Correcting GR7 Integral 3_197_2

Correcting GR7 Integral 3_197_2 PhL 1.30.11 1. Verify Bateman page 115 (6) starting from Bateman page 59 (10). 1 2. Convert lower endpoint from 1 to u 2 3. Try to make the F function look like that in GR7 page 317 3 4. Derive the PhL version of the GR7 result: 5 5. Comparing the GR7 and the Bateman ET II results 6 6. Email to Dan Zwillinger 8 Overview In this document I derive from scratch an integral in GR7, and I find that there are in fact two errors which I then report in an email to Dan Z which appears as the last section (mailed 1.31.11). I needed this integral as one step in the process of computing a certain Mehler sech2 integral, as detailed elsewhere. 1. Verify Bateman page 115 (6) starting from Bateman page 59 (10). Bateman page 115 (6) makes this claim: Let's try to derive this from the "fundamental" integral representation page 59 (10) which is this which I will write out as B(b,c-b) F(a,b;c;z) = !Syntax Error, Idt tb-1 (1-t)c-b-1 (1-tz)-a Now do the pain: t = x-1 dt = -x-2dx = -t2dx !Syntax Error, Idt = !Syntax Error, I(-x-2dx) = !Syntax Error, Idx x-2 tb-1 = x1-b (1-t)c-b-1 = tc-b-1(t-1-1)c-b-1 = x1+b-c (x-1)c-b-1 (1-tz)-a = t-a (t-1-z)-a = xa (x-z)-a So we then have B(b,c-b) F(a,b;c;z) = !Syntax Error, Idx x-2 x1-b x1+b-c (x-1)c-b-1 xa (x-z)-a = !Syntax Error, Idx [x-2 x1-b x1+b-c xa ] (x-1)c-b-1 (x-z)-a = !Syntax Error, Idx [x-c xa ] (x-1)c-b-1 (x-z)-a = !Syntax Error, Idx xa-c (x-1)c-b-1 (x-z)-a which then says B(b,c-b) F(a,b;c;z-1) = !Syntax Error, I(x-1)c-b-1 xa-c (x-z-1)-a dx and this agrees with the page 115 result quoted above. Thus the thing is verified. 2. Convert lower endpoint from 1 to u We start with our verified result B(b,c-b) F(a,b;c; z) = !Syntax Error, I(x-1)c-b-1 xa-c (x-z)-a dx (2.1) Now do the pain: y = ux x = y/u dx = dy/u !Syntax Error, Idx = !Syntax Error, Idy u-1 (x-1)c-b-1 = u1+b-c (ux-u)c-b-1 = u1+b-c (y-u)c-b-1 xa-c = uc-a (ux)a-c = uc-a ya-c (x-z)-a = ua (ux-uz)-a = ua(y-uz)-a Then we have B(b,c-b) F(a,b;c;z) = !Syntax Error, I dx (x-1)c-b-1 xa-c (x-z)-a = !Syntax Error, Idy u-1 u1+b-c (y-u)c-b-1 uc-a ya-c ua(y-uz)-a = !Syntax Error, Idy [u-1 u1+b-c uc-a ua ] (y-u)c-b-1 ya-c (y-uz)-a = !Syntax Error, Idy [ub ] (y-u)c-b-1 ya-c (y-uz)-a So we have shown that B(b,c-b) F(a,b;c;z) = ub !Syntax Error, Idy ya-c (y-u)c-b-1 (y-uz)-a We now set z' = -uz so z = -z'/u and we have B(b,c-b) F(a,b;c; -z'/u) = ub !Syntax Error, Idy ya-c (y-u)c-b-1 (y+z')-a and we remove the primes to