cos(nx) over sqrt(a-bcosx)

Integral: cos(nx) over sqrt(a-bcosx) PhL 1.18.10 New 0. New summary (now that things have come to light finally) 1 New 1. An immediate application: free space Green's Function in Cylindrical Coordinates 2 New 2. Evaluation of !Syntax Error, Idψ cos(nψ) / 3 New 3. Evaluation of the integral !Syntax Error, Idψ sin(nψ) / 4 New 4. Evaluation of !Syntax Error, Idψ cos(nψ -ψn) / 5 New 5. Alternate forms for New 2 and New 3 and New 4 results. 5 0. Old Summary: 6 1. One place in which this integral arises. 7 2. Another place it arises. 8 3. Casting into elliptic integral form 9 4. Searching the literature for this integral 9 5. Doing the integral 10 6. Test for special case n = 0. 11 7. Test for special case n = 1. 12 8 What does Bateman Vol 4 have to say about this integral? 16 9. Doing the integral by using a certain integral representation of Q [ 5.2.10] 16 New 0. New summary (now that things have come to light finally) From our "transforms.doc" we can write the cosine Fourier series this way for f(θ) even in θ: f(θ) = a0/2 + Σn=1∞ an cos(nθ) = (1/2) Σn=0∞ εn an cos(nθ) // expansion an = (1/π) !Syntax Error, Idθ f(θ) cos(nθ) where εn = 2-δn,0 // projection !Syntax Error, I dθ cos(nθ)cos(n'θ) = (π/εn) δnn' // orthogonality Σn=0+∞εn cos[n(θ-θ')] = 2π δ(θ-θ') // completeness and I will translate the first two lines to say f(x) = (1/2) Σn=0∞ εn An cos(nx) // expansion An = (2/π) !Syntax Error, Idx f(x) cos(nx) // projection We wish now to apply this to the function f(x) = 1/ and we find that An = (2/π) !Syntax Error, Idx f(x) cos(nx) = (2/π) !Syntax Error, Idx cos(nx)/ The integral shown here is the one of interest in this document! We show in Section 9 below that this integral is given by this amazingly simple result !Syntax Error, Idx cos(nx)/ = Qn-1/2(a/b) so that we then have An = (2/π) Qn-1/2(a/b) which is a MUCH simpler result that the finite sum of hypergeometric functions I was getting before I realized this simple form for the result. Therefore, we know that 1/ = (1/π) Σn=0∞ εn Qn-1/2(a/b) cos(nx) // expansion !Syntax Error, Idx cos(nx)/ = Qn-1/2(a/b) // projection New 1. An immediate application: free space Green's Function in Cylindrical Coordinates If we have two points r = (ρ,φ,z) and r' = (ρ',φ',z') in cylindrical coordinates, it is very easy first of all to show that [ using x = ρcosφ, y = ρsinφ ] 1/R ≡ 1/|r-r'| = 1/ We can then define a = ρ2 + ρ'2 + (z-z')2 a/b = (ρ2 + ρ'2 + (z-z')2)/(2ρρ') b = 2ρρ' = 1/ x = φ-φ' so that 1/R = 1/ = (1/π) Σn=0∞ εn Qn-1/2(a/b) cos(nx) = (1/π) Σn=0∞ εn cos[n(φ-φ')] Qn-1/2[(ρ2 + ρ'2 + (z-z')2)/(2ρρ')] This then is a form of the "free space Green's function" (aka the "fundamental solution" by Stak) in cylindrical coordinates. This equation appears as (5) in the 2007 Selvaggi paper "toroidal electrostatics.pdf" which I have today been reading. [5.2.10] It also appears in the 1953 famous monograph of Chester Snow in this form where Chester defines his ε'n so that 2ε'n = εn and he uses x instead of z and 1 instead of prime. I think this