Mehler-Fock Transform

The Mehler-Fock Transform PhL 12.29.15 This plays a crucial role in bowl doc, but I seem never to have derived it. It would be a good addition to bowl doc it not too hard. Here is the full thing g(y) = !Syntax Error, Idτ Pm-1/2+iτ(y) f(τ) // expansion f(τ) = (τ/π) sh(πτ) Γ(1/2-m+iτ) Γ(1/2-m-iτ) !Syntax Error, Idy Pm-1/2+iτ(y) g(y) // projection (2.4.1) So I presume the interval for y is (1,∞) and for τ is (0,∞). Mimic of bowl doc Section 1.5 1 Plans to determine the normalizing factor 3 Plan A: integrate orthog over dτ' 3 Plan B: insert int rep #1 for both P's in orthog (μ=0 only) 6 Plan C: insert int rep #2 for both P's in orthog 7 Plan D: try int rep #3 (μ=0 only) 8 Plan E: find out what the normalization factor must be 9 Plan F: Try completeness with int rep #1 and known f(τ,μ) 11 Plan G: Integrate both dies of completeness over z 11 Plan H: Rosenthal thesis int rep #4 12 Plan I: Use Rosen int rep #4 to replace just one of the P functions in orthog 13 Plan J: Use Rosen int rep #4 to replace both of the P functions in orthog 14 Plan K: Use Rosen int rep #4 to replace one of the P functions in completeness 15 Plan L: Use Rosen int rep #4 to replace one of the P functions in dz integration of completeness 15 Mimic of bowl doc Section 1.5 The differential operator must be the Legendre one somehow. Here is the ODE from GR7 So in my theory I would say (don't know which sign) L = (1-z2)∂z2 - 2z∂z + [ν(ν+1)-μ2/(1-z2)] or L = - (z2-1)∂z2 - 2z∂z + [ν(ν+1)+μ2/(z2-1)] The general form is L = -∂z[p(z)∂z] + q(z) = -p(z)∂z2 - (∂zp)∂z + q So let's try p(z) = (z2-1) ∂zp = 2z q = [ν(ν+1)+μ2/(z2-1)] Then the general form says L = -p(z)∂z2 - (∂zp)∂z + q = - (z2-1)∂z2 - 2z∂z + [ν(ν+1)+μ2/(z2-1)] and this matches the Legendre, so we have accomplished the first step. What's next? Let's rewrite using τ so that ν(ν+1) = (iτ-1/2)(iτ+1/2) = -τ2 - 1/4 = -(τ2+1/4) Then we have L = - (z2-1)∂z2 - 2z∂z + [μ2/(z2-1) - (τ2+1/4)] Think of μ as a bystander variable and τ is the item of interest, like the discrete n. How do you find the weight function s(x) ? The eigenvalue problem is this Lφn(x) = λns(x)φn(x) which becomes L φ(τ,x) = λ(τ) s(x) φ(τ,x) Aha! Start over now that I recognize the eigenvalue. In this restart we write L = - (z2-1)∂z2 - 2z∂z +μ2/(z2-1) p(z) = (z2-1) ∂zp = 2z q = μ2/(z2-1) Now the EV equation is this Lφ(ν,z) = -ν(ν+1)s(z)φ(ν,z) But I know that LPνμ(z) = -ν(ν+1)Pνμ(z) and then LPμiτ-1/2(z) = +(τ2+1/4) Pμiτ-1/2(z) same for Qμiτ-1/2(z) So luckily s(x) = 1 and we can forget about that. so we know that λ(τ) = (τ2+1/4) = the eigenvalue! Question: are both P and Q in the eigenfunction set? I know that Q diverges at z = 1 so am tempted to thing it is not allowed. So for now, assume only the P are eigenfunctions. Now we come to orthogonality !Syntax Error, Idx s(x) n(x) φm(x) = δn,m which becomes !Syntax Error, Idz [Pμiτ-1/2(z)]* Pμiτ'-1/2(z) = K δ(τ-τ') I think P-ν-1 = Pν says we can remove the * for real μ (z is real) so looking for !Syntax Error, Idz Pμiτ-1/2(z) Pμiτ'-1/2(z) = Kμ(τ) δ(τ-τ') // orthogonality Plans to determine the normalizing factor Plan A: integrate orthog over dτ' Setup: Suppose the above were true. Then integrate both sides on dτ' from 0 to infinity !Syntax Error, Idτ'!Syntax Error, Idz Pμiτ-1/2(z) Pμiτ'-1/2(z) = !Syntax Error, Idτ' Kμ(τ) δ(τ-τ') or !Syntax Error, Idz Pμiτ-1/2(z) !Syntax Error, Idτ' Pμiτ'-1/2(z) = Kμ(τ) Now I have the dτ' integral here from PBM Therefore, !Syntax Error, Idτ' Pμiτ'-1/2(z) = (z-1)-(μ+1)/2 (1/Γ(1/2-μ)) (z+1)μ/2 I then have Kμ(τ) = !Syntax Error, Idz Pμiτ-1/2(z) [ (z-1)-(μ+1)/2 (1/Γ(1/2-μ)) (z+1)μ/2 ] = (1/Γ(1/2-μ)) !Syntax Error, Idz Pμiτ-1/2(z)(z-1)-(μ+1)/2(z+1)μ/2 Search: Does this integral exist somewhere? (z-1)-(μ+1)/2(z+1)μ/2 = (z-1)-(μ+1)/2(z+1)μ/2 * (z-1)μ/2(z-1)-μ/2 = (z2-1)μ/2 (z-1)-μ-1/2 so then I am looking for !Syntax Error, Idz Pμiτ-1/2(z) (z2-1)μ/2 (z-1)-μ-1/2 How about this Set λ-1 = -μ-1/2 so λ = -μ+1/2 so need μ< 1/2 . Just try it. λ = 1/2-μ λ+μ = 1/2 λ+μ+ν = 1-μ < 0 ouch cannot have μ = 0 -λ-μ-ν = -λ - 1/2 = μ-1/2 - 1/2 = μ-1 1-λ - μ+ν = μ+1/2 - μ - ν = 1/2-ν 1-μ+ν = itself -μ-ν = -1/2 1-λ-μ = μ+1/2 - μ = 1/2 so integral from GR7 is then Γ(1/2-μ)Γ(μ-1)Γ(1/2-ν) / (Γ(1-μ+ν)Γ(-1/2) Γ(1/2)) So here is what I then have Kμ(τ) = (1/Γ(1/2-μ)) Γ(1/2-μ)Γ(μ-1)Γ(1/2-ν) / (Γ(1-μ+ν)Γ(-1/2) Γ(1/2)) now so rewrite the above as Kμ(τ) = - (1/Γ(1/2-μ)) Γ(1/2-μ)Γ(μ-1)Γ(1/2-ν) / (Γ(1-μ+ν)2 ) = - (1/2π) Γ(μ-1)Γ(1/2-ν)/Γ(1-μ+ν) = - (1/2) Γ(μ-1)Γ(1/2-ν)/Γ(1-μ+ν) ouch pole at μ = 0 Now set ν = iτ-1/2 . Then 1/2-ν = 1/2 - (iτ-1/2) = 1-iτ 1-μ+ν = 1-μ+( iτ-1/2 ) = iτ + 1/2 - μ so then Kμ(τ) = - (1/2) Γ(μ-1)Γ(1-iτ)/Γ( iτ + 1/2 - μ) At least I have something! But I was hoping to get Kμ(τ) = 1/f(τ,μ) where f(τ,μ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) If his were true, I would need Kμ(τ)f(τ,μ) = 1 or - (1/2) Γ(μ-1)Γ(1-iτ)/Γ( iτ + 1/2 - μ) * (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) = 1 or - (1/2) Γ(μ-1)Γ(1-iτ) * (τ/π) sh(πτ) Γ(1/2-μ-iτ) = 1 Stop. This is singular when μ = 0 due to Γ(μ-1) and I now see that way back above. Plan B: insert int rep #1 for both P's in orthog (μ=0 only) But you have to prove this somehow. So we need an integral rep for P. How about P-1/2+iτ(z) = (/π ) ch(πτ) !Syntax Error, Idx cos(τx) / int rep #1 (L.3.5) Insert this twice to get (doing μ = 0 only here) !Syntax Error, Idz Piτ-1/2(z) Piτ'-1/2(z) = !Syntax Error, Idz[ (/π ) ch(πτ) !Syntax Error, Idx cos(τx) /] [(/π ) ch(πτ') !Syntax Error, Idx' cos(τ'x') /] = (/π )2 !Syntax Error, Idx cos(τx)!Syntax Error, Idx' cos(τ'x')!Syntax Error, Idz 1/[ ] I suspect that dz integral gives f(x) δ(x-x'). If I assume that then I can continue = (/π )2!Syntax Error, Idx cos(τx)!Syntax Error, Idx' cos(τ'x') f(x)δ(x-x') = (/π )2!Syntax Error, Idx cos(τx)cos(τ'x) f(x) Now I suspect that f(x) = f0 = constant and then we get = (/π )2 f0!Syntax Error, I!Syntax Error, Idx cos(τx)cos(τ'x) Now I need to quote the Fourier Cosine Series Transform f(z) = !Syntax Error, Idk cos(kz) fk // expansion fk = !Syntax Error, Idz cos(kz) f(z) // projection !Syntax Error, Idz cos(kz) cos(k'z) = (π/2)δ(k-k') // orthogonality !Syntax Error, Idk cos(kz) cos(kz') = (π/2)δ(z-z') // completeness Using the last two lines, I can continue the above = (/π )2f0(π/2) δ(τ-τ') = (2/π2)(π/2)f0 δ(τ-τ') = (1/π) f0 δ(τ-τ') So now then do I do this integral !Syntax Error, Idz 1/[ ] Oops, it diverges at the high end, so this path is a no go, wrong integral rep for P! Also, this is an integral of a positive-always integrand, so cannot possibly give 0 when x ≠ x'. Plan C: insert int rep #2 for both P's in orthog !Syntax Error, Idz Pμiτ-1/2(z) Pμiτ'-1/2(z) = K δ(τ-τ') // orthogonality Need an integral rep for P. Try this one Set ν = iτ-1/2 so then ν+1/2 = iτ. Then we have Pμiτ-1/2(chα) = !Syntax Error, I dt ch(iτt) / (chα-cht)μ+1/2 int rep #2 = !Syntax Error, I dt cos(τt) / (chα-cht)μ+1/2 If I try this rep, I get !Syntax Error, Idz Pμiτ-1/2(z) Pμiτ'-1/2(z) = !Syntax Error, Id(chα) ( )2!Syntax Error, I dt cos(τt) / (chα-cht)μ+1/2 !Syntax Error, I dt' cos(τ't') / (chα-cht')μ+1/2 = !Syntax Error, Idz (z2-1)μ/2 !Syntax Error, I dt cos(τt) / (z-cht)μ+1/2 !Syntax Error, I dt' cos(τ't') / (z-cht')μ+1/2 This seems a huge mess, so no go. Plan D: try int rep #3 (μ=0 only) I am in over my head, it is not so simple as I hoped. I look on the web. Here is one idea I see By doing parts, he fixes my convergence problem! Let's assume this last line is valid. Then Piτ-1/2(z) = ch(πτ)/( πτ) !Syntax Error, Idξ shξ sin(τξ) (z + chξ)-3/2 int rep #3 where z is still nicely isolated. Then I would need to find this integral !Syntax Error, Idz (z + chξ)-3/2 (z + chξ')-3/2 = !Syntax Error, Idz This is a doable integral according to Maple So just for fun, let's pursue this a bit !Syntax Error, Idz Piτ-1/2(z) Piτ'-1/2(z) = = !Syntax Error, Idz [ ch(πτ)/( πτ) !Syntax Error, Idξ shξ sin(τξ) (z + chξ)-3/2] [ ch(πτ')/( πτ') !Syntax Error, Idξ' shξ' sin(τ'ξ') (z + chξ')-3/2] [ch(πτ)/( πτ)][ ch(πτ')/( πτ')] * !Syntax Error, Idξ shξ sin(τξ)!Syntax Error, Idξ' shξ' sin(τ'ξ') !Syntax Error, Idz(z + chξ)-3/2(z + chξ')-3/2 But then I am stuck with three integrals to worry about, and the dz one is a bit ugly. So this seems a hopeless path. Once again, the dz integral cannot give δ(ξ-ξ') because it is the integral of an all positive integrand, so when ξ ≠ ξ' you won't get 0. I think I am doing non-uniform convergence somehow. Plan E: find out what the normalization factor must be OK, I give up on this effort for now. Let's go back to the claimed transform itself, g(y) = !Syntax Error, Idτ Pμ-1/2+iτ(y) f(τ) // expansion