International Journal of Mathematics and Mathematical Sciences

International Journal of Mathematics and Mathematical Sciences / 2010 / Article

Research Article | Open Access

Volume 2010 |Article ID 915958 | https://doi.org/10.1155/2010/915958

Howard Jacobowitz, Scott D. Metzler, "Geometric Sensitivity of a Pinhole Collimator", International Journal of Mathematics and Mathematical Sciences, vol. 2010, Article ID 915958, 18 pages, 2010. https://doi.org/10.1155/2010/915958

Geometric Sensitivity of a Pinhole Collimator

Academic Editor: Harvinder S. Sidhu
Received25 Aug 2009
Revised07 Dec 2009
Accepted19 Feb 2010
Published01 Apr 2010

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.

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 β€œCAT” 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 by

𝑑𝑔=2sin3πœƒ16β„Ž2,(1.1) where 𝑑 is the diameter of the pinhole, β„Ž is the perpendicular distance, and πœƒ 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 πœƒ=πœ‹/2

π‘”ξ‚€πœ‹πœƒ=2=12βˆ’12𝑑1+2β„Ž2ξ‚Ήβˆ’1/2β‰ˆ12βˆ’12ξ‚Έ11βˆ’2𝑑2β„Ž2ξ‚Ή=𝑑216β„Ž2.(1.2) 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 πœƒβˆ’πœ‹/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 πœƒβ‰ πœ‹/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:

1𝑔=4πœ‹β„Ž2ξ€œ0𝑑/2ξ€œπœŒπ‘‘πœŒ02πœ‹π‘‘π›½sin3πœƒπ‘Ž,(1.3) where 𝜌 and 𝛽 are polar coordinates on the aperture plane and πœƒπ‘Ž is the incidence angle of the photon at a particular value of 𝜌 and 𝛽, and is given by [3]

cot2πœƒπ‘Ž=cot2πœŒπœƒβˆ’2β„ŽπœŒcotπœƒcos(π›½βˆ’πœ™)+2β„Ž2,(1.4) where πœ™ is the third coordinate of the point source. For integrals over 2πœ‹ the difference π›½βˆ’πœ™ can be replaced with only 𝛽 without loss of generality. This yields

sin2πœƒπ‘Ž=ξ‚Έcsc2πœŒπœƒβˆ’2β„ŽπœŒcotπœƒcos𝛽+2β„Ž2ξ‚Ήβˆ’1.(1.5)

Applying this to (1.3),

1𝑔=4πœ‹β„Ž2ξ€œ0𝑑/2ξ€œπœŒπ‘‘πœŒ02πœ‹ξ‚Έπ‘‘π›½csc2πœŒπœƒβˆ’2β„ŽπœŒcotπœƒcos𝛽+2β„Ž2ξ‚Ήβˆ’3/2.(1.6)

When there is penetration the above would be modified to

1𝑔=4πœ‹β„Ž2ξ€œβˆž0ξ€œπœŒπ‘‘πœŒ02πœ‹ξ‚Έπ‘‘π›½csc2πœŒπœƒβˆ’2β„ŽπœŒcotπœƒcos𝛽+2β„Ž2ξ‚Ήβˆ’3/2π‘’βˆ’πœ‡Ξ”πΏ,(1.7) 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

𝑔=𝑔NP+14πœ‹β„Ž2ξ€œβˆžπ‘‘/2ξ€œπœŒπ‘‘πœŒ02πœ‹ξ‚Έπ‘‘π›½csc2πœŒπœƒβˆ’2β„ŽπœŒcotπœƒcos𝛽+2β„Ž2ξ‚Ήβˆ’3/2π‘’βˆ’πœ‡Ξ”πΏ=𝑔NP+𝑔PEN,(1.8) 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 the 𝛽 integral in (1.6).

2. Two Approximations for 𝑔𝑁𝑃

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

2.1. The Expansion for 𝑑 Small

Let us, in this section, denote 𝑔NP by 𝑔. Upon substituting π‘Ÿ=𝑑/2β„Ž and πœ‚=𝜌/β„Ž into (1.6) we obtain

1𝑔(π‘Ÿ,πœƒ)=ξ€œ4πœ‹02πœ‹ξ€œπ‘Ÿ0ξ€Ίcsc2πœƒβˆ’2πœ‚cotπœƒcos𝛽+πœ‚2ξ€»βˆ’3/2=1πœ‚π‘‘πœ‚π‘‘π›½,ξ€œ4πœ‹02πœ‹ξ‚΅ξ€œπ‘Ÿ0ξ‚Άπœ‚πœ™(𝛽,πœ‚,πœƒ)π‘‘πœ‚π‘‘π›½.(2.1) Clearly, 𝑔(0,πœƒ)=0. Since