came from a website called the Hathi Trust http://catalog.hathitrust.org/Record/001687950 The above equation comes from page 229 of " Hypergeometric and Legendre functions with applications ...a" which has full view at the Hathi site, but you cannot download anything more than a page at a time. The book on Cap and Ind calculations is there, but you cannot view it! We should compare this 1/R result to Jackson page 62 in sphericals 1/R = Σn=0∞ (r<n /r>n+1) Pn[ cosθ cosθ' + sinθ sinθ' cos(φ-φ')] There is something that should be said about this comparison, but I don't yet know what it is. In this case outside the Pn we have a non-oscillatory factor, with two oscillatory appearing in the P argument. But in the Q form above, we have an outside oscillatory factor and inside we have the other two. I think you can always write the 1/R in several different ways for a given coordinate system, I have seen this where we had 1/R ~ JJ one way and ~ IK a different way. An addition theorem is involved in all these things. And this brings in the group theory stuff as usual. New 2. Evaluation of !Syntax Error, Idψ cos(nψ) / Consider the following integral In = !Syntax Error, Idψ cos(nψ) / We can change the integration variable to x = ψ-ψ' to get (and also shift the range) In = !Syntax Error, Idx cos(n[x+ψ']) / We then expand cos(n[x+ψ']) = cos(nx)cos(nψ') – sin(nx)sin(nψ') The 2nd term results in an even integral with an odd integrand and so makes no contribution. We then conclude that In = cos(nψ') !Syntax Error, Idx cos(nx) / = 2 cos(nψ') !Syntax Error, Idx cos(nx) / But above we claimed that !Syntax Error, Idx cos(nx)/ = Qn-1/2(a/b) so we can set a = r2+r'2 and b = 2rr' which tells us = 1/. We then get In = 2 cos(nψ') 1/ * Qn-1/2[(r2+r'2)/(2rr')] Here then is what we have shown: !Syntax Error, Idψ cos(nψ) / = 2 cos(nψ') Qn-1/2[(r2+r'2)/(2rr')] / New 3. Evaluation of the integral !Syntax Error, Idψ sin(nψ) / Lets redo the above in the case of sin instead of cos. In = !Syntax Error, Idψ sin(nψ) / We can change the integration variable to x = ψ-ψ' to get (and also shift the range) In = !Syntax Error, Idx sin(n[x+ψ']) / We then expand sin(n[x+ψ']) = sin(nx)cos(nψ') + cos(nx)sin(nψ') The 1st term results in an even integral with an odd integrand and so makes no contribution. We then conclude that In = sin(nψ') !Syntax Error, Idx cos(nx) / = 2 sin(nψ') !Syntax Error, Idx cos(nx) / But above we claimed that !Syntax Error, Idx cos(nx)/ = Qn-1/2(a/b) so we can set a = r2+r'2 and b = 2rr' which tells us = 1/. We then get In = 2 sin(nψ') 1/ * Qn-1/2[(r2+r'2)/(2rr')] Here then is what we have shown: !Syntax Error, Idψ sin(nψ) / = 2 sin(nψ') Qn-1/2[(r2+r'2)/(2rr')] / New 4. Evaluation of !Syntax Error, Idψ cos(nψ -ψn) / Here ψn are some constant phase shifts. We start off writing cos(nψ-ψn) = cos(nψ)cos(ψn) + sin(nψ)sin(ψn) Let's call our New 4 integral J4, and those in the previous two sections J2 and J3. Then we can see that J4 = cos(ψn) J2 + sin(ψn) J3 = cos(ψn) 2 cos(nψ') Qn-1/2[(r2+r'2)/(2rr')] / + sin(ψn) 2 