f(τ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) !Syntax Error, Idy Pμ-1/2+iτ(y) g(y) // projection (2.4.1) My theory above says no s(z) weight functions What happens if I use these in the above transform. OK do the "two ways" thing: g(y) = !Syntax Error, Idτ Pμ-1/2+iτ(y) f(τ) = !Syntax Error, Idτ Pμ-1/2+iτ(y)[ (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) !Syntax Error, Idy' Pμ-1/2+iτ(y') g(y')] = !Syntax Error, Idy' g(y') { !Syntax Error, Idτ Pμ-1/2+iτ(y)Pμ-1/2+iτ(y') (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) } So the implication here is that !Syntax Error, Idτ Pμ-1/2+iτ(y)Pμ-1/2+iτ(y') (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) = δ(y-y') Now go the other way f(τ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) !Syntax Error, Idy Pμ-1/2+iτ(y) g(y) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) !Syntax Error, Idy Pμ-1/2+iτ(y) !Syntax Error, Idτ' Pμ-1/2+iτ'(y) f(τ') = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) !Syntax Error, Idτ' f(τ') { !Syntax Error, Idy Pμ-1/2+iτ(y) Pμ-1/2+iτ'(y) } with the implication that (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ)!Syntax Error, Idy Pμ-1/2+iτ(y) Pμ-1/2+iτ'(y) = δ(τ-τ') So rewrite this implications: !Syntax Error, Idτ [(τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) ] Pμ-1/2+iτ(y)Pμ-1/2+iτ(y') = δ(y-y') // comp !Syntax Error, Idy [(τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) ] Pμ-1/2+iτ(y) Pμ-1/2+iτ'(y) = δ(τ-τ') // orthog So why is there a kernel in there when I thought not from my simple theory? OK, not function of z. Now define f(τ,μ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) Then here is what things must look like: !Syntax Error, Idτ f(τ,μ) Pμ-1/2+iτ(y)Pμ-1/2+iτ(y') = δ(y-y') // completeness !Syntax Error, Idy f(τ,μ) Pμ-1/2+iτ(y) Pμ-1/2+iτ'(y) = δ(τ-τ') // orthogonality Now you can remove f(τ,μ) from the second line to write !Syntax Error, Idτ f(τ,μ) Pμ-1/2+iτ(y)Pμ-1/2+iτ(y') = δ(y-y') // completeness !Syntax Error, IdyPμ-1/2+iτ(y) Pμ-1/2+iτ'(y) = δ(τ-τ')/ f(τ,μ) // orthogonality Comment: Notice that ANY f(τ,μ) makes "both directions work", and so "both directions" does not determine f(τ,μ) Plan F: Try completeness with int rep #1 and known f(τ,μ) OK. I have now written up the above in Section J.5 of bowl doc. But now I would like to use P-1/2+iτ(z) = (/π ) ch(πτ) !Syntax Error, Idx cos(τx) / int rep #1 (L.3.5) to show that !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) Pμiτ-1/2(z') = δ(z-z') with f(τ,μ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) . I had trouble in the other direction, maybe I can do this direction? !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) Pμiτ-1/2(z') = !Syntax Error, Idτ f(τ,μ) [(/π ) ch(πτ) !Syntax Error, Idx cos(τx) /] [(/π ) ch(πτ) !Syntax Error, Idx' cos(τx') /] = (/π )2 !Syntax Error, Idx (1/) !Syntax Error, Idx' (1/) !Syntax Error, Idτ f(τ,μ) cos(τx) cos(τx') so again I am faced with a painful dτ integral, namely !Syntax Error, Idτ (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) cos(τx) cos(τx') Plan G: Integrate both dies of completeness over z Assume that !