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## 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

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# (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)
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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)
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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)
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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 $ω↦ωpHin(ω)Hjm(ω)$ is rational and therefore is holomorphic on $C−z1,z2,z3,z4$ 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 $p−2(n+m)$. The condition $lim|ω|→∞ωp+1Hin(ω)Hjm(ω)=0$ is fulfilled under the assumption $2(n+m)−p≥2$. Using Jordan’s lemma,

(C.4)
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# Calculation of $I(ζ)=∫−∞∞dz(1−z2)2+4ζ2z2$

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

(C.6)
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We consider successively the three cases , and $ζ=1$.

When $ζ<1$, P has two roots,

(C.7)
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where $i=1,2$. The poles of Q are the square roots of zi. These are , and $−α¯0$ where

(C.8)
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Factorizing Q gives

(C.9)
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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)
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Figure C.1 Closed path of integration for I.

Since all poles are simple, their residues are calculated by

(C.11)
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where . One obtains

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(p.293) and

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By summing the two above expressions,

(C.12)
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And finally,

(C.13)
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When and the two roots of P are

(C.14)
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with $i=1,2$. The four poles of Q are then $α1$, , and $−α2$ where

(C.15)
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(C.16)
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and Q becomes

(C.17)
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(p.294) By choosing the same integration path (see Fig. C.1), the two residues to be computed are

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and

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Hence,

(C.18)
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And finally,

(C.19)
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The last case $ζ=1$ is straightforward. Since $Q(z)=1/(1+z2)2$, the change of variable gives

(C.20)
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The same result is therefore obtained for all cases,

(C.21)
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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)
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By the change of variable $z=ω/ωi$,

(C.23)
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(p.295) And finally,

(C.24)
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## Application to $∫ω2|Hi|2dω$

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

(C.25)
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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)
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(C.27)
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(C.28)
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And finally,

(C.29)
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## Application to $∫ıωHidω$

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

(C.30)
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where the limits of integration are arbitrary and finite. By the change of variable $z=ω/ωi$,

(C.31)
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The first integral of the right-hand side is convergent when the limits become infinite and its value is $I(ζ)$. Therefore,

(C.32)
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(p.296) But the second integral is clearly divergent at infinite. However, the integrand is an odd function, so

(C.33)
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We shall write

(C.34)
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which must be understood in the sense of principal values.

# Calculation of $Jn(r,ζ1,ζ2)=∫−∞∞ız2n+1dz(1+2ıζ2rz−r2z2)((1−z2)2+4ζ12z2)$

C.6

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

(C.35)
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where

(C.36)
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Since $Pζi$ may be factorized as

(C.37)
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where

(C.38)
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we get

(C.39)
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The six poles of Q are therefore , , 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)
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(p.297) where

(C.41)
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and

(C.42)
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Figure C.2 Closed path of integration for Jn.

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

(C.43)
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and

(C.44)
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By adding $Res(Q,α‾1)$ and $Res(Q,−α1)$ and applying (C.40),

(C.45)
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(p.298) As, $Re(α1)=1−ζ12$ and $Im(α1)=ζ1$

(C.46)
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Since

(C.47)
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(C.48)
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it yields

(C.49)
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(C.50)
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Finally,

(C.51)
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and

(C.52)
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## Application to $∫−∞∞ıωHj|Hi|2dω$

(C.53)
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By introducing the ratio $r=ωi/ωj$ and performing the change of variable $z=ω/ωi$, it yields

(C.54)
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We recognize the expression of $J0(r,ζi,ζj)$. Therefore by substituting eqn (C.51), it leads to

(C.55)
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(p.299) After simplification,

(C.56)
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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)
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where the integral of the right-hand side is $J1(r,ζi,ζj)$. By substituting eqn (C.52),

(C.58)
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After simplification,

(C.59)
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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)
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(C.61)
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(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)
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(C.63)
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(C.64)
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The integral $J‾0(1/r,ζi,ζj)$ may be recognized,

(C.65)
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Expanding $J¯0(1/r,ζi,ζj)$ in the above equality reestablishes eqn (C.59).

# Calculation of $Kn=∫−∞∞z2ndz((1−ω12z2)2+4ζ12ω12z2)((1−ω22z2)2+4ζ22ω22z2)$

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)
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Factorizing P gives

(C.67)
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where

(C.68)
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Thus, all zeros of P are localized in the upper complex half-plane $Re(z)>0$. By developing the polynomial P, we get

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

(C.69)
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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)
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(C.71)
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(C.72)
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Now the denominator is

(C.73)
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where

(C.74)
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By separating $4ζ1ζ2ω1ω2(ζ1ω1+ζ2ω2)(ζ1ω2+ζ2ω1)$ from the above and developing and simplifying the other terms, we find $ζ1ζ2(ω12−ω22)2$. So,

(C.75)
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Furthermore,

(C.76)
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(C.77)
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(C.78)
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Finally,

(C.79)
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(p.302)
(C.80)
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(C.81)
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## Application to $∫−∞∞ω2(n+1)|HjHi|2dω$

Since $ω↦ω2(n+1)|Hj(ω)Hi(ω)|2$ 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)
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After developing,

(C.83)
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One finally recognizes the integral $K2−n$,

(C.84)
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Or, after substituting the values of $K2−n$,

(C.85)
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(C.86)
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(C.87)
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These integrals are valid for any non-negative value of $ζi$ and $ζj$ if $ζi2+ζj2≠0$.

# Calculation of $L(α,β)=∫−∞∞exp−α2x+β2dx$

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 $β↦I(α,β)$ is holomorphic on $C$ and its derivative is

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Therefore $β↦L(α,β)$ is constant on $C$. Thus,

(C.88)
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Let us calculate $L(α,0)$ by squaring it:

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

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

(C.90)
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for all $α,β∈C$ such that $Reα2>0$.