sin(nψ') Qn-1/2[(r2+r'2)/(2rr')] / = [cos(nψ')cos(ψn) + sin(nψ')sin(ψn)] 2 Qn-1/2[(r2+r'2)/(2rr')] / = 2 cos(nψ'-ψn) Qn-1/2[(r2+r'2)/(2rr')] / We see that we basically get the New 2 result with the phase shifts added. The phase shifts make no difference in the integral, they just "pass through". New 5. Alternate forms for New 2 and New 3 and New 4 results. In my Sneddon raw notes for Section 6.3, I derive two alternate forms for the integrals shown in the New sections above. π!Syntax Error, IJn(rx)Jn(r'x) dx = 2 (rr')-n !Syntax Error, Ids s2n / [ ] = Qn-1/2[ (r2+r'2)/(2rr')] / Thus, we can write !Syntax Error, Idψ cos(nψ) / = 2 cos(nψ') Qn-1/2[(r2+r'2)/(2rr')] / = 2 cos(nψ') π!Syntax Error, IJn(rx)Jn(r'x) dx = 4 cos(nψ') (rr')-n !Syntax Error, Ids s2n / [ ] and !Syntax Error, Idψ sin(nψ) / = 2 sin(nψ') Qn-1/2[(r2+r'2)/(2rr')] / = 2 sin(nψ') π!Syntax Error, IJn(rx)Jn(r'x) dx = 4sin(nψ') (rr')-n !Syntax Error, Ids s2n / [ ] The New 4 result is just the New 2 one with phase shifts added and we get !Syntax Error, Idψ cos(nψ-ψn) / = 2 cos(nψ'-ψn) Qn-1/2[(r2+r'2)/(2rr')] / = 2 cos(nψ'-ψn) π!Syntax Error, IJn(rx)Jn(r'x) dx = 4 cos(nψ'-ψn) (rr')-n !Syntax Error, Ids s2n / [ ] If we take the extreme left and right elements of the above equality chain, we obtain the result quoted in Sneddon page 70 equation 3.4.6 !! This is the magic integral which allows the Copson solution of the Dirichlet disk to be solved! 0. Old Summary: The integral of interest is this: ( n = integer) I = !Syntax Error, Idx cos(nx)/ = (4/ ) !Syntax Error, Idx cos(2nx)/ = !Syntax Error, Idx cos(nx)/ = 2 !Syntax Error, Idx cos(nx)/ where we have recast it into elliptic form (this is derived below). Consider the Fourier Analysis of the function 1/ on (0,2π) , f(x) = 1/ = a0/2 + Σn=1∞ an cos(nx) // expansion I = πan = !Syntax Error, Idx cos(nx)/ // projection The integral I we study here is basically the Fourier coefficient for this expansion. The result is, as shown below, I = (2/ ) Σs=0n (-1)s (2n, 2n-2s) B[ s+1/2, n-s+1/2 ] F [s+1/2, 1/2; n+1; -2b/(a-b) ] and here are two special cases also figured out below, where k2 = -2b/(a-b), n=0 I = 4/(a-b) K(k) n=1 I = (4/ ) K(k) k2 = -2b/(a-b) n=2 I = (4/ ) [(a/b) K(k) + (1 - (a/b) E(k) ] I have confirmed the n=1 result with Maple, and the n=2 with Wolfram's integrator. the n=0 case comes from GR7 p409 where I take b→ -b. The ±b is the same here since changing the order in which you add things makes no difference. For general n, we get a sum of n+1 F functions with c = 1,2,3...n+1. You have to use the Gauss F shifter formulas to reduce the higher c functions to those having c = 1, and this is very painful to do by hand. Maple V does not seem able to do it at all, it does not even know how to express an F in terms of standard special functions (well, not true, see convert). In general, I think for any n you get some linear combination of K and E functions. For some reason, this Fourier Expansion is a very messy one. Expansion onto Legendre's is nicer. The two fundamental F functions at the bottom of