π‘”π‘Ÿ1(π‘Ÿ,πœƒ)=ξ€œ4πœ‹02πœ‹π‘Ÿπœ™(𝛽,π‘Ÿ,πœƒ)𝑑𝛽,(2.2) we have π‘”π‘Ÿ(0,πœƒ)=0. Differentiating (2.12) three more times with respect to π‘Ÿ we obtain

π‘”π‘Ÿπ‘Ÿ1(π‘Ÿ,πœƒ)=ξ€œ4πœ‹02πœ‹ξ€·π‘Ÿπœ™π‘Ÿξ€Έπ‘”(𝛽,0,πœƒ)+πœ™(𝛽,π‘Ÿ,πœƒ)𝑑𝛽,π‘Ÿπ‘Ÿπ‘Ÿ1(π‘Ÿ,πœƒ)=ξ€œ4πœ‹02πœ‹ξ€·π‘Ÿπœ™π‘Ÿπ‘Ÿ(𝛽,0,πœƒ)+2πœ™π‘Ÿξ€Έπ‘”(𝛽,π‘Ÿ,πœƒ)𝑑𝛽,π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ1(π‘Ÿ,πœƒ)=ξ€œ4πœ‹02πœ‹ξ€·π‘Ÿπœ™π‘Ÿπ‘Ÿπ‘Ÿ(𝛽,0,πœƒ)+3πœ™π‘Ÿπ‘Ÿξ€Έ(𝛽,π‘Ÿ,πœƒ)𝑑𝛽,(2.3) and then setting π‘Ÿ=0 in each of the resulting equations we obtain

π‘”π‘Ÿπ‘Ÿ1(0,πœƒ)=ξ€œ4πœ‹02πœ‹π‘”πœ™(𝛽,0,πœƒ)𝑑𝛽,π‘Ÿπ‘Ÿπ‘Ÿ1(0,πœƒ)=ξ€œ2πœ‹02πœ‹πœ™π‘Ÿπ‘”(𝛽,0,πœƒ)𝑑𝛽,π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ3(0,πœƒ)=ξ€œ4πœ‹02πœ‹πœ™π‘Ÿπ‘Ÿ(𝛽,0,πœƒ)𝑑𝛽.(2.4) For convenience let us write

𝑒=csc2πœƒ,𝑣=cotπœƒ,(2.5) so πœ™ becomes

ξ€·πœ™(𝛽,π‘Ÿ,πœƒ)=π‘’βˆ’2π‘£π‘Ÿcos𝛽+π‘Ÿ2ξ€Έβˆ’3/2.(2.6) Then

πœ™(𝛽,0,πœƒ)=π‘’βˆ’3/2,πœ™π‘Ÿ(𝛽,0,πœƒ)=3π‘’βˆ’5/2πœ™π‘£cos𝛽,π‘Ÿπ‘Ÿ(𝛽,0,πœƒ)=15π‘’βˆ’7/2𝑣2cos2π›½βˆ’3π‘’βˆ’5/2.(2.7) We substitute into the Taylor expansion

𝑔(π‘Ÿ,πœƒ)=𝑔(0,πœƒ)+π‘”π‘Ÿ1(0,πœƒ)π‘Ÿ+𝑔2!π‘Ÿπ‘Ÿ(0,πœƒ)π‘Ÿ2+1𝑔3!π‘Ÿπ‘Ÿπ‘Ÿ(0,πœƒ)π‘Ÿ3+1𝑔4!π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿ(0,πœƒ)π‘Ÿ4+β‹―(2.8) and obtain

1𝑔(π‘Ÿ,πœƒ)=2ξ‚΅1ξ€œ4πœ‹02πœ‹π‘’βˆ’3/2ξ‚Άπ‘Ÿπ‘‘π›½2+1ξ‚΅33!ξ€œ2πœ‹02πœ‹6π‘’βˆ’5/2ξ‚Άπ‘Ÿπ‘£cos𝛽𝑑𝛽3+1ξ‚΅14!ξ€œ4πœ‹02πœ‹3ξ€·15π‘’βˆ’7/2𝑣2cos2π›½βˆ’3π‘’βˆ’5/2ξ€Έξ‚Άπ‘Ÿπ‘‘π›½4+β‹―.(2.9) Since

ξ€œ02πœ‹ξ€œcos𝛽𝑑𝛽=0,02πœ‹cos2𝛽𝑑𝛽=πœ‹(2.10) and π‘’βˆ’1=sin2πœƒ we obtain

π‘Ÿπ‘”(π‘Ÿ,πœƒ)=24sin33πœƒ+ξ€·325sin5πœƒcos2πœƒβˆ’2sin5πœƒξ€Έπ‘Ÿ4+β‹―.(2.11) Finally we replace cos2πœƒ by 1βˆ’sin2πœƒ and π‘Ÿ by 𝑑/2β„Ž and obtain