get B(b,c-b) F(a,b;c; -z/u) = ub !Syntax Error, Idy ya-c (y-u)c-b-1 (y+z)-a which gives us this integral evaluation !Syntax Error, Idy ya-c (y-u)c-b-1 (y+z)-a = u-b B(b,c-b) F(a,b;c; -z/u) (2.2) 3. Try to make the F function look like that in GR7 page 317 Ignore this section, skip to the next section! Results in this section might be wrong! This just gives irrelevant forms of the integral. That integral is this: I will start with (2.2) and do various edits. First, replace y with x and z with β !Syntax Error, Idy ya-c (y-u)c-b-1 (y+z)-a = u-b B(b,c-b) F(a,b;c; -z/u) !Syntax Error, Idx xa-c (x-u)c-b-1 (x+β)-a = u-b B(b,c-b) F(a,b;c; -β/u) (3a) Now I will try these replacements in my formula (trying to match the GR F function) a = λ b = μ c = λ-μ Then we find a-c = λ - λ+μ = μ c-b-1 = λ-μ - μ -1 = λ - 1 - 2μ -a = -λ This gives !Syntax Error, Idx xa-c (x-u)c-b-1 (x+β)-a = u-b B(b,c-b) F(a,b;c; - β/u) !Syntax Error, Idx xμ (x-u)λ-1-2μ (x+β)-λ = u-μ B(μ,λ-2μ) F(λ,μ;λ-μ; - β/u) (3.2) (3b) Alternatively, I could have done this: a = μ b = λ c = λ-μ Then we find a-c = μ - λ+μ = -λ c-b-1 = λ-μ - λ -1 = -μ-1 -a = -μ This gives !Syntax Error, Idx xa-c (x-u)c-b-1 (x+β)-a = u-b B(b,c-b) F(a,b;c; - β/u) !Syntax Error, Idx x-λ (x-u)-μ-1 (x+β)-μ = u-λ B(λ,-μ) F(μ,λ; λ-μ; - β/u) (3.3) (3c) Summary of results and comparison with GR7: !Syntax Error, Idx xμ (x-u)λ-1-2μ (x+β)-λ = u-μ B(μ,λ-2μ) F(λ,μ;λ-μ; - β/u) (3.2) !Syntax Error, Idx x-λ (x-u)-μ-1 (x+β)-μ = u-λ B(λ,-μ) F(λ,μ; λ-μ; - β/u) (3.3) !Syntax Error, Idx x-λ (x-u)μ-1 (x+β)ν =?= u-λ (β+u)u+νB(λ-μ-ν,μ) F(λ,μ; λ-μ; - β/u) (GR7) I think the GR7 result is wrong! Let's continue: (3d) What does the Kummer 1 formula do (Bateman p 105) Notice that it leaves c and z unchanged. If we apply this to F(λ,μ; λ-μ; - β/u) we have c-a = λ-μ - λ = -μ c-b = λ-μ - μ = λ-2μ c-a-b = λ-μ-λ-μ = -2μ F(λ,μ; λ-μ; - β/u) = (1+β/u)-2μ F(-μ,λ-2μ; λ-μ; - β/u) But this is some very different F function from what appears in the Bateman formula. My conclusion is that the GR7 result is just plain wrong! 4. Derive the PhL version of the GR7 result: Well, we already have this result from (2.2) above !Syntax Error, Idy ya-c (y-u)c-b-1 (y+z)-a = u-b B(b,c-b) F(a,b;c; -z/u) (2.2) which I rewrite as !Syntax Error, Idx xa-c (x-u)c-b-1(x+β)-a = u-b B(b,c-b) F(a,b;c; -β/u) (4.1) The LHS of the GR7 integral is this !Syntax Error, Idx x-λ (x-u)μ-1 (x+β)ν = So I need to solve these equations (left column) for a,b,c: [ notice that c = λ-ν, not λ-μ ! ] -a = ν => a = -ν a = -ν a-c = -λ => -ν-c = -λ => c = λ-ν c = λ-ν c-b = μ => b = c-μ = λ-μ-ν b = λ-μ-ν c-a-b = μ+ν Then my result 4.1 becomes !Syntax Error, Idx x-λ (x-u)μ-1 (x+β)ν = u-(λ-μ-ν) B(λ-μ-ν,μ) F(-ν, λ-μ-ν; λ-ν; -β/u) (4.2) This agrees with page 286 of GR of my hardcopy GR4! Now we can get an alternate version using our Kummer 1 which then says F(-ν, λ-μ-ν; λ-ν; -β/u) = F(a,b;c; -β/u) = (1+β/u)μ+ν F(λ,μ; λ-ν; -β/u) = u-ν-μ (u+β)μ+ν F(λ,μ; λ-ν; -β/u) Then my alternate form is !Syntax Error, Idx x-λ (x-u)μ-1 (x+β)ν = u-(λ-μ-ν) B(λ-μ-ν,μ) F(-ν, λ-μ-ν; λ-ν; -β/u) = u-(λ-μ-ν) B(λ-μ-ν,μ) u-ν-μ (u+β)μ+ν F(λ,μ; λ-ν; -β/u) = u-λ+μ+ν u-ν-μ (β+u)μ+ν B(λ-μ-ν,μ) F(λ,μ; λ-ν; -β/u) = u-λ (β+u)μ+ν B(λ-μ-ν,μ) F(λ,μ; λ-ν; -β/u) This then agrees with Bateman GR7 except for one symbol I show in red! GR7 has it wrong! Does this appear in the errata? His latest errata is April 2008 and no, it does not appear there. I should report this if I can get more confirmation. Well I just got it from GR4. I will soon send in a comment. 5. Comparing the GR7 and the Bateman ET II results Here is the GR7 result and here I have finally found the Bateman result on page 146 of my Russian ET II. The Russian page has the wrong heading on it, that is why I was having trouble finding the result. Here is what ET II really says Let's copy the ET II result as follows: Γ(μ)-1 !Syntax Error, Idx x-λ (x+α)ν (x-y)μ-1 = yμ+ν-λ Γ(λ-μ-ν)/Γ(λ-ν) * F(-ν,λ-μ-ν; λ-ν; -α/y) !Syntax Error, Idx x-λ (x+α)ν (x-y)μ-1 = yμ+ν-λ Γ(λ-μ-ν) Γ(μ)/Γ(λ-ν) * F(-ν,λ-μ-ν; λ-ν; -α/y) !Syntax Error, Idx x-λ (x+α)ν (x-y)μ-1 = yμ+ν-λ B(λ-μ-ν,μ) * F(-ν,λ-μ-ν; λ-ν; -α/y) We then have to do the Kummer thing F(-ν, λ-μ-ν; λ-ν; -α/y) = F(a,b;c; -α/y) = (1+α/y)μ+ν F(λ,μ; λ-ν; -α/y) = y-ν-μ (y+α)μ+ν F(λ,μ; λ-ν; -α/y) so the Bateman result becomes !Syntax Error, Idx x-λ (x+α)ν (x-y)μ-1 = yμ+ν-λ B(λ-μ-ν,μ) * F(-ν,λ-μ-ν; λ-ν; -α/y) = yμ+ν-λ B(λ-μ-ν,μ) y-ν-μ (y+α)μ+ν F(λ,μ; λ-ν; -α/y) = y-λ (y+α)μ+ν B(λ-μ-ν,μ) F(λ,μ; λ-ν; -α/y) So this is what Bateman has to say. We change now y→u and α → β to get = u-λ (u+β)μ+ν B(λ-μ-ν,μ) F(λ,μ; λ-ν; -β/u) So we have now transcribed Bateman to read !Syntax Error, Idx x-λ (x+β)ν (x-u)μ-1 = u-λ (β+u)μ+ν B(λ-μ-ν,μ) F(λ,μ; λ-ν; -β/u) which confirms the typo Now while we are here, let's look at the Bateman conditions: | arg(α/y)| < π |(α/y)| < 1 => | arg(β/u)| < π |(β/u)| < 1 Amazingly, the GR7 result has the left one upside down. 6. Email to Dan Zwillinger See separated doc "dan z email 2.2.11.doc" .