the Gauss ladder are these: F(1/2, 1/2, 1, z) = (2/π)K(k) F(-1/2, 1/2, 1, z) = (2/π)E(k) 1. One place in which this integral arises. In "ring and disk Green's problems", and in electrostatic in general, you often integrate over a surface to get a potential, and you want to enforce that potential to be 0. In the problem of a point charge inside the hole of a metal plate, we get this equation V(r') = q/r1 + !Syntax Error, Ir dr!Syntax Error, Idθ σ(r,θ) / |r - r'| = 0 We want to solve this for some σ(r,θ) that makes the equation be true. One approach is to expand σ(r,θ) = Σn=0∞ σn(r) cos(nθ) and then you have this resulting equation V(r',θ') = q/r1 + Σn=0∞ !Syntax Error, Ir dr σn(r)!Syntax Error, Idθ cos(nθ) / = 0 This is where our integral arises. Define f = r'/r and this becomes V(r') = q/r1 + Σn=0∞ !Syntax Error, Idr σn(r) { !Syntax Error, Idθ cos(nθ) / } and there is our integral of interest in the curly brackets. By changing to θ" = θ-θ', expanding the cos, looking at symmetry, we end up with I(f) = cos(nθ') Fn(f) Fn(f) = !Syntax Error, Idθ cosnθ/ V(r',θ') = q/r1 + Σn=0∞ cos(nθ') !Syntax Error, Idr σn(r) Fn(f). Let a = 1+f2 > 1 and b = 2f > 0 Fn(a,b) = !Syntax Error, Idx cos(nx)/ = I, our integral of interest We have then, for this application, that a+b = (1+f)2 > 0 a-b = (1-f)2 > 0 just for the record The resulting integral is so ugly that probably this is a bad way to expand the surface charge, although it seems the simplest expansion you could use for a polar coordinates problem. 2. Another place it arises. Someone hands you the function f(x) = 1/ on the interval (0,2π). They ask you: "what is the Fourier expansion of this function" ? You go to Schaum p 131 and set L = π and the answer is this: f(x) = 1/ = a0/2 + Σn=1∞ an cos(nx) where πan = !Syntax Error, Idx cos(nx)/ = I So our integral of interest is merely the normal everyday Fourier projection of this particular function. 3. Casting into elliptic integral form [ This was my original approach ] From above we have, I = !Syntax Error, Idx cos(nx)/ = 2 !Syntax Error, Idx cos(nx)/ Now consider the following algebra, cos(x) = 1-2sin2(x/2) let x' = x/2 cos(x) = 1 - 2 sin2(x') a - bcos(x) = a - b [1 - 2 sin2(x')] = (a-b) + 2b sin2(x') = (a-b) [ 1 - k2 sin2(x')] k2 = -2b/(a-b) The constant is 4 because got 2 from range (0,π), and another 2 from dx = 2dx', I = (4/ ) !Syntax Error, Idx' cos(2nx')/ I = (4/ ) !Syntax Error, Idx cos(2nx)/ k2 = (b-a)/2b = [ 1 - (a/b)]/2 a = 1+f2 f = r'/r b = 2f a-b = 1+f2 -2f = (1-f)2 k2 = -2b/(a-b) = -4f/(1-f)2 So our integral is then I = (4/ ) J where J = !Syntax Error, Idx cos(2nx)/ I would say this was in "elliptical integral form", and this might be helpful for searching. Notice that the integral J can only depend on a and b only through k2. So we might write I = !Syntax Error, Idx cos(nx)/ = (4/ ) J 4. Searching the literature for this integral GR definite elementary trig: Section 3.67 and 3.84 are the only possible places in the index list on page xiii Section 3.67 (p 386): Right off the bat we have this