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) Pμiτ-1/2(z') = δ(z-z') Integral both sides 1 to infinity to get !Syntax Error, Idz' !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) Pμiτ-1/2(z') = !Syntax Error, Idz' δ(z-z') or !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) !Syntax Error, Idz' Pμiτ-1/2(z') = 1 Now I know that integral (I think). It is not a Mehler integral. But this integral only converges for the Mehler value if ν, so no one has this integral. But it certainly looks like one that could be done. Comment: The real problem here is that the functions Pμiτ-1/2(z) are just completely different from the normal Legendre functions and all the usual methods just don't apply! The integrals in the references are not geared to this conical function. Plan H: Rosenthal thesis int rep #4 I just looked at Rosenthal's thesis, and he seems interested in this integral rep, though he quotes it from page 156 of HTF 1. Here is his quote Can I connect these two? Well, the first step is to negate μ everywhere in my quote above, Here is my negation of the GR7 quote above Pνμ(z) = Γ(1/2-μ) (z2-1)-μ/2 / [ Γ(ν-μ+1)Γ(-μ-ν) ] * !Syntax Error, Idt cosh[(ν+1/2)t] / (z+cht)-μ+1/2 Now set ν = iτ-1/2 so that -μ+ν+1 = -μ + (iτ-1/2)+1 = -μ+iτ + 1/2 = 1/2-μ+iτ -μ-ν = -μ - iτ + 1/2 = 1/2 - μ -iτ cosh[(ν+1/2)t] = cosh[(iτ)t] = cos(τt) Then my translation of the HTF 1 result is Piτ-1/2μ(y) = Γ(1/2-μ) (y2-1)-μ/2 / [ Γ(1/2-μ+iτ)Γ( 1/2 - μ -iτ ] * !Syntax Error, Idt cos(τt) * (y+cht)μ-1/2 int rep # 4 Comments: (1) this agrees with Rosenthal except I think he is missing a minus sign I show in red, let's assume this is just a local typo in his thesis. (2) Now recall that f(τ,μ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) So the good news is that you see exactly that product of gamma functions showing up in his integral rep!! That I think is very significant. Let's now look at the adjusted conditions: Re(z) > -1 just fine ν = iτ - 1/2 Re(ν-μ)> - 1 -1/2 - μ > -1 -μ > -1/2 μ < 1/2 ok Re(-μ-ν) > 0 -μ + 1/2 > 0 μ < 1/2 same condition! So I like these conditions very much. He even writes Re(μ) < 1/2 and adds y > 1. OK, so we are sitting on this interesting integral representation, Piτ-1/2μ(z) = (z2-1)-μ/2 !Syntax Error, Idt cos(τt) * (z+cht)μ-1/2 ir #4 Can this somehow help me in either of these problems: !Syntax Error, Idz Pμiτ-1/2(z) Pμiτ'-1/2(z) = δ(τ-τ') / f(τ,μ) // orthogonality !