𝑔NP=116sin3πœƒξ‚€π‘‘β„Žξ‚2+3512sin5πœƒξ€·3βˆ’5sin2πœƒξ€Έξ‚€π‘‘β„Žξ‚4𝑑+π’ͺ6ξ€Έ.(2.12)

2.2. The Expansion for πœƒβˆ’πœ‹/𝟐 Small

To start we let

𝑓(πœƒ,πœ‚,𝛽)=csc2πœƒβˆ’2πœ‚cotπœƒcos𝛽+πœ‚2ξ€Έβˆ’3/2(2.13) and compute its Taylor expansion at πœƒ=πœ‹/2,

𝑓(πœƒ,πœ‚,𝛽)=1+πœ‚2ξ€Έβˆ’3/2ξ€·βˆ’31+πœ‚2ξ€Έβˆ’5/2ξ‚€πœ‹πœ‚cosπ›½πœƒβˆ’2+1ξ‚€ξ€·2!151+πœ‚2ξ€Έβˆ’7/2πœ‚2cos2ξ€·π›½βˆ’31+πœ‚2ξ€Έβˆ’5/2πœ‹ξ‚ξ‚€πœƒβˆ’22ξ‚€πœ‹+π’ͺπœƒβˆ’23.(2.14)

Since

𝑔NP=1ξ€œ4πœ‹02πœ‹ξ€œ0𝑑/2β„Žπ‘“(πœƒ,πœ‚,𝛽)πœ‚π‘‘πœ‚π‘‘π›½(2.15) and since we know that only even powers of πœƒβˆ’πœ‹/2 occur, we obtain

𝑔NP=1211βˆ’βˆš1+(𝑑/2β„Ž)2ξƒͺβˆ’38(𝑑/2β„Ž)2ξ€·1+(𝑑/2β„Ž)2ξ€Έ5/2ξ‚€πœ‹πœƒβˆ’22ξ‚€πœ‹+π’ͺπœƒβˆ’24.(2.16)

3. The Contour Integral

We have

ξ€œπΌ=02πœ‹ξ‚Έπ‘‘π›½csc2πœŒπœƒβˆ’2β„ŽπœŒcotπœƒcos𝛽+2β„Ž2ξ‚Ήβˆ’3/2,𝑔(3.1)NP=14πœ‹β„Ž2ξ€œ0𝑑/2πΌπœŒπ‘‘πœŒ.(3.2)𝐼 may be recast as a contour integral by setting 𝑧=𝑒𝑖𝛽: ξ€œπΌ=Ξ“π‘‘π‘§ξ‚ƒπ‘–π‘§βˆ’π‘ξ‚€π‘§2𝑧2π‘Žβˆ’2𝑏𝑧+1ξ‚ξ‚„βˆ’3/2=ξ‚€2𝑏3/21π‘–ξ€œΞ“ξ‚Έ(π‘§βˆ’π‘₯0)(π‘₯1βˆ’π‘§)π‘§ξ‚Ήβˆ’3/2𝑑𝑧𝑧,(3.3) where

Ξ“istheunitcircle,π‘Ž=csc2πœŒπœƒ+2β„Ž2,πœŒπ‘=2β„Žπ‘₯cotπœƒ,0=π‘Žπ‘ξƒ¬ξ‚™1βˆ’π‘1βˆ’2π‘Ž2ξƒ­,π‘₯1=π‘Žπ‘ξƒ¬ξ‚™1+𝑏1βˆ’2π‘Ž2ξƒ­.(3.4) We note that π‘Ž>𝑏 and so 0<π‘₯0<1<π‘₯1.

In this section we take 𝜌,πœƒ, and β„Ž to be constant. We write

𝑧𝐹(𝑧)=ξ€·π‘§βˆ’π‘₯0π‘₯ξ€Έξ€·1ξ€Έξƒ­ξ‚Έπ‘§βˆ’π‘§((π‘§βˆ’π‘₯0)(π‘₯1ξ‚Ήβˆ’π‘§))1/21𝑧=1ξ€·π‘§βˆ’π‘₯0π‘₯ξ€Έξ€·1ξ€Έξ‚Έπ‘§βˆ’π‘§(π‘§βˆ’π‘₯0)(π‘₯1ξ‚Ήβˆ’π‘§)1/2,ξ‚€2𝐼=𝑏3/21π‘–ξ€œΞ“πΉ(𝑧)𝑑𝑧.(3.5) We need to be careful because of the square roots. We use the usual convention that

lim𝑦→0Β±βˆšβˆ’1+𝑖𝑦=±𝑖.(3.6) We wish to show that 𝐹(𝑧) is holomorphic (and single valued) in

ξ€½{π‘§βˆΆ|𝑧|<1}βˆ’π‘₯∢0<π‘₯<π‘₯0ξ€Ύ.(3.7) To isolate the square root factor in 𝐹 we set

𝑧𝑔(𝑧)=ξ€·π‘§βˆ’π‘₯0π‘₯ξ€Έξ€·1ξ€Έ.βˆ’π‘§(3.8) We take |𝑧|≀1. If in addition Re𝑧<0 and |Im𝑧|β‰ͺ|π‘§βˆ’π‘₯0| then Re(π‘§βˆ’π‘₯0)(π‘₯1βˆ’π‘§)<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>π‘₯0 (no condition on Im𝑧 now) then Re(π‘§βˆ’π‘₯0)(π‘₯1βˆ’π‘§)>0 and 𝑔(𝑧) has a single valued square root in Re𝑧>π‘₯0.

