An Integral Needed for Recently-Developed Diffraction-Tomography Algorithms

Integrals needed for recently-developed diffraction-tomography algorithms

M. R. Howells, 5-16-00

Background The new diffraction-tomography reconstruction technique proposed by M. Anastasio and X. Pan (JOSA A, 17, 391-400 (2000) promises a much-reduced computational load. A careful study of the mathematical arguments is therefore indicated. One of the steps involves evaluation of the integral

exp{2pnmrsin(f - q ) - i[kf + 2pnm rcos(f -q )]}df т0 where k is an integer and the other quantities are real variables. Following C. Metz and X. Pan (IEEE Trans. Med. Imag. 14, 643-658 (1995)), we adopt the notation s = 2pnmr, t= 2pnm r and observe that the integrand is periodic with period 2. This implies that the value of the integral is unchanged if the same arbitrary angle is added to both limits. Such “rigid shifts” of the limits will be useful in what follows. We choose to make a change of variable a = p 2 - (f - q ) giving

p +q -2p 2 ж p exp -ikж +qцц exp{scosa + i[ka + tsina]} d(-a) . и и2 шш pт +q 2

We now drop the constant in front, interchange the limits and change d(-a) to da . Then applying a rigid shift, we can get back to the original limits and at the same time eliminate  from the integrand. It is therefore sufficient to study the function defined by Metz and Pan as

2p 1 gk (s,t) = exp{scosa ұ i[tsina - ka ]}da . 2p т0

We shift the range of integration to -p ® +p and note that t sina - ka is an odd function of , while scosa is an even one. Thus either sign can be used in front of i and it is equally valid to use the average of exp{+i[tsina - ka]} and exp{-i[tsina - ka]} . This leads to an equivalent definition of

gk (s,t) , 2p 1 g (s,t) = escosa cos(tsina - ka ) da k 2p т0

Statement of what we need to prove

Metz and Pan show that gk (s,t) is given by the following expression (their equation (A7).

k мй s+ t щ I s2 - t2 if t s п 2 2 k( ) Ј плк s - t ыъ gk (s,t) = н k Metz and Pan (A7) й t + s щ 2 2 п J t - s if t і s пк 2 2 ъ k( ) ол t - s ы

-k where Jk (x) is the kth order Bessel function of the first kind and Ik(x) = i Jk(ix) is the kth order modified Bessel function of the first kind. The argument leading to the last equation is elegantly made by induction starting from the following simpler integral.

2 2 2p мI s - t if t s 1 scosa п 0( ) Ј g (s,t) = e cos(tsina ) da = . 0 2p н 2 2 т J 0 ( t - s ) if t і s 0 оп

However, the derivation of this latter result is not given except via a reference to the tabulation of Gradshteyn and Ryzhik ("Table of Integrals Series and Products", Academic Press, San Diego California (1994)). Therefore we provide the following derivation.

Derivation We start from the first definition of gk (s,t) with shifted limits

+p 1 scosa +itsina g0 (s,t) = e da . 2p -тp

Assuming for the moment that t Ј s, we expand the exponential function in powers of scosa + itsina = s2 - t2 cos(a - b) . k +p м 2 2 k ь 1 п Ҙ ( s - t ) cos(a - b)п g (s,t) = da . 0 2p не k! э -тp k =0 оп юп

We now change the variable to a - b , restore the limits to -p ® +p , and integrate term by term. This can be accomplished by the use of fairly standard definite integrals of cosk x . We start by integrating by parts

k k-1 I k = тcos x dx = тcos xcosx dx k-1 2 k-2 = sinxcos x + (k -1)тsin xcos x dx k-1 = sinxcos x + (k -1)Ik-2 - (k -1)Ik which gives us the recurrence relation

sinxcosk-1 x k - 1 I = + I . k k k k-2

Now since we are interested in the definite integral from -p to +p , the first term vanishes at both limits and we can write

+ p +p +p k -1 (k - 1)(k - 3) cosk x dx = cosk-2 x dx = cosk-4 x dx k k(k - 2) -тp -тp -тp and so on. For k odd this leads to

