# Geometric Sensitivity of a Pinhole Collimator

Authors: Howard Jacobowitz; Scott D. Metzler
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

Geometric sensitivity for single photon emission computerized tomography
(SPECT) is given by a double integral over the detection plane. It
would be useful to be able to explicitly evaluate this quantity. This paper
shows that the inner integral can be evaluated in the situation where
there is no gamma ray penetration of the material surrounding the pinhole
aperture. This is done by converting the integral to an integral in
the complex plane and using Cauchy's theorem to replace it by one which
can be evaluated in terms of elliptic functions.


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## Body

## 1. Introduction

Nuclear-medicine imaging provides images that assess how the body is functioning [1, 2], as opposed to anatomical modalities (e.g., X-ray computed tomography, commonly known as a CT or “CAT” scans) that provide little or no information about function, but great detail of the body's structure. Nuclear-medicine images the biodistribution of radiolabeled molecules that are typically intravenously injected into the patient in tracer (i.e., nonpharmacological) quantities. The compounds may have different biochemical properties that affect the biodistribution and, hence, the choice of pharmaceuticals used to assess the disease state of a patient.Two common nuclear-medicine techniques are single photon emission computed tomography (SPECT) and positron emission tomography (PET). Molecules labeled with a SPECT tracer emit a single photon. PET tracers emit a positron, which annihilates with a nearby electron to produce two nearly back-to-back photons at 511 keV, the rest energy of an electron. The line of these two photons contains the emission point of the positron. Many photon pairs are detected in coincidence and reconstructed into a three-dimensional (3D) image of the tracer's distribution.Since SPECT tracers emit only one photon, the detection of the photon itself gives little information about the location of the emission since neither the direction nor origin of the emission is known. Consequently, SPECT uses collimation that allows only photons along certain lines to be detected. The origin is still unknown, but the 3D image may be reconstructed when photons from numerous angles are detected [4, 5]. Pinhole collimation is often used when a small organ (e.g., the thyroid) is imaged by a large detector with a SPECT tracer. Figure 1 is a diagram of the apparatus. It is not hard to see that the pinhole collimation's geometric efficiency, namely, the fraction of emitted photons that pass through the circular aperture of the collimator, is approximately given byFigure 1
Perspective View of the Knife-edge Pinhole Collimator. A point source atz=h and angle θ emits a photon that intersects the z=0 plane at (ρcosβ,ρsinβ,0) with an incident angle θa. The acceptance angle of the collimator is α and the diameter of the aperture is d. ΔL is the path length of the photon in the attenuating medium. Reproduced from [3] with permission (© 2001 IEEE).(1.1)g=d2sin3θ16h2,
where d is the diameter of the pinhole, h is the perpendicular distance, and θ is the incidence angle of the photon on the aperture plane at the center of aperture. This formula becomes exact in the limit that d goes to zero. Also, an exact formula is easily obtained when θ=π/2(1.2)g(θ=π2)=12-12[1+(d2h)2]-1/2≈12-12[1-12(d2h)2]=d216h2.
We derive stronger versions of these two results in the next section. But the main goal of this paper is to simplify the integral representation for the geometric sensitivity without taking d or θ-π/2 to be small. We are concerned with the case where there is no penetration in the sense that not all gamma rays are stopped by the attenuating material surrounding the pinhole aperture. We take θ≠π/2. The geometric sensitivity, as defined above, is given by the integral over the aperture of the flux times the sine of the incidence angle at that point on the aperture:(1.3)g=14πh2∫0d/2ρdρ∫02πdβsin3θa,
where ρ and β are polar coordinates on the aperture plane and θa is the incidence angle of the photon at a particular value of ρ and β, and is given by [3](1.4)cot2θa=cot2θ-2ρhcotθcos(β-ϕ)+ρ2h2,
where ϕ is the third coordinate of the point source. For integrals over 2π the difference β-ϕ can be replaced with only β without loss of generality. This yields(1.5)sin2θa=[csc2θ-2ρhcotθcosβ+ρ2h2]-1.Applying this to (1.3),(1.6)g=14πh2∫0d/2ρdρ∫02πdβ[csc2θ-2ρhcotθcosβ+ρ2h2]-3/2.When there is penetration the above would be modified to(1.7)g=14πh2∫0∞ρdρ∫02πdβ[csc2θ-2ρhcotθcosβ+ρ2h2]-3/2e-μΔL,
where the path length through attenuating material, ΔL, is zero for ρ<d/2 and is otherwise given by (11) in [3].Thus, (1.7) can be rewritten as(1.8)g=gNP+14πh2∫d/2∞ρdρ∫02πdβ[csc2θ-2ρhcotθcosβ+ρ2h2]-3/2e-μΔL=gNP+gPEN,
where gPEN is the efficiency due to penetration and gNP, the efficiency under the assumption of no penetration, is given by (1.6).In this paper we use complex variable methods to calculate theβ integral in (1.6).