interesting integral !Syntax Error, Isinαx cosβx / = combination of beta function and HGF I think this is what we can use. 5. Doing the integral Now consider this fact eimx = (cos(x) + i sin(x))m = (C + iS)m = Σk=0m (m,k) Ck(iS)m-k = Σk=0m (i)m-k(m,k) CkSm-k Let's define a new mirror summation index j = m-k which runs from 0 to m. Then eimx = Σj=0m (i)j (m,m-j) Cm-jSj The factor (i)j is real when j = even, so we can then say cos(mx) = Σj=0,evenm (i)j (m,m-j) Cm-jSj If m is odd, the sum stops of course at m-1. We know (i)j = (i)2N = (-1)N = (-1)j/2 so cos(mx) = Σj=0,evenm (-1)j/2 (m,m-j) Cm-jSj Now make a new summation index s = j/2 so we then have cos(mx) = Σs=0m/2 (-1)s (m, m-2s) Cm-2sS2s where again the sum ends one early if m/2 is fractional. Now let's replace m with 2n to get cos(2nx) = Σs=0n (-1)s (2n, 2n-2s) C2n-2sS2s and now there is no longer confusion about where the sum ends. We can now install this into our integral: J = J(n,k2) = !Syntax Error, Idx cos(2nx)/ = !Syntax Error, Idx { Σs=0n (-1)s (2n, 2n-2s) C2n-2sS2s })/ = Σs=0n (-1)s (2n, 2n-2s) !Syntax Error, Idx [ cos(x)]2n-2s [ sin(x)]2s / and now we have our GR p 386 integral with α = 2s and β = 2n-2s. !Syntax Error, Isinαx cosβx / = (1/2) B[ (α+1)/2, (β+1)/2 ] F [(α+1)/2, 1/2; (α+β+2)/2; k2 ) which is something we can at least comprehend. So = (1/2) B[ (2s+1)/2, (2n-2s+1)/2 ] F [(2s+1)/2, 1/2; (2s+2n-2s+2)/2; k2 ] = (1/2) B[ s+1/2, n-s+1/2 ] F [s+1/2, 1/2; n+1; k2 ] This sure looks familiar to me! But I can't find it in the Legendre table. AS 556 shows other cases, but in all cases the a,b parameters are "correlated". So I guess we just leave it "as is". Then we have J = Σs=0n (-1)s (2n, 2n-2s) (1/2) B[ s+1/2, n-s+1/2 ] F [s+1/2, 1/2; n+1; k2 ] so there are no less than 6 gamma functions here, perhaps it could be simplified, but let's just leave it. We than have our result I = !Syntax Error, Idx cos(nx)/ = ( 4/ ) J(n,k2) = (2/ ) Σs=0n (-1)s (2n, 2n-2s) B[ s+1/2, n-s+1/2 ] F [s+1/2, 1/2; n+1; -2b/(a-b) ] so I think that is "not bad". This then is the General Formula for I. 6. Test for special case n = 0. I = !Syntax Error, Idx 1/ = (2/ ) Σs=0n (-1)s (2n, 2n-2s) B[ s+1/2, n-s+1/2 ] F [s+1/2, 1/2; n+1; -2b/(a-b) ] = (2/ ) (0, 0) B[1/2, +1/2 ] F [1/2, 1/2; 1; -2b/(a-b) ] = (2/ ) Γ(1/2)2/Γ(1) F [1/2, 1/2; 1; -2b/(a-b) ] = (2/ ) π F [1/2, 1/2; 1; -2b/(a-b) ] Now look at AS page 591 top which says KAS(m) = (π/2) F(1/2,1/2;1;m) m = k2 = -2b/(a-b) so our answer was I = (2/ ) π * (2/π) K(k2) = (4/ ) K(k) Here is Maple's evaluation of this integral and we agree. 7. Test for special case n = 1. I = !Syntax Error, Idx cos(x)/ = (4/ ) J = (2/ ) Σs=0n (-1)s (2n, 2n-2s) B[ s+1/2, n-s+1/2 ] F [s+1/2, 1/2; n+1; -2b/(a-b) ] = (2/ ) Σs=01 (-1)s (2, 2-2s) B[ s+1/2, 1-s+1/2 ] F [s+1/2, 1/2; 2; -2b/(a-b) ] so there are going to be two terms: = (2/ ) { (2, 2) B[ 1/2, 3/2 ] F [1/2, 1/2; 2; -2b/(a-b) ] - (2, 0) B[ 3/2, 1/2 ] F [3/2, 1/2; 2; -2b/(a-b) ] } Now B[ 3/2, 1/2 ] = Γ(3/2)Γ(1/2)/Γ(2) = (π/2), (2,0) = 1 so we get = (2/ ) (π/2){ F [1/2, 1/2; 2; -2b/(a-b) ] - F [3/2, 1/2; 2; -2b/(a-b) ] } Notice that c = 2 now, so both these functions are "new". Here is one of the Gauss relations from Bateman p 103 [ AS and GR agree that there are no z factors in this formula ] Reduction of the F's using Gauss shifting formulas. Our targets are these functions F(1/2,1/2,1,z) = (2/π)K F(-1/2,1/2,1,z) = (2/π)E which both have c = 1, but our two starting formulas have c = 2. So we want to select a Gauss formula that has three F's in it where one is at c+1 and two are at c, then we will set c = 1. Our candidate is Bateman (38) which says c(1-z)F - cF(a-1) + (c-b) z F(c+1) = 0 and we go ahead and set c = 1 (1-z)F - F(a-1) + (1-b) z F(c=2) = 0 and we now need to write this out, where of course z = k2 (1-z)F(a, b, 1, z) - F(a-1, b, 1, z) + (1-b) z F(a, b, 2, z) = 0 Our first F of interest is F [1/2, 1/2, 2, z]. To get it on the right, we select a = b = 1/2 so (1-z)F(1/2, 1/2, 1, z) - F(-1/2, 1/2, 1, z) + (1/2) z F(1/2, 1/2, 2, z) = 0 Multiply through by (2/z) to get (2/z)(1-z)F(1/2, 1/2, 1, z) - (2/z) F(-1/2, 1/2, 1, z) + F(1/2, 1/2, 2, z) = 0 and solve for the K of interest: F(1/2, 1/2, 2, z) = - (2/z)(1-z)F(1/2, 1/2, 1, z) + (2/z) F(-1/2, 1/2, 1, z) = (2/z)[ - (1-z) (2/π)K + (2/π)E ] = (4/πz)[ -(1-z) K + E ] Now go back to our general shifter formula (1-z)F(a, b, 1, z) - F(a-1, b, 1, z) + (1-b) z F(a, b, 2, z) = 0 Our second F of interest is F [3/2, 1/2; 2; z ] . This time we want a = 3/2 and b = 1/2. At least this will lower our c by one step. So we then have (1-z)F(3/2, 1/2, 1, z) - F(1/2, 1/2, 1, z) + (1/2) z F(3/2, 1/2, 2, z) = 0 As before, multiply through by (2/z) to get (2/z) (1-z)F(3/2, 1/2, 1, z) - (2/z)F(1/2, 1/2, 1, z) + F(3/2, 1/2, 2, z) = 0 and solve for the K of interest F(3/2, 1/2, 2, z) = - (2/z) (1-z)F(3/2, 1/2, 1, z) + (2/z)F(1/2, 1/2, 1, z) (*) But now we still have an F that we don't know and which needs more work: F(3/2, 1/2, 1, z). Let's try Bateman (36) on page 103 (c-a-b)F - (c-a)F(a-1) + b(1-z) F(b+1) = 0 Here c does not change, so set c = 1 everywhere (1-a-b)F(a, b, 1, z) - (1-a)F(a-1, b, 1, z) + b(1-z) F(a, b+1, 1, z) = 0 Try b = 1/2 and a = 1/2 and we get (0)F(1/2, 1/2, 1, z) - (1/2)F(-1/2, 1/2, 1, z) + (1/2)(1-z) F(1/2, 3/2, 1, z) = 0 - F(-1/2, 1/2, 1, z) + (1-z) F(1/2, 3/2, 1, z) = 0 Solve for the one we want (1-z) F(1/2, 3/2, 1, z) = F(-1/2, 1/2, 1, z) Now insert this into (*) above to get F(3/2, 1/2, 2, z) = - (2/z) (1-z)F(3/2, 1/2, 1, z) + (2/z)F(1/2, 1/2, 1, z) = - (2/z) F(-1/2, 1/2, 1, z) + (2/z)F(1/2, 1/2, 1, z) = (2/z) [ - (2/π)E + (2/π)K ] = (4/πz) [ K - E] So far then we have determined that F [1/2, 1/2, 2, z] = (4/πz)[ -(1-z) K + E ] F(3/2, 1/2, 2, z) = (4/πz) [ K - E] Our integral is then I = !Syntax Error, Idx cos(x)/ = (2/ ) (π/2){ F [1/2, 1/2; 2; -2b/(a-b) ] - F [3/2, 1/2; 2; -2b/(a-b) ] } = (2/ ) (π/2) (4/πz) { [ -(1-z) K + E ] - [ K - E] } = (2/ ) (2/z) { [ -(2-z) K + 2E ] } = (2/ ) (2/z) 2 { [ -(1-z/2) K + E ] } = (8/ ) { [ -(1-z/2) K + E ] }/z z = k2 = k2 = -2b/(a-b) = (4/ ) { [ -(1-z/2) K + E ] } (2/z) = (4/ ) [ -(2/z - 