+ p +p k (k - 1)(k - 3)K 4 Ч2 cos x dx = cosdx = 0 . т k(k - 2)K 5Ч3 т - p -p

On the other hand if k is even we get

+ p +p (k - 1)(k - 3)K 5 3 (k- 1)(k - 3)K 5 3 k! cosk x dx = Ч cos2 x dx = Ч 2p = 2p 2 k т k(k - 2)K 6 Ч4 т k(k - 2)K 6 Ч4 Ч2 [(k 2)!] 2 - p -p

Substituting this in the above equation for g0 (s,t) we find k ж 2 2 k s - t ц 2 2 Ҙ s - t Ҙ з ч 1 ( ) k! и 2 ш . g (s,t) = 2p = 0 2p k! 2 k 2 еk=0 [(k 2)!] 2 еk=0 [(k 2)!]

So finally setting k = 2l we get

2l 2 2 l жi s - t ц (-1) з Ҙ 2 ч и ш 2 2 2 2 g0 (s,t) = = J0 (i s - t ) = I0 ( s - t ) (l)!(l)! lе=0 where we have used one of the standard definitions of the Bessel functions of integral order (see for example G. N. Watson "A treatise on the theory of Bessel functions", Cambridge University Press, 1941, p15). Evidently for s Ј t we also have

2 2 g0 (s,t) = J0 ( s -t ) .

Another application of the function gk (s,t) Equation (4) in the paper by Anastasio and Pan reads

+Ҙ +Ҙ 2p 1 -2pi n x +n z -ikf P (n ,n ,f) = p(x,z,f ) e ( a z ) dxdzdf k a z 2p т т т . x = -Ҙ z= -Ҙ f = 0

The function Pk is similar to the Fourier transform of p which is the projection (Radon transform) of the unknown object a oriented at angle 

+Ҙ p(x, z,f) = тa(r,q,z) dh where x = rcos(f - q), h = rsin(f - q) . -Ҙ

Thus Ҙ Ҙ Ҙ 2p 1 -2pin z -i(2pn x+kf) P (n ,n ,f) = e z dz a(r,q, z) e a dxdhdf k a z 2p т т т т z=0 x=-Ҙ h=-Ҙ f =0 Ҙ Ҙ 2p 2p 1 -i 2pn rcos(f -q)+kf = e-2pin zzdz a(r,q, z) e ( a ) rdrdqdf 2p т т т т z=-Ҙ r=0 q =0 f=0

The problem is to carry out the  integral, which does not involve the unknown object. Changing the variable to u = f -q , we find

Ҙ Ҙ 2p 2p 1 -i 2pn rcos(f-q )+ku P (n ,n ,f) = e-2pin zzdz a(r,q, z) e-ikq e ( a ) rdrdqdu k a z 2p т т т т z= -Ҙ r= 0 q = 0 u= 0 and a further variable change to a = p 2 - u leads to

Ҙ Ҙ 2p p 2-2p p 1 -2pin z -i(2pn rsina -ka) -ik P (n ,n ,f) = e z dz a(r,q ,z) e-ikq e a e 2 rdrdqd(-a) k a z 2p z=т-Ҙ r=т0 qт=0 a =тp 2 Ҙ Ҙ 2p 2p 1 -2pin z -i(2pn rsina -ka ) k = e z dz a(r,q,z) e-ikq e a (-1) rdrdqda 2p z=т-Ҙ r=т0 qт=0 aт=0

Evidently the  integral is equal to gk (s,t) with s = 0 and t= 2pnar which certainly implies s Ј t . Thus by Metz and Pan equation (A7) quoted above the  integral is equal to Jk (2pnar) . Thus we finally have

Ҙ Ҙ 2p k -2pin zz -ikq Pk(na,nz,f) = (-1) т e dz т т a(r,q, z) e Jk(2pnar) rdrdq z= -Ҙ r= 0 q = 0 which is equation (5) of Anastasio and Pan.