## 2. Two Approximations forgNP

The special form ofgNP makes it easy to derive approximations for d/h small and for θ-π/2 small which generalize (1.1) and (1.2). Before doing this, we observe that only even powers of d and of θ-π/2 can occur in the expansion of gNP. We see this by writing the integral as the sum of integrals from 0 to π and from π to 2π.


### 2.1. The Expansion ford Small

Let us, in this section, denotegNP by g. Upon substituting r=d/2h and η=ρ/h into (1.6) we obtain(2.1)g(r,θ)=14π∫02π∫0r[csc2θ-2ηcotθcosβ+η2]-3/2ηdηdβ,=14π∫02π(∫0rηϕ(β,η,θ)dη)dβ.
Clearly, g(0,θ)=0. Since(2.2)gr(r,θ)=14π∫02πrϕ(β,r,θ)dβ,
we have gr(0,θ)=0. Differentiating (2.12) three more times with respect to r we obtain(2.3)grr(r,θ)=14π∫02π(rϕr(β,0,θ)+ϕ(β,r,θ))dβ,grrr(r,θ)=14π∫02π(rϕrr(β,0,θ)+2ϕr(β,r,θ))dβ,grrrr(r,θ)=14π∫02π(rϕrrr(β,0,θ)+3ϕrr(β,r,θ))dβ,
and then setting r=0 in each of the resulting equations we obtain(2.4)grr(0,θ)=14π∫02πϕ(β,0,θ)dβ,grrr(0,θ)=12π∫02πϕr(β,0,θ)dβ,grrrr(0,θ)=34π∫02πϕrr(β,0,θ)dβ.
For convenience let us write(2.5)u=csc2θ,v=cotθ,
so ϕ becomes(2.6)ϕ(β,r,θ)=(u-2vrcosβ+r2)-3/2.
Then(2.7)ϕ(β,0,θ)=u-3/2,ϕr(β,0,θ)=3u-5/2vcosβ,ϕrr(β,0,θ)=15u-7/2v2cos2β-3u-5/2.
We substitute into the Taylor expansion(2.8)g(r,θ)=g(0,θ)+gr(0,θ)r+12!grr(0,θ)r2+13!grrr(0,θ)r3+14!grrrr(0,θ)r4+⋯
and obtain(2.9)g(r,θ)=12(14π∫02πu-3/2dβ)r2+13!(32π∫02π6u-5/2vcosβdβ)r3+14!(14π∫02π3(15u-7/2v2cos2β-3u-5/2)dβ)r4+⋯.
Since(2.10)∫02πcosβdβ=0,∫02πcos2βdβ=π
and u-1=sin2θ we obtain(2.11)g(r,θ)=r24sin3θ+332(5sin5θcos2θ-2sin5θ)r4+⋯.
Finally we replace cos2θ by 1-sin2θ and r by d/2h and obtain(2.12)gNP=116sin3θ(dh)2+3512sin5θ(3-5sin2θ)(dh)4+𝒪(d6).