It follows from Cauchy's theorem that the integral of 𝐹(𝑧) over the unit circle Ξ“ is equal to the integral over the β€œ barbell” 𝛾 around 0 and π‘₯0. More explicitly, 𝛾 is the boundary of

𝐡=π΅πœ–(0)βˆͺπ΅πœ–ξ€·π‘₯0ξ€Έβˆͺ𝑅,(3.9) where π΅πœ–(π‘₯) is the ball of radius πœ– and centered at π‘₯ and 𝑅 is the rectangle

𝑅=𝑧=π‘₯+π‘–π‘¦βˆΆ0≀π‘₯≀π‘₯0ξ€Ύ,βˆ’π›Ώβ‰€π‘¦β‰€π›Ώ(3.10) with 𝛿β‰ͺπœ–, see Figure 2.

The curve 𝛾 decomposes into smooth pieces which we describe using the angle 𝛼=arcsin(𝛿/πœ–)

π›Ύπ‘Ÿ=ξ€½||π‘§βˆΆπ‘§βˆ’π‘₯0||=πœ–,βˆ’πœ‹+𝛼≀argπ‘§βˆ’π‘₯0ξ€Ύ,π›Ύβ‰€πœ‹βˆ’π›Όπ‘™π›Ύ={π‘§βˆΆ|𝑧|=πœ–,𝛼≀arg(𝑧)≀2πœ‹βˆ’π›Ό},Β±=ξ€½π‘§βˆΆπ‘§=π‘₯±𝑖𝛿,πœ–cos𝛼≀π‘₯≀π‘₯0ξ€Ύ.βˆ’πœ–cos𝛼(3.11)

To bound the integral over 𝛾𝑙 we set 𝑧=πœ–π‘’π‘–πœƒ

||||ξ€œπ›Ύπ‘™πΉ||||=||||ξ€œ(𝑧)𝑑𝑧𝛼2πœ‹βˆ’π›ΌπΉξ€·πœ–π‘’π‘–πœƒξ€Έπœ–π‘’π‘–πœƒ||||ξ€·πœ–π‘‘πœƒ=π’ͺ3/2ξ€Έ.(3.12) To evaluate βˆ«π›Ύπ‘ŸπΉ(𝑧)𝑑𝑧 we set 𝑧=π‘₯0+πœ–π‘’π‘–πœƒ. Then

𝐹π‘₯0+πœ–π‘’π‘–πœƒξ€Έ=πΆπœ–βˆ’3/2π‘’βˆ’(3/2)π‘–πœƒξ€·πœ–+π’ͺβˆ’1/2ξ€Έ(3.13) with

π‘₯𝐢=01/2ξ€·π‘₯1βˆ’π‘₯0ξ€Έ3/2,(3.14) and so

ξ€œπ›Ύπ‘Ÿξ€œπΉ(𝑧)𝑑𝑧=πœ‹βˆ’π›Όβˆ’πœ‹+π›Όξ€·πœ–βˆ’3/2πΆπ‘’βˆ’(3/2)π‘–πœƒξ€·πœ–+π’ͺβˆ’1/2ξ€Έξ€Έπœ–π‘–π‘’π‘–πœƒ=ξ€œ(1+π’ͺ(πœ–))π‘‘πœƒπœ‹βˆ’π›Όβˆ’πœ‹+π›ΌπΆπ‘–πœ–βˆ’1/2π‘’βˆ’(1/2)π‘–πœƒξ€·πœ–+π’ͺ1/2ξ€Έπ‘‘πœƒ.(3.15) Since 𝛼=𝑂(πœ–), we may set 𝛼=0 and absorb the error in the π’ͺ(πœ–1/2) term. We then do the integration and obtain

ξ€œπ›Ύπ‘ŸπΉ(𝑧)𝑑𝑧=4π‘–πΆβˆšπœ–ξ€·πœ–+π’ͺ1/2ξ€Έ.(3.16) We want to compute the integral of 𝐹 over 𝛾+. We need to be careful about the square root. Recall that we are assuming

πœ–cos𝛼<π‘₯<π‘₯0βˆ’πœ–cos𝛼,𝛿β‰ͺπœ–.(3.17)

