#### 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 βCATβ 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 511βkeV, 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

where is the diameter of the pinhole, is the perpendicular distance, and is the incidence angle of the photon on the aperture plane at the center of aperture. This formula becomes exact in the limit that goes to zero. Also, an exact formula is easily obtained when

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 or 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 . 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:

where and are polar coordinates on the aperture plane and is the incidence angle of the photon at a particular value of and , and is given by [3]

where is the third coordinate of the point source. For integrals over the difference can be replaced with only without loss of generality. This yields

Applying this to (1.3),

When there is penetration the above would be modified to

where the path length through attenuating material, , is zero for and is otherwise given by (11) in [3].

Thus, (1.7) can be rewritten as

where is the efficiency due to penetration and , the efficiency under the assumption of no penetration, is given by (1.6).

In this paper we use complex variable methods to calculate the integral in (1.6).

#### 2. Two Approximations for

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

##### 2.1. The Expansion for Small

Let us, in this section, denote by . Upon substituting and into (1.6) we obtain

Clearly, . Since

we have . Differentiating (2.12) three more times with respect to we obtain

and then setting in each of the resulting equations we obtain

For convenience let us write

so becomes

Then

We substitute into the Taylor expansion

and obtain

Since

and we obtain

Finally we replace by and by and obtain

##### 2.2. The Expansion for Small

To start we let

and compute its Taylor expansion at ,

Since

and since we know that only even powers of occur, we obtain

#### 3. The Contour Integral

We have

may be recast as a contour integral by setting : where

We note that and so .

In this section we take and to be constant. We write

We need to be careful because of the square roots. We use the usual convention that

We wish to show that is holomorphic (and single valued) in

To isolate the square root factor in we set

We take . If in addition and then . This implies that never takes on negative real values and so has a single valued square root in . Similarly, if and (no condition on now) then and has a single valued square root in .

It follows from Cauchy's theorem that the integral of over the unit circle is equal to the integral over the β barbellβ around and . More explicitly, is the boundary of

where is the ball of radius and centered at and is the rectangle

with , see Figure 2.

The curve decomposes into smooth pieces which we describe using the angle

To bound the integral over we set

To evaluate we set . Then

with

and so

Since , we may set and absorb the error in the term. We then do the integration and obtain

We want to compute the integral of over . We need to be careful about the square root. Recall that we are assuming

Lemma 3.1. *On **
and on *

*Proof. *This is a consequence of the branching of the square root function. Let
We see that
Recall that is replaced by on and that . So on both and while on and on . Thus on
while on
Since
the lemma follows.

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

We let and and use to obtain

Now let

We have

We may let in the lower limit, but not in the upper limit. So we now have

Here . We may let and redefine to be .

We get the same value for the integral over . 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 , continuously differentiable on the interval , we have

as can easily be seen by an integration by parts.

We apply this to as in (3.29). This yields

We now take twice this quantity and add to it the results of (3.14) and (3.16),

We substitute the value for from (3.14) and let

Finally, we evaluate this integral using Mathematica and obtain

where and are the complete elliptic functions of the first and second kinds

We now have

with

#### 4. Limiting Cases

We validate (3.36) by considering two limiting cases.

##### 4.1. Expansion for Small

We look for an expansion of under the assumption that is small and rederive (2.12). The dependence of on is given in (3.39). The two-term Taylor expansion of as a function of is

with

Since we have It is convenient to work with the quantity , so we note, using (4.1),

We start with

Substituting the expression for from (3.38) and for from (4.5) we obtain, after some manipulation

We now use the expansions (as given by Mathematica)

So

From the expression for in (4.6), we derive

and this leads to

We now substitute for and using (4.2) and also for using (4.4)

Substituting this into (3.37) and integrating, we again derive (2.12).

##### 4.2. An Expansion for Near

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

So from (3.39) we have

Thus

These expansions let us write (4.7) as

Using (4.15) and (4.17) we obtain a substitute for (4.1)

This gives, in light of (4.14),

Finally, we integrate term by term.

#### 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 and on the value of . 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 , the numerical integration (solid lines) is shown with the results of the residue calculation (open circles) in Figure 3 as a function of . Figure 4 shows the numerical integration and residue results for fixed values of as a function of . 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.