### 2.2. The Expansion forθ-π/2 Small

To start we let(2.13)f(θ,η,β)=(csc2θ-2ηcotθcosβ+η2)-3/2
and compute its Taylor expansion at θ=π/2,(2.14)f(θ,η,β)=(1+η2)-3/2-3(1+η2)-5/2ηcosβ(θ-π2)+12!(15(1+η2)-7/2η2cos2β-3(1+η2)-5/2)(θ-π2)2+𝒪(θ-π2)3.Since(2.15)gNP=14π∫02π∫0d/2hf(θ,η,β)ηdηdβ
and since we know that only even powers of θ-π/2 occur, we obtain(2.16)gNP=12(1-11+(d/2h)2)-38(d/2h)2(1+(d/2h)2)5/2(θ-π2)2+𝒪(θ-π2)4.


## 2.1. The Expansion ford Small

Let us, in this section, denotegNP by g. Upon substituting r=d/2h and η=ρ/h into (1.6) we obtain(2.1)g(r,θ)=14π∫02π∫0r[csc2θ-2ηcotθcosβ+η2]-3/2ηdηdβ,=14π∫02π(∫0rηϕ(β,η,θ)dη)dβ.
Clearly, g(0,θ)=0. Since(2.2)gr(r,θ)=14π∫02πrϕ(β,r,θ)dβ,
we have gr(0,θ)=0. Differentiating (2.12) three more times with respect to r we obtain(2.3)grr(r,θ)=14π∫02π(rϕr(β,0,θ)+ϕ(β,r,θ))dβ,grrr(r,θ)=14π∫02π(rϕrr(β,0,θ)+2ϕr(β,r,θ))dβ,grrrr(r,θ)=14π∫02π(rϕrrr(β,0,θ)+3ϕrr(β,r,θ))dβ,
and then setting r=0 in each of the resulting equations we obtain(2.4)grr(0,θ)=14π∫02πϕ(β,0,θ)dβ,grrr(0,θ)=12π∫02πϕr(β,0,θ)dβ,grrrr(0,θ)=34π∫02πϕrr(β,0,θ)dβ.
For convenience let us write(2.5)u=csc2θ,v=cotθ,
so ϕ becomes(2.6)ϕ(β,r,θ)=(u-2vrcosβ+r2)-3/2.
Then(2.7)ϕ(β,0,θ)=u-3/2,ϕr(β,0,θ)=3u-5/2vcosβ,ϕrr(β,0,θ)=15u-7/2v2cos2β-3u-5/2.
We substitute into the Taylor expansion(2.8)g(r,θ)=g(0,θ)+gr(0,θ)r+12!grr(0,θ)r2+13!grrr(0,θ)r3+14!grrrr(0,θ)r4+⋯
and obtain(2.9)g(r,θ)=12(14π∫02πu-3/2dβ)r2+13!(32π∫02π6u-5/2vcosβdβ)r3+14!(14π∫02π3(15u-7/2v2cos2β-3u-5/2)dβ)r4+⋯.
Since(2.10)∫02πcosβdβ=0,∫02πcos2βdβ=π
and u-1=sin2θ we obtain(2.11)g(r,θ)=r24sin3θ+332(5sin5θcos2θ-2sin5θ)r4+⋯.
Finally we replace cos2θ by 1-sin2θ and r by d/2h and obtain(2.12)gNP=116sin3θ(dh)2+3512sin5θ(3-5sin2θ)(dh)4+𝒪(d6).