Lemma 3.1. On 𝛾+π‘₯𝐹(𝑧)=𝑖0π‘₯βˆ’π‘₯ξ€Έξ€·1βˆ’π‘₯ξ€Έξ€Έβˆ’3/2π‘₯1/2+π’ͺ(𝛿)(3.18) and on π›Ύβˆ’π‘₯𝐹(𝑧)=βˆ’π‘–ξ€·ξ€·0π‘₯βˆ’π‘₯ξ€Έξ€·1βˆ’π‘₯ξ€Έξ€Έβˆ’3/2π‘₯1/2+π’ͺ(𝛿).(3.19)

Proof. This is a consequence of the branching of the square root function. Let 𝑧𝐻(𝑧)=ξ€·π‘§βˆ’π‘₯0π‘₯ξ€Έξ€·1ξ€Έ=π‘§ξ€·βˆ’π‘§π‘§βˆ’π‘₯0π‘₯ξ€Έξ€·1βˆ’π‘§ξ€Έ||π‘§βˆ’π‘₯0||2||π‘₯1||βˆ’π‘§2,ξ€·β„Ž(𝑧)=π‘§π‘§βˆ’π‘₯0π‘₯ξ€Έξ€·1βˆ’π‘§ξ€Έ.(3.20) We see that β„Žξ€·(π‘₯+𝑖𝛿)=π‘₯π‘₯βˆ’π‘₯0π‘₯ξ€Έξ€·1ξ€Έξ€·π‘₯βˆ’π‘₯+𝑖𝛿2ξ€Έξ€·π›Ώβˆ’1+π’ͺ2ξ€Έ.(3.21) Recall that 𝛿 is replaced by βˆ’π›Ώ on π›Ύβˆ’ and that 0<π‘₯<π‘₯0<π‘₯1. So Re𝐻(𝑧)<0 on both 𝛾+ and π›Ύβˆ’ while Im𝐻(𝑧)<0 on 𝛾+ and Im𝐻(𝑧)>0 on π›Ύβˆ’. Thus on 𝛾+𝐻(𝑧)1/2||||π‘₯=βˆ’π‘–(π‘₯βˆ’π‘₯0)(π‘₯1||||βˆ’π‘₯)1/2ξ‚΅π‘₯+π’ͺ(𝛿)=βˆ’π‘–(π‘₯0βˆ’π‘₯)(π‘₯1ξ‚Άβˆ’π‘₯)1/2+π’ͺ(𝛿),(3.22) while on π›Ύβˆ’π»(𝑧)1/2ξ‚΅π‘₯=𝑖(π‘₯0βˆ’π‘₯)(π‘₯1ξ‚Άβˆ’π‘₯)1/2+π’ͺ(𝛿).(3.23) Since 1𝐹(𝑧)=𝐹(π‘₯)+π’ͺ(𝛿)=ξ€·π‘₯βˆ’π‘₯0π‘₯ξ€Έξ€·1ξ€Έβˆ’π‘₯𝐻(𝑧)1/21+π’ͺ(𝛿)=βˆ’ξ€·π‘₯0π‘₯βˆ’π‘₯ξ€Έξ€·1ξ€Έπ»βˆ’π‘₯(𝑧)1/21+π’ͺ(𝛿)=βˆ’ξ€·π‘₯0π‘₯βˆ’π‘₯ξ€Έξ€·1π‘₯βˆ’π‘₯βˆ’π‘–(π‘₯0βˆ’π‘₯)(π‘₯1ξ‚Άβˆ’π‘₯)1/2ξƒͺ+π’ͺ(𝛿),(3.24) the lemma follows.

We also need to be careful about the orientation. The curve 𝛾 is oriented counter-clockwise. This means that the integration over 𝛾+ goes in the direction of π‘₯0 to 0. We need to reverse its sign in order to write it in the conventional way

ξ€œπ›Ύ+ξ€œπΉ(𝑧)𝑑𝑧=βˆ’π‘₯0βˆ’πœ–cosπ›Όπœ–cosπ›Όξ€œπΉ(π‘₯+𝑖𝛿)𝑑π‘₯=βˆ’π‘₯0βˆ’πœ–cosπ›Όπœ–cos𝛼𝑖π‘₯1/2𝑑π‘₯ξ€·π‘₯0ξ€Έβˆ’π‘₯3/2ξ€·π‘₯1ξ€Έβˆ’π‘₯3/2+π’ͺ(πœ–).(3.25) We let π‘₯=π‘₯0𝑑 and Ξ”=πœ–π‘₯0βˆ’1cos𝛼 and use 𝛿β‰ͺπœ– to obtain