1) K + (2/z)E ] So we know that 2/z = 1 - (a/b) // scratch so the result of MY n=1 case is this I = (4/ ) [(a/b) K(k) + (1 - (a/b) E(k) ] Here is the Wolfram indefinite integral for this n = 1 case: If I evaluate this at π and 0 I get = 2 { a F(π/2 | k2) + (b-a) E(π/2 | k2) }/ b = (2/) { a/b K( k2) + (b-a)/b E(k2) } = (2/) { (a/b) K( k2) + [ 1 - (a/b)] E(k2) } Then my integral is this (goes to 2π, so add factor 2) I = !Syntax Error, Idx cos(nx)/ = (4/) { (a/b) K( k2) + [ 1 - (a/b)] E(k2) } and this agrees with my calculation. 8 What does Bateman Vol 4 have to say about this integral? I = !Syntax Error, Idx cos(nx)/ But this is a Fourier Series transform, not a Fourier Transform, so this won't appear in Bateman. Does anyone publish a table of Fourier series transforms? I cannot find tables with "uncommon" functions. I thought in passing about doing the Laplace transform and then evaluating, but that was a red herring, you still have to do the integral! 9. Doing the integral by using a certain integral representation of Q [ 5.2.10] Consider the integral representation (10) shown on page 156 of Bateman for Q as a (0,π) integral, but consider it in the case that ν = n-1/2 where n is an integer. In this situation we have n = ν+1/2 and cos(νπ) = cos(nπ-π/2) = -sin(nπ) = 0 so there is no second term! Then we have this: Qn-1/2μ(z) = eiπμ (2π)-1/2(z2-1)μ/2 Γ(μ+1/2) !Syntax Error, Idt (z-cost)-μ-1/2 cos(nt) dt There is a little smudge in Bateman, but this same integral representation appears in GR7 p 961 Now set μ = 0 to get Qn-1/2(z) = (2π)-1/2 Γ(1/2) !Syntax Error, Idx cos(nx) (z-cos(x))-1/2 = (2π)-1/2 π1/2 !Syntax Error, Idx cos(nx) (z-cos(x))-1/2 = (1/) !Syntax Error, Idx cos(nx) (z-cos(x))-1/2 which is beginning to look a lot like our desired integral! Let's go back to our desired integral I = !Syntax Error, Idx cos(nx)/ = 2 !Syntax Error, Idx cos(nx)/ = 2 (1/) !Syntax Error, Idx cos(nx)/ = 2 (1/) Qn-1/2(a/b) = 2 Qn-1/2(a/b) Wow! This means that an(a,b) = I/π = (2/π) Qn-1/2(a/b) As a check on this, rewrite this way ( bear with me for a moment) an(α,β) = I/π = (2/π) Qn-1/2(α/β) then set α = b2+ ρ2 β = 2cρ α/β = (b2+ ρ2)/(2cρ) an(α,β) = (2/π) (1/) Qn-1/2[(b2+ ρ2)/(2cρ)] This is the same as the result I obtained in this document: Iris Green's by Dirichlet method Attempt #3.doc page 21 b2 = c2 + d2 So that clinches the deal! This is a MUCH better form for the result! So here is our final result: I = !Syntax Error, Idx cos(nx)/ = 2 Qn-1/2(a/b) Comments: This result says, apart from defining the constant, that Qn-1/2(a/b)/ is the Fourier Series Coefficient of the function 1/. I have looked very hard at both my GR editions and I cannot find this integral anywhere in there. If is of course only valid for integral n. Bateman does not give tables of Fourier Series transforms, so not in the ET volumes. Another interesting way to write this is as follows: Qn-1/2(coshξ) = (2π)-1/2 Γ(1/2) !Syntax Error, Idx cos(nx) (cosh(ξ)-cos(x))-1/2 and this has the definite "look" of something going on in toroidal coordinates.