## 2.2. The Expansion forθ-π/2 Small

To start we let(2.13)f(θ,η,β)=(csc2θ-2ηcotθcosβ+η2)-3/2
and compute its Taylor expansion at θ=π/2,(2.14)f(θ,η,β)=(1+η2)-3/2-3(1+η2)-5/2ηcosβ(θ-π2)+12!(15(1+η2)-7/2η2cos2β-3(1+η2)-5/2)(θ-π2)2+𝒪(θ-π2)3.Since(2.15)gNP=14π∫02π∫0d/2hf(θ,η,β)ηdηdβ
and since we know that only even powers of θ-π/2 occur, we obtain(2.16)gNP=12(1-11+(d/2h)2)-38(d/2h)2(1+(d/2h)2)5/2(θ-π2)2+𝒪(θ-π2)4.


## 3. The Contour Integral

We have(3.1)I=∫02πdβ[csc2θ-2ρhcotθcosβ+ρ2h2]-3/2,(3.2)gNP=14πh2∫0d/2Iρdρ.I may be recast as a contour integral by setting z=eiβ:
(3.3)I=∫Γdziz[-b2z(z2-2abz+1)]-3/2=(2b)3/21i∫Γ[(z-x0)(x1-z)z]-3/2dzz,
where(3.4)Γistheunitcircle,a=csc2θ+ρ2h2,b=2ρhcotθ,x0=ab[1-1-b2a2],x1=ab[1+1-b2a2].
We note that a>b and so 0<x0<1<x1.In this section we takeρ,θ, and h to be constant. We write(3.5)F(z)=[z(z-x0)(x1-z)][z((z-x0)(x1-z))]1/2[1z]=1(z-x0)(x1-z)[z(z-x0)(x1-z)]1/2,I=(2b)3/21i∫ΓF(z)dz.
We need to be careful because of the square roots. We use the usual convention that(3.6)limy→0±-1+iy=±i.
We wish to show that F(z) is holomorphic (and single valued) in(3.7){z:|z|<1}-{x:0<x<x0}.
To isolate the square root factor in F we set(3.8)g(z)=z(z-x0)(x1-z).
We take |z|≤1. If in addition Rez<0 and |Imz|≪|z-x0| then Re(z-x0)(x1-z)<0. This implies that g(z) never takes on negative real values and so has a single valued square root in Rez<0. Similarly, if |z|≤1, and Rez>x0 (no condition on Imz now) then Re(z-x0)(x1-z)>0 and g(z) has a single valued square root in Rez>x0.It follows from Cauchy's theorem that the integral ofF(z) over the unit circle Γ is equal to the integral over the “ barbell”  γ around 0 and x0. More explicitly, γ is the boundary of(3.9)B=Bϵ(0)∪Bϵ(x0)∪R,
where Bϵ(x) is the ball of radius ϵ and centered at x and R is the rectangle(3.10)R={z=x+iy:0≤x≤x0,-δ≤y≤δ}
with δ≪ϵ, see Figure 2.Figure 2
The curveγ.The curveγ decomposes into smooth pieces which we describe using the angle α=arcsin(δ/ϵ)(3.11)γr={z:|z-x0|=ϵ,-π+α≤argz-x0≤π-α},γl={z:|z|=ϵ,α≤arg(z)≤2π-α},γ±={z:z=x±iδ,ϵcosα≤x≤x0-ϵcosα}.To bound the integral overγl we set z=ϵeiθ(3.12)|∫γlF(z)dz|=|∫α2π-αF(ϵeiθ)ϵeiθdθ|=𝒪(ϵ3/2).
To evaluate ∫γrF(z)dz we set z=x0+ϵeiθ. Then(3.13)F(x0+ϵeiθ)=Cϵ-3/2e-(3/2)iθ+𝒪(ϵ-1/2)
with(3.14)C=x01/2(x1-x0)3/2,
and so(3.15)∫γrF(z)dz=∫-π+απ-α(ϵ-3/2Ce-(3/2)iθ+𝒪(ϵ-1/2))ϵieiθ(1+𝒪(ϵ))dθ=∫-π+απ-αCiϵ-1/2e-(1/2)iθ+𝒪(ϵ1/2)dθ.
Since α=O(ϵ), we may set α=0 and absorb the error in the 𝒪(ϵ1/2) term. We then do the integration and obtain(3.16)∫γrF(z)dz=4iCϵ+𝒪(ϵ1/2).
We want to compute the integral of F over γ+. We need to be careful about the square root. Recall that we are assuming(3.17)ϵcosα<x<x0-ϵcosα,δ≪ϵ.Lemma 3.1.
Onγ+(3.18)F(z)=i((x0-x)(x1-x))-3/2x1/2+𝒪(δ)