ξ€œπ›Ύ+ξ€œπΉ(𝑧)𝑑𝑧=βˆ’Ξ”1βˆ’Ξ”π‘–π‘₯03/2𝑑1/2𝑑π‘₯ξ€·π‘₯0βˆ’π‘₯0𝑑3/2ξ€·π‘₯1βˆ’π‘₯0𝑑3/2+π’ͺ(πœ–).(3.26) Now let

π‘₯𝜏=1π‘₯0>1.(3.27) We have

ξ€œπ›Ύ+𝑖𝐹(𝑧)𝑑𝑧=βˆ’π‘₯03/2ξ€œΞ”1βˆ’Ξ”π‘‘1/2𝑑𝑑(πœβˆ’π‘‘)3/2(1βˆ’π‘‘)3/2+π’ͺ(πœ–).(3.28) We may let Ξ”β†’0 in the lower limit, but not in the upper limit. So we now have

ξ€œπ›Ύ+𝑖𝐹(𝑧)𝑑𝑧=βˆ’π‘₯03/2ξ€œ01βˆ’Ξ”π‘‘1/2𝑑𝑑(πœβˆ’π‘‘)3/2(1βˆ’π‘‘)3/2+π’ͺ(πœ–).(3.29) Here Ξ”=πœ–π‘₯0βˆ’1cos𝛼. We may let 𝛿→0 and redefine Ξ” to be πœ–π‘₯0βˆ’1.

We get the same value for the integral over π›Ύβˆ’. 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 function 𝑔(𝑑), continuously differentiable on the interval [0,1], we have

ξ€œ01βˆ’Ξ”π‘”(𝑑)(1βˆ’π‘‘)3/2𝑑𝑑=2𝑔(1)Ξ”1/2ξ€œβˆ’2𝑔(0)βˆ’210(1βˆ’π‘‘)βˆ’1/2π‘”ξ…žξ€·Ξ”(𝑑)𝑑𝑑+𝑂1/2ξ€Έ(3.30) as can easily be seen by an integration by parts.

We apply this to 𝑔(𝑑)=𝑑1/2/(πœβˆ’π‘‘)3/2 as in (3.29). This yields

ξ€œπ›Ύ+𝐹(𝑧)𝑑𝑧=βˆ’2𝑖(πœβˆ’1)3/2π‘₯03/2Ξ”1/2+𝑖π‘₯03/2ξ‚΅ξ€œ10π‘‘βˆ’1/2(πœβˆ’π‘‘)βˆ’3/2+3𝑑1/2(πœβˆ’π‘‘)βˆ’5/2ξ‚Άξ€·πœ–π‘‘π‘‘+π’ͺ1/2ξ€Έ.(3.31) We now take twice this quantity and add to it the results of (3.14) and (3.16),

ξ€œ|𝑧|=1ξ€œπΉ(𝑧)𝑑𝑧=𝛾=𝐹(𝑧)𝑑𝑧4π‘–πΆβˆšπœ–ξƒ©βˆš+2βˆ’2𝑖π‘₯0ξ€·π‘₯1βˆ’π‘₯0ξ€Έβˆ’3/2βˆšπœ–+𝑖π‘₯03/2ξ€œ10(1βˆ’π‘‘)βˆ’1/2ξ€·(πœβˆ’π‘‘)βˆ’3/2π‘‘βˆ’1/2+3𝑑1/2(πœβˆ’π‘‘)βˆ’5/2ξ€Έξ‚Άξ€·πœ–π‘‘π‘‘+π’ͺ1/2ξ€Έ.(3.32) We substitute the value for 𝐢 from (3.14) and let πœ–β†’0

ξ€œ|𝑧|=1𝐹(𝑧)𝑑𝑧=2𝑖π‘₯03/2ξ€œ10(1βˆ’π‘‘)βˆ’1/2ξ€·(πœβˆ’π‘‘)βˆ’3/2π‘‘βˆ’1/2+3𝑑1/2(πœβˆ’π‘‘)βˆ’5/2𝑑𝑑.(3.33) Finally, we evaluate this integral using Mathematica and obtain

ξ€œ|𝑧|=1𝐹(𝑧)𝑑𝑧=2𝑖π‘₯03/2ξ‚Ά[][])2(2𝜏𝐸1/πœβˆ’(βˆ’1+𝜏)𝐾1/𝜏(πœβˆ’1)2√𝜏,(3.34) where 𝐾 and 𝐸 are the complete elliptic functions of the first and second kinds

𝐾[π‘š]=ξ€œ10ξ€·1βˆ’π‘šπ‘‘2ξ€Έβˆ’1/2ξ€·1βˆ’π‘‘2ξ€Έβˆ’1/2𝐸[π‘š]=ξ€œπ‘‘π‘‘,10ξ€·1βˆ’π‘šπ‘‘2ξ€Έ1/2ξ€·1βˆ’π‘‘2ξ€Έβˆ’1/2𝑑𝑑.(3.35)