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) Pμiτ-1/2(z') = δ(z-z') // completeness where f(τ,μ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) (J.5.15) Plan I: Use Rosen int rep #4 to replace just one of the P functions in orthog It seems that I only want to replace ONE of the P functions with this integral rep. Let's try orthogonality and see what happens: !Syntax Error, Idz Pμiτ-1/2(z) Pμiτ'-1/2(z) = !Syntax Error, Idz [ (z2-1)-μ/2 !Syntax Error, Idt cos(τt) * (z+cht)μ-1/2] Pμiτ'-1/2(z) = !Syntax Error, Idz (z2-1)-μ/2!Syntax Error, Idt cos(τt) * (z+cht)μ-1/2] Pμiτ'-1/2(z) = !Syntax Error, Idt cos(τt) !Syntax Error, Idz (z2-1)-μ/2 (z+cht)μ-1/2 Pμiτ'-1/2(z) Do I have any chance on this integral? I don't see it in GR7. If it came out being cos(τ't) then I would be in fat city! Plan J: Use Rosen int rep #4 to replace both of the P functions in orthog Suppose I use the expansion twice? Then we get !Syntax Error, Idz Pμiτ-1/2(z) Pμiτ'-1/2(z) = !Syntax Error, Idz * (z2-1)-μ/2 !Syntax Error, Idt cos(τt) * (z+cht)μ-1/2 (z2-1)-μ/2 !Syntax Error, Idt' cos(τ't') * (z+cht')μ-1/2 Now I am faced with this integral !Syntax Error, Idz (z2-1)-μ (z+cht)μ-1/2 (z+cht')μ-1/2 which has this form !Syntax Error, Idz (z2-1)-μ [(z+a)(z+b)]μ-1/2 Do I have any chance on this integral? How about for μ = 0. Then it is this !Syntax Error, Idz [(z+a)(z+b)]-1/2 and we have that same convergence issue for μ = 0. Plan K: Use Rosen int rep #4 to replace one of the P functions in completeness OK, lets switch over to the completeness condition instead: !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) Pμiτ-1/2(z') = δ(z-z') // completeness and insert his expansion for the first P function, so we get !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) Pμiτ-1/2(z') = !Syntax Error, Idτ f(τ,μ)[ (z2-1)-μ/2 !Syntax Error, Idt cos(τt) * (z+cht)μ-1/2] Pμiτ-1/2(z') Now f(τ,μ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) = (τ/π) sh(πτ) Γ(1/2-μ) Then we continue = !Syntax Error, Idτ (τ/π) sh(πτ) Γ(1/2-μ)[ (z2-1)-μ/2 !Syntax Error, Idt cos(τt) * (z+cht)μ-1/2] Pμiτ-1/2(z') = Γ(1/2-μ)(z2-1)-μ/2 !Syntax Error, Idτ τ sh(πτ) !Syntax Error, Idt cos(τt) * (z+cht)μ-1/2] Pμiτ-1/2(z') = Γ(1/2-μ)(z2-1)-μ/2 !Syntax Error, Idt (z+cht)μ-1/2!Syntax Error, Idτ τ sh(πτ)cos(τt) Pμiτ-1/2(z') and I am then facing a Mehler integral. It is not one that I have done. It is not in the Boeing book either. Plan L: Use Rosen int rep #4 to replace one of the P functions in dz integration of completeness Go back to R's integral rep, Piτ-1/2μ(z) = Γ(1/2-μ) (z2-1)-μ/2 / [ Γ(1/2-μ+iτ)Γ( 1/2 - μ -iτ ] * !Syntax Error, Idt cos(τt) * (z+cht)μ-1/2 int rep #4 Go back to // f(τ,μ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) Pμiτ-1/2(z') = δ(z-z') Now integrate both sides over z' as I did earlier in this doc., !Syntax Error, Idz !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) Pμiτ-1/2(z') = 1 or !