and on γ-(3.19)F(z)=-i((x0-x)(x1-x))-3/2x1/2+𝒪(δ).Proof.
This is a consequence of the branching of the square root function. Let(3.20)H(z)=z(z-x0)(x1-z)=z(z¯-x0)(x1-z¯)|z-x0|2|x1-z|2,h(z)=z(z¯-x0)(x1-z¯).
We see that
(3.21)h(x+iδ)=x(x-x0)(x1-x)+iδ(x2-1)+𝒪(δ2).
Recall that δ is replaced by -δ on γ- and that 0<x<x0<x1. So ReH(z)<0 on both γ+ and γ- while ImH(z)<0 on γ+ and ImH(z)>0 on γ-. Thus on γ+(3.22)H(z)1/2=-i|x(x-x0)(x1-x)|1/2+𝒪(δ)=-i(x(x0-x)(x1-x))1/2+𝒪(δ),
while on γ-(3.23)H(z)1/2=i(x(x0-x)(x1-x))1/2+𝒪(δ).
Since
(3.24)F(z)=F(x)+𝒪(δ)=1(x-x0)(x1-x)H(z)1/2+𝒪(δ)=-1(x0-x)(x1-x)H(z)1/2+𝒪(δ)=-1(x0-x)(x1-x)(-i(x(x0-x)(x1-x))1/2)+𝒪(δ),
the  lemma follows.We also need to be careful about the orientation. The curveγ is oriented counter-clockwise. This means that the integration over γ+ goes in the direction of x0 to 0. We need to reverse its sign in order to write it in the conventional way(3.25)∫γ+F(z)dz=-∫ϵcosαx0-ϵcosαF(x+iδ)dx=-∫ϵcosαx0-ϵcosαix1/2dx(x0-x)3/2(x1-x)3/2+𝒪(ϵ).
We let x=x0t and Δ=ϵx0-1cosα and use δ≪ϵ to obtain(3.26)∫γ+F(z)dz=-∫Δ1-Δix03/2t1/2dx(x0-x0t)3/2(x1-x0t)3/2+𝒪(ϵ).
Now let(3.27)τ=x1x0>1.
We have(3.28)∫γ+F(z)dz=-ix03/2∫Δ1-Δt1/2dt(τ-t)3/2(1-t)3/2+𝒪(ϵ).
We may let Δ→0 in the lower limit, but not in the upper limit. So we now have(3.29)∫γ+F(z)dz=-ix03/2∫01-Δt1/2dt(τ-t)3/2(1-t)3/2+𝒪(ϵ).
Here Δ=ϵx0-1cosα. We may let δ→0 and redefine Δ to be ϵx0-1.We get the same value for the integral overγ-. This is because the integrand is the negative of that in (3.29) but the orientation is reversed and so the two negatives cancel.For any functiong(t), continuously differentiable on the interval [0,1], we have(3.30)∫01-Δg(t)(1-t)3/2dt=2g(1)Δ1/2-2g(0)-2∫01(1-t)-1/2g′(t)dt+O(Δ1/2)
as can easily be seen by an integration by parts.We apply this tog(t)=t1/2/(τ-t)3/2 as in (3.29). This yields(3.31)∫γ+F(z)dz=-2i(τ-1)3/2x03/2Δ1/2+ix03/2(∫01t-1/2(τ-t)-3/2+3t1/2(τ-t)-5/2dt)+𝒪(ϵ1/2).
We now take twice this quantity and add to it the results of (3.14) and (3.16),(3.32)∫|z|=1F(z)dz=∫γF(z)dz=4iCϵ+2(-2ix0(x1-x0)-3/2ϵ+ix03/2∫01(1-t)-1/2((τ-t)-3/2t-1/2+3t1/2(τ-t)-5/2)dt)+𝒪(ϵ1/2).
We substitute the value for C from (3.14) and let ϵ→0(3.33)∫|z|=1F(z)dz=2ix03/2∫01(1-t)-1/2((τ-t)-3/2t-1/2+3t1/2(τ-t)-5/2)dt.
Finally, we evaluate this integral using Mathematica and obtain(3.34)∫|z|=1F(z)dz=(2ix03/2)2(2τE[1/τ]-(-1+τ)K[1/τ])(τ-1)2τ,
where K and E are the complete  elliptic functions of the first and second kinds(3.35)K[m]=∫01(1-mt2)-1/2(1-t2)-1/2dt,E[m]=∫01(1-mt2)1/2(1-t2)-1/2dt.We now have(3.36)I=(2b)3/21i∫ΓF(z)dz=(2b)3/22x03/22(2τE(1/τ)-(τ-1)K(1/τ))(τ-1)2τ=(2b)3/24τ1/4(τ-1)2(2τE(1/τ)-(τ-1)K(1/τ)),(3.37)gNP=14πh2∫0d/2ρIdρ
with(3.38)b=2ρhcotθ,a=csc2θ+ρ2h2,(3.39)τ=x1x0=1x02,x0=ab[1-1-b2a2],x1=ab[1+1-b2a2].