We now have

ξ‚€2𝐼=𝑏3/21π‘–ξ€œΞ“=ξ‚€2𝐹(𝑧)𝑑𝑧𝑏3/22π‘₯03/22(2𝜏𝐸(1/𝜏)βˆ’(πœβˆ’1)𝐾(1/𝜏))(πœβˆ’1)2√𝜏=ξ‚€2𝑏3/24𝜏1/4(πœβˆ’1)2𝑔(2𝜏𝐸(1/𝜏)βˆ’(πœβˆ’1)𝐾(1/𝜏)),(3.36)NP=14πœ‹β„Ž2ξ€œ0𝑑/2πœŒπΌπ‘‘πœŒ(3.37) with

πœŒπ‘=2β„Žcotπœƒ,π‘Ž=csc2πœŒπœƒ+2β„Ž2,π‘₯(3.38)𝜏=1π‘₯0=1π‘₯02,π‘₯0=π‘Žπ‘ξƒ¬ξ‚™1βˆ’π‘1βˆ’2π‘Ž2ξƒ­,π‘₯1=π‘Žπ‘ξƒ¬ξ‚™1+𝑏1βˆ’2π‘Ž2ξƒ­.(3.39)

4. Limiting Cases

We validate (3.36) by considering two limiting cases.

4.1. Expansion for Small 𝑑

We look for an expansion of 𝑔NP under the assumption that 𝑑/β„Ž is small and rederive (2.12). The dependence of π‘₯0 on 𝜌 is given in (3.39). The two-term Taylor expansion of π‘₯0 as a function of 𝜌 is

π‘₯0=𝐴𝜌+𝐡𝜌3ξ€·πœŒ+π’ͺ5ξ€Έ(4.1) with

𝐴=cosπœƒsinπœƒβ„Ž,𝐡=𝐴3βˆ’π΄β„Ž2csc2πœƒ.(4.2) Since 1𝜏=π‘₯02,(4.3) we have 1𝜏=𝐴2𝜌2ξ€·πœŒ+π’ͺ4ξ€Έ,(4.4)1(πœβˆ’1)2=1𝜏2+2𝜏3ξ€·πœŒ+π’ͺ8ξ€Έ.(4.5) It is convenient to work with the quantity 𝜌2𝜏, so we note, using (4.1),

𝜌21𝜏=𝐴2βˆ’2𝐡𝐴3𝜌2ξ€·πœŒ+π’ͺ4ξ€Έ.(4.6) We start with

ξ‚€2𝐼=𝑏3/24𝜏1/4(πœβˆ’1)2ξ‚€ξ‚€12πœπΈπœξ‚ξ‚€1βˆ’(πœβˆ’1)𝐾𝜏.(4.7) Substituting the expression for 𝑏 from (3.38) and for (πœβˆ’1)2 from (4.5) we obtain, after some manipulation

ξƒ©ξ‚€β„ŽπΌ=cotπœƒ3/24ξ€·πœŒ2πœξ€Έ3/41πœξ€·πœŒ+π’ͺ4ξ€Έξƒͺξ‚€ξ‚€12πœπΈπœξ‚ξ‚€1βˆ’(πœβˆ’1)𝐾𝜏.(4.8)

We now use the expansions (as given by Mathematica)

πœ‹πΈ(𝑑)=2βˆ’πœ‹π‘‘8𝑑+π’ͺ2ξ€Έ,πΎπœ‹(𝑑)=2+πœ‹π‘‘8𝑑+π’ͺ2ξ€Έ.(4.9) So

ξ‚€β„ŽπΌ=cotπœƒ3/24ξ€·πœŒ2πœξ€Έ3/4ξ‚€πœ‹2+πœ‹+πœ‹8πœπœξ‚ξ€·πœŒ+π’ͺ4ξ€Έ.(4.10) From the expression for 𝜌2𝜏 in (4.6), we derive

1ξ€·πœŒ2πœξ€Έ3/4=𝐴3/2ξ‚€31+2𝐡𝐴𝜌2ξ€·πœŒ+π’ͺ4,(4.11) and this leads to

ξ‚€β„ŽπΌ=4cotπœƒ3/2𝐴3/2ξ‚΅πœ‹2+πœ‹+πœ‹8𝜏𝜏+3𝐡𝜌2πœ‹ξ‚Άξ€·πœŒ4𝐴+π’ͺ4ξ€Έ.(4.12) We now substitute for 𝐴 and 𝐡 using (4.2) and also for πœβˆ’1 using (4.4)

𝐼=sin3πœƒξ‚΅2πœ‹+3πœ‹2ξ€·3sin2πœƒβˆ’5sin4πœƒξ€ΈπœŒ2β„Ž2ξ‚Άξ€·πœŒ+π’ͺ4ξ€Έ.(4.13) Substituting this into (3.37) and integrating, we again derive (2.12).