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z') !Syntax Error, Idz Pμiτ-1/2(z) = 1 or !Syntax Error, Idτ Pμiτ-1/2(z') [ f(τ,μ)!Syntax Error, Idz Pμiτ-1/2(z) ] = 1 So does int rep #4 help with this dz integral I could not do above? f(τ,μ)!Syntax Error, Idz Pμiτ-1/2(z) = = f(τ,μ) !Syntax Error, Idz { Γ(1/2-μ) (z2-1)-μ/2 / [ Γ(1/2-μ+iτ)Γ( 1/2 - μ -iτ ] * !Syntax Error, Idt cos(τt) * (z+cht)μ-1/2} = (τ/π) sh(πτ) !Syntax Error, Idz { (z2-1)-μ/2 !Syntax Error, Idt cos(τt) (z+cht)μ-1/2 } = (τ/π) sh(πτ) !Syntax Error, Idt cos(τt) !Syntax Error, Idz { (z2-1)-μ/2 (z+cht)μ-1/2 } Even if I could do this dz integral, I then have two more integrals dt and then dτ to deal with. I call this just too messy. So every road I try has a blockade in the form of a messy integral. Go back to, !Syntax Error, Idτ f(τ,μ) Pμiτ-1/2(z) Pμiτ-1/2(z') = δ(z-z') f(τ,μ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) I wonder if Maple would get zero if I numerically integrate this for some z ≠ z' ? Let's just try μ = 0 !Syntax Error, Idτ f(τ,0) Piτ-1/2(z) Piτ-1/2(z') = δ(z-z') f(τ,0) = (τ/π) sh(πτ) Γ(1/2+iτ) Γ(1/2-iτ) = (τ/π) sh(πτ) π/cosh(πτ) = τ tanh(πτ) since Γ(1/2+iτ)Γ(1/2-iτ) = π/cosh(πτ). Then you want to show that !Syntax Error, Idτ τ tanh(πτ) Piτ-1/2(z) Piτ-1/2(z') = δ(z-z') Question: Does this integral converge? I don't have data on the P's asym. I did some Maple plots and I now grok how the above equation works out. I am giving up on trying to prove Mehler-Fock but my notes in J.5 of bowl doc are OK. It is now 9 PM or so on 12/29/15. Next day 12/30/15 I added headings to all the above efforts. I just paged through all of Sneddon. His only integrals with ∞ upper endpoint involve Bessel functions. He does nothing with Mehler. I need a source that is focused on this transform. Jstor article few clips. They use the following L operator, which basically describes transforming a function ψ(τ), so I would call this a Mehler integral. He then has a Mahler integral operator, so to speak. So the above defines the L operator. They also define this Ax operator Then we get this fact which must be the eigenvalue equation somehow. Ax does not look Legendre to me. I have L = - (z2-1)∂z2 - 2z∂z +μ2/(z2-1) but he is doing μ = 0 so then I have L = - (z2-1)∂z2 - 2z∂z. Then (z2-1)-1L = -∂z2 - 2z(z2-1)-1 ∂z Now change to z = chξ so that dz = shξdξ and then ∂z = (1/shξ)∂ξ then the above reads (1/sh2ξ) L = - ((1/shξ)∂ξ)((1/shξ)∂ξ) - 2chξ (1/sh2ξ) (1/shξ)∂ξ = - (1/shξ)2∂ξ2 + (1/shξ)(1/sh2ξ) chξ∂ξ - 2chξ (1/sh2ξ) (1/shξ)∂ξ = - (1/shξ)2∂ξ2 - (1/shξ)(1/sh2ξ) chξ∂ξ and then L = - ∂ξ2 - (1/shξ)chξ∂ξ = - ∂ξ2 - cothξ ∂ξ which is what he shows above. Thus, Ax is really Aξ = L and so the above really is the Legendre equation and those really are the eigenvalues. Fine. Now if you apply the Ax operator k times (k = positive integer) they show that which seems correct to me. Next, he rolls out the same Rosenthal integral rep and parts it Next step is to claim where he uses that second parts rep thing and applies Ax k times. They then say so that F means a full fourier transform. I