## 4. Limiting Cases

We validate (3.36) by considering two limiting cases.


### 4.1. Expansion for Smalld

We look for an expansion ofgNP under the assumption that d/h is small and rederive (2.12). The dependence of x0 on ρ is given in (3.39). The two-term Taylor expansion of x0 as a function of ρ is(4.1)x0=Aρ+Bρ3+𝒪(ρ5)
with(4.2)A=cosθsinθh,B=A3-Ah2csc2θ.
Since
(4.3)τ=1x02,
we have
(4.4)1τ=A2ρ2+𝒪(ρ4),(4.5)1(τ-1)2=1τ2+2τ3+𝒪(ρ8).
It is convenient to work with the quantity ρ2τ, so we note, using (4.1),(4.6)ρ2τ=1A2-2BA3ρ2+𝒪(ρ4).
We start with(4.7)I=(2b)3/24τ1/4(τ-1)2(2τE(1τ)-(τ-1)K(1τ)).
Substituting the expression for b from (3.38) and for (τ-1)2 from (4.5) we obtain, after some manipulation(4.8)I=((hcotθ)3/24(ρ2τ)3/41τ+𝒪(ρ4))(2τE(1τ)-(τ-1)K(1τ)).We now use the expansions (as given by Mathematica)(4.9)E(t)=π2-πt8+𝒪(t2),K(t)=π2+πt8+𝒪(t2).
So(4.10)I=(hcotθ)3/24(ρ2τ)3/4(π2+π8τ+πτ)+𝒪(ρ4).
From the expression for ρ2τ in (4.6), we derive(4.11)1(ρ2τ)3/4=A3/2(1+32BAρ2+𝒪(ρ4)),
and this leads to(4.12)I=4(hcotθ)3/2A3/2(π2+π8τ+πτ+3Bρ2π4A)+𝒪(ρ4).
We now substitute for A and B using (4.2) and also for τ-1 using (4.4)(4.13)I=sin3θ(2π+3π2(3sin2θ-5sin4θ)ρ2h2)+𝒪(ρ4).
Substituting this into (3.37) and integrating, we again derive (2.12).