4.2. An Expansion for πœƒ Near πœ‹/𝟐

We show how (4.7) also gives the expansion of Section 2.2. From (3.38) we see that

πœŒπ‘Ž=1+2β„Ž2+ξ‚€πœ‹πœƒβˆ’22ξ‚€πœ‹+π’ͺπœƒβˆ’24,(4.14)𝑏=βˆ’2πœŒβ„Žξ‚€πœ‹πœƒβˆ’2ξ‚ξ‚€πœ‹+π’ͺπœƒβˆ’23.(4.15) So from (3.39) we have

π‘₯0=𝑏𝑏2π‘Ž1+24π‘Ž2ξ‚€πœ‹+π’ͺπœƒβˆ’23ξ‚Ά.(4.16) Thus

𝜏=4π‘Ž2𝑏2ξ‚€πœ‹βˆ’2+π’ͺπœƒβˆ’22,1(4.17)(πœβˆ’1)2=1𝜏2ξ‚΅21+πœξ‚€πœ‹+π’ͺπœƒβˆ’24ξ‚Ά,1πœξ‚€ξ‚€12πœπΈπœξ‚ξ‚€1βˆ’(πœβˆ’1)𝐾𝜏=πœ‹ξ‚ξ‚2+9πœ‹ξ‚€πœ‹8𝜏+π’ͺπœƒβˆ’24.(4.18) These expansions let us write (4.7) as

ξ‚€2𝐼=𝑏3/24𝜏3/4ξ‚΅ξ‚€21+πœπœ‹ξ‚ξ‚€2+9πœ‹ξ‚ξ‚€πœ‹8𝜏+π’ͺπœƒβˆ’24ξ‚Ά=27/2𝑏2πœξ€Έ3/4ξ‚΅ξ‚€21+πœπœ‹ξ‚ξ‚€2+9πœ‹ξ‚ξ‚€πœ‹8𝜏+π’ͺπœƒβˆ’24ξ‚Ά.(4.19) Using (4.15) and (4.17) we obtain a substitute for (4.1)

1𝑏2πœξ€Έ3/4=4βˆ’3/4π‘Žβˆ’3/2ξ‚΅1+3𝑏28π‘Ž2ξ‚Άξ‚€πœ‹+π’ͺπœƒβˆ’24.(4.20) This gives, in light of (4.14),

𝐼=2πœ‹π‘Žβˆ’3/2+158πœ‹π‘2π‘Žβˆ’7/2ξ‚€πœ‹+π’ͺπœƒβˆ’24ξ‚΅πœŒ=2πœ‹1+2β„Ž2ξ‚Άβˆ’3/2ξ‚΅πœŒβˆ’3πœ‹1+2β„Ž2ξ‚Άβˆ’5/2ξ‚€πœ‹πœƒβˆ’22+152πœ‹πœŒ2β„Ž2ξ‚΅πœŒ1+2β„Ž2ξ‚Άβˆ’7/2ξ‚€πœ‹πœƒβˆ’22ξ‚€πœ‹+π’ͺπœƒβˆ’24.(4.21) Finally, we integrate term by term.

1𝑔=4πœ‹β„Ž2ξ€œ0𝑑/2=1𝜌𝐼(𝜌)π‘‘πœŒ4πœ‹β„Ž22πœ‹β„Ž2𝑑1βˆ’1+2β„Ž2ξ‚Άβˆ’1/2ξƒͺβˆ’3πœ‹β„Ž213𝑑1βˆ’1+2β„Ž2ξ‚Άβˆ’3/2ξƒͺξ‚€πœ‹πœƒβˆ’22+152πœ‹β„Ž22βˆ’152+5(𝑑/2β„Ž)2𝑑151+2β„Ž2ξ‚Άβˆ’5/2ξƒͺξ‚€πœ‹πœƒβˆ’22ξƒͺξ‚€πœ‹+π’ͺπœƒβˆ’24=12𝑑1βˆ’1+2β„Ž2ξ‚Άβˆ’1/2ξƒͺβˆ’38𝑑1+2β„Ž2ξ‚Άβˆ’5/2𝑑2β„Ž2ξ‚€πœ‹πœƒβˆ’22ξ‚€πœ‹+π’ͺπœƒβˆ’24.(4.22)

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/β„Ž and on the value of πœƒ. 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/β„Ž, the numerical integration (solid lines) is shown with the results of the residue calculation (open circles) in Figure 3 as a function of πœƒ. Figure 4 shows the numerical integration and residue results for fixed values of πœƒ as a function of 𝜌/β„Ž. All corresponding values of the residue calculation and the numerical integration agree.

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.

Acknowledgment

Scott D. Metzler was supported by the National Institute for Biomedical Imaging and Bioengineering of the National Institutes of Health under Grant R01-EB-6558.

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Copyright © 2010 Howard Jacobowitz and Scott D. Metzler. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


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