ω2)2

By separating 4ζ1ζ2ω1ω2(ζ1ω1+ζ2ω2)(ζ1ω2+ζ2ω1) from the above and developing and simplifying the other terms, we find ζ1ζ2(ω12ω22)2. So,

(C.75)
a0(a1a2a3a0a32a12a4)=4ω1ω2ζ1ζ2(ω12ω22)2+4ω1ω2(ζ1ω1+ζ2ω2)(ζ1ω2+ζ2ω1)

Furthermore,

(C.76)
a2a3a1a4=2(ω12+4ζ1ζ2ω1ω2+ω22)ω1ω2(ζ1ω2+ζ2ω1)(ζ1ω1+ζ2ω2)ω12ω22=2ω1ω2ζ1ω23+4ζ1ζ2ω1ω2(ζ1ω2+ζ2ω1)+ζ2ω13
(C.77)
a0a3=2ω1ω2ζ1ω2+ζ2ω1
(C.78)
a0a1=2ζ1ω1+ζ2ω2

Finally,

(C.79)
K0=π2ζ1ζ2×ζ1ω23+4ζ1ζ2ω1ω2(ζ1ω2+ζ2ω1)+ζ2ω13(ω12ω22)2+4ω1ω2(ζ1ω1+ζ2ω2)(ζ1ω2+ζ2ω1)
(p.302)
(C.80)
K1=π2ζ1ζ2×ζ1ω2+ζ2ω1(ω12ω22)2+4ω1ω2(ζ1ω1+ζ2ω2)(ζ1ω2+ζ2ω1)
(C.81)
K2=π2ζ1ζ2×ζ1ω1+ζ2ω2ω1ω2(ω12ω22)2+4ω1ω2(ζ1ω1+ζ2ω2)(ζ1ω2+ζ2ω1)

Application to ω2(n+1)|HjHi|2dω

Since ωω2(n+1)|Hj(ω)Hi(ω)|2 is an even function, the integration bounds may be limited to 0 and ∞. The change of variable z=1/ω then leads to

(C.82)
ω2(n+1)|HjHi|2dω=20z2(n+1)Hi1zHj1z2dzz2

After developing,

(C.83)
ω2(n+1)|HjHi|2dω=2mi2mj20z42ndz((1ωi2z2)2+4ζi2ωi2z2)((1ωj2z2)2+4ζj2ωj2z2)

One finally recognizes the integral K2n,

(C.84)
ω2(n+1)|HjHi|2dω=K2n(ωi,ωj,ζi,ζj)mi2mj2

Or, after substituting the values of K2n,

(C.85)
ω2|HjHi|2dω=π2ζiζjmi2mj2×ζiωi+ζjωjωiωj(ωi2ωj2)2+4ωiωj(ζiωi+ζjωj)(ζiωj+ζjωi)
(C.86)
ω4|HjHi|2dω=π2ζiζjmi2mj2×ζiωj+ζjωi(ωi2ωj2)2+4ωiωj(ζiωi+ζjωj)(ζiωj+ζjωi)
(C.87)
ω6|HjHi|2dω=π2ζiζjmi2mj2×ζiωj3+4ζiζjωiωj(ζiωj+ζjωi)+ζjωi3(ωi2ωj2)2+4ωiωj(ζiωi+ζjωj)(ζiωj+ζjωi)

These integrals are valid for any non-negative value of ζi and ζj if ζi2+ζj20.

Calculation of L(α,β)=expα2x+β2dx

Let α‎ and β‎ be two complex numbers. The convergence of the integral is ensured by the condition Reα2>0.

(p.303) Let us fix α‎. The function βI(α,β) is holomorphic on C and its derivative is

βL(α,β)=βexpα2x+β2dx=α2x+βexpα2x+β2dx=expα2x+β2x=x=+=0

Therefore βL(α,β) is constant on C. Thus,

(C.88)
L(α,β)=L(α,0)

Let us calculate L(α,0) by squaring it:

expα2x2dx2=expα2(x2+y2)dxdy=02πdφ0expα2r2rdr=2πexpα2r22α2r=0r=+=πα2

The first line is obtained by Fubini’s theorem and the second line is a change of variable in polar coordinates. Thus,

(C.89)
L(α,0)=πα

By combining (C.88) and (C.89), it yields

(C.90)
expα2x+β2dx=πα

for all α,βC such that Reα2>0.