### 4.2. An Expansion forθ Near π/2

We show how (4.7) also gives the expansion of Section 2.2. From (3.38) we see that(4.14)a=1+ρ2h2+(θ-π2)2+𝒪(θ-π2)4,(4.15)b=-2ρh(θ-π2)+𝒪(θ-π2)3.
So from (3.39) we have(4.16)x0=b2a(1+b24a2+𝒪(θ-π2)3).
Thus(4.17)τ=4a2b2-2+𝒪(θ-π2)2,(4.18)1(τ-1)2=1τ2(1+2τ+𝒪(θ-π2)4),1τ(2τE(1τ)-(τ-1)K(1τ))=π2+9π8τ+𝒪(θ-π2)4.
These expansions let us write (4.7) as(4.19)I=(2b)3/24τ3/4((1+2τ)(π2+9π8τ)+𝒪(θ-π2)4)=27/2(b2τ)3/4((1+2τ)(π2+9π8τ)+𝒪(θ-π2)4).
Using (4.15) and (4.17) we obtain a substitute for (4.1)(4.20)1(b2τ)3/4=4-3/4a-3/2(1+3b28a2)+𝒪(θ-π2)4.
This gives, in light of (4.14),(4.21)I=2πa-3/2+158πb2a-7/2+𝒪(θ-π2)4=2π(1+ρ2h2)-3/2-3π(1+ρ2h2)-5/2(θ-π2)2+152πρ2h2(1+ρ2h2)-7/2(θ-π2)2+𝒪(θ-π2)4.
Finally, we integrate term by term.(4.22)g=14πh2∫0d/2ρI(ρ)dρ=14πh2(2πh2(1-(1+(d2h)2)-1/2)-3πh213(1-(1+(d2h)2)-3/2)(θ-π2)2+152πh2(215-2+5(d/2h)215(1+(d2h)2)-5/2)(θ-π2)2)+𝒪(θ-π2)4=12(1-(1+(d2h)2)-1/2)-38(1+(d2h)2)-5/2(d2h)2(θ-π2)2+𝒪(θ-π2)4.


## 4.1. Expansion for Smalld

We look for an expansion ofgNP under the assumption that d/h is small and rederive (2.12). The dependence of x0 on ρ is given in (3.39). The two-term Taylor expansion of x0 as a function of ρ is(4.1)x0=Aρ+Bρ3+𝒪(ρ5)
with(4.2)A=cosθsinθh,B=A3-Ah2csc2θ.
Since
(4.3)τ=1x02,
we have
(4.4)1τ=A2ρ2+𝒪(ρ4),(4.5)1(τ-1)2=1τ2+2τ3+𝒪(ρ8).
It is convenient to work with the quantity ρ2τ, so we note, using (4.1),(4.6)ρ2τ=1A2-2BA3ρ2+𝒪(ρ4).
We start with(4.7)I=(2b)3/24τ1/4(τ-1)2(2τE(1τ)-(τ-1)K(1τ)).
Substituting the expression for b from (3.38) and for (τ-1)2 from (4.5) we obtain, after some manipulation(4.8)I=((hcotθ)3/24(ρ2τ)3/41τ+𝒪(ρ4))(2τE(1τ)-(τ-1)K(1τ)).We now use the expansions (as given by Mathematica)(4.9)E(t)=π2-πt8+𝒪(t2),K(t)=π2+πt8+𝒪(t2).
So(4.10)I=(hcotθ)3/24(ρ2τ)3/4(π2+π8τ+πτ)+𝒪(ρ4).
From the expression for ρ2τ in (4.6), we derive(4.11)1(ρ2τ)3/4=A3/2(1+32BAρ2+𝒪(ρ4)),
and this leads to(4.12)I=4(hcotθ)3/2A3/2(π2+π8τ+πτ+3Bρ2π4A)+𝒪(ρ4).
We now substitute for A and B using (4.2) and also for τ-1 using (4.4)(4.13)I=sin3θ(2π+3π2(3sin2θ-5sin4θ)ρ2h2)+𝒪(ρ4).
Substituting this into (3.37) and integrating, we again derive (2.12).