wonder where this is all leading! They are defining a very special thread! The have somehow then shown in all this that L is continuous and that is the end of their section 2. We now start into section 3: Section 3. Define the following ψφ operator. So this is basically just this definition, ψφ(x) = πtanh(πτ) !Syntax Error, Id(chx)Piτ-1/2(chx) φ(x). They now define Ax' as the "adjoint" of Ax I guess in the Stak sense, so Then using this they claim, and then I could probably verify this just based on what is shown above. The second claim is k parts integrations I think. We then end up with and then with more work of iteration they get This work has all been to prove some Lemma 3.1 which is a technical issue. Lemma 3.2 claims this, where the first = is just a definition from above Most of their work is "technical" checking limits and convergence and all that stuff. The final conclusion is this, and somehow this is the regular MF transform. Wow, what a pile of work! They then repeat it all for the generalized MF. OK, they just summarize the generalized part. I could quote this reference for a good proof! I said it was hard, and here they show it is hard. They do NOT ever talk about orthogonality or completeness. Rosenthal has a later 1974 paper where he refers to Sneddon's integral transforms book and shows the class of functions which work in MF. but this R paper has nothing useful to me. Next I note an "Index Transforms" book on google of Semen and Yakubovich which is a new form for me. I had g(y) = !Syntax Error, Idτ Pμ-1/2+iτ(y) f(τ) Notice the index has iτ/2 ! This source gets off on the K function stuff. a VERY long book section with stuff not helpful to me. pathway. Footnote: MacDonald's function is just the Bessel K function, modified second kind. I finally found something with delta functions!!! Saved the PDF. Here it is But the fancy P functions are not clearly defined! They are defined here OK, here is the connection So go back the previous and set λ = μ !Syntax Error, IdyPμ-1/2+iτ(y) Pμ-1/2+iτ'(y) = δ(τ-τ')/ f(τ,μ) // orthogonality (mine) so they are then claiming !Syntax Error, Idx Pμ-1/2+iτ(x) Pμ-1/2+iτ'(x) = 2 π2δ(τ-τ')[1/τ sh(2πτ) ] * 1/fourgamma fourgamma = Γ(1/2+iτ)Γ(1/2-iτ)Γ(1/2-μ+iκ)Γ(1/2-μ-iκ) = But recall Γ(1/2+iτ)Γ(1/2-iτ) = π/cosh(πτ) so we then have fourgamma = π/cosh(πτ) * Γ(1/2-μ+iκ)Γ(1/2-μ-iκ) So the right side of the above is then 2 π2δ(τ-τ')[1/τ sh(2πτ) ] * 1/[ π/cosh(πτ) * Γ(1/2-μ+iκ)Γ(1/2-μ-iκ)] = 2 π2δ(τ-τ')[1/τ 2sh(πτ)ch(πτ) ] * 1/[ π/ch(πτ) * Γ(1/2-μ+iκ)Γ(1/2-μ-iκ)] = π2δ(τ-τ')[1/τ sh(πτ) ] * 1/[ π * Γ(1/2-μ+iκ)Γ(1/2-μ-iκ)] = π2δ(τ-τ')[1/(πτ sh(πτ)) ] * 1/[Γ(1/2-μ+iκ)Γ(1/2-μ-iκ)] = π2δ(τ-τ') / [πτ sh(πτ)Γ(1/2-μ+iκ)Γ(1/2-μ-iκ)] = δ(τ-τ') / [π-1τ sh(πτ)Γ(1/2-μ+iκ)Γ(1/2-μ-iκ)] But recall from earlier that f(τ,μ) = (τ/π) sh(πτ) Γ(1/2-μ+iτ) Γ(1/2-μ-iτ) so their result is = V Hurray! Their paper is this