## 4.2. An Expansion forθ Near π/2

We show how (4.7) also gives the expansion of Section 2.2. From (3.38) we see that(4.14)a=1+ρ2h2+(θ-π2)2+𝒪(θ-π2)4,(4.15)b=-2ρh(θ-π2)+𝒪(θ-π2)3.
So from (3.39) we have(4.16)x0=b2a(1+b24a2+𝒪(θ-π2)3).
Thus(4.17)τ=4a2b2-2+𝒪(θ-π2)2,(4.18)1(τ-1)2=1τ2(1+2τ+𝒪(θ-π2)4),1τ(2τE(1τ)-(τ-1)K(1τ))=π2+9π8τ+𝒪(θ-π2)4.
These expansions let us write (4.7) as(4.19)I=(2b)3/24τ3/4((1+2τ)(π2+9π8τ)+𝒪(θ-π2)4)=27/2(b2τ)3/4((1+2τ)(π2+9π8τ)+𝒪(θ-π2)4).
Using (4.15) and (4.17) we obtain a substitute for (4.1)(4.20)1(b2τ)3/4=4-3/4a-3/2(1+3b28a2)+𝒪(θ-π2)4.
This gives, in light of (4.14),(4.21)I=2πa-3/2+158πb2a-7/2+𝒪(θ-π2)4=2π(1+ρ2h2)-3/2-3π(1+ρ2h2)-5/2(θ-π2)2+152πρ2h2(1+ρ2h2)-7/2(θ-π2)2+𝒪(θ-π2)4.
Finally, we integrate term by term.(4.22)g=14πh2∫0d/2ρI(ρ)dρ=14πh2(2πh2(1-(1+(d2h)2)-1/2)-3πh213(1-(1+(d2h)2)-3/2)(θ-π2)2+152πh2(215-2+5(d/2h)215(1+(d2h)2)-5/2)(θ-π2)2)+𝒪(θ-π2)4=12(1-(1+(d2h)2)-1/2)-38(1+(d2h)2)-5/2(d2h)2(θ-π2)2+𝒪(θ-π2)4.


## 5. Numerical Confirmation

The result in (3.36) is further validated in this section with numerical integration of (3.1). In this original formulation, the integral depends on the ratio of ρ/h and on the value of θ. We analytically considered one limiting case for each of these quantities in the previous section. We consider these and other values for these quantities in this section using numerical methods. For fixed values of ρ/h, the numerical integration (solid lines) is shown with the results of the residue calculation (open circles) in Figure 3 as a function of θ. Figure 4 shows the numerical integration and residue results for fixed values of θ as a function of ρ/h. All corresponding values of the residue calculation and the numerical integration agree.Figure 3
Numerical integration of (3.1) (solid lines) and residue calculation of (3.36) (open circles) as a function of θ for fixed values of ρ/h: 0.0, 0.2, 0.4, 0.6, 0.8, and 1.0.Figure 4
Numerical integration of (3.1) (solid lines) and residue calculation of (3.36) (open circles) as a function of ρ/h for fixed values of θ: 50, 60, 70, 80, and 90 degrees.


## 6. Conclusions

The accuracy of collimator efficiency is important for design and for quantitative reconstruction programs. Analytic forms obviate the need for numerical models, which do not readily offer insight into the interplay of collimator parameters. This paper approaches the problem by recasting one of the integrals in terms of contour integration. This integration was completed, but the authors were unable to execute the second integral to yield a complete analytic solution. The result was studied for special cases with known outcomes for validation. Future efforts may be able to build on this validated result to find a complete analytic solution.



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*Source: 915958-2010-04-01.xml*