more study of the Goldstine p 77 integral

More study of the 1/sqrt(R) integral PhL 11.3.16 Here I summarize work done in two other docs, and derive new forms as well. It was a royal pain, but I think everything is down pat. Ultimately you just differentiate to test a result, perhaps with Maple help. But I wanted to derive results, not test results handed to me. 1. Derive the arcsinh formula 1 2. Obtain the arcsin formula 2 3. Obtain the arccos formula 5 4. Summary of formulas derived above 5 5. Goldstein page 77 Application 5 6. Other forms of the integral 7 1. Derive the arcsinh formula Here is a statement of the arcsinh formula of interest from GR7 : I have derived the first form involving the log. I understand that the denominator in the log result is not significant and the basic fact is this: I = !Syntax Error, Idx/ = (1/) ln [ 2cx + b + 2] + constant where the constant can depend on a,b,c but not on x. Question: How can one use this last result to derive the sh-1 version of the result stated above, namely, I = !Syntax Error, Idx/ = (1/) sh-1 [ (2cx +b)/] + constant It must be possible to show that sh-1 [ (2cx +b)/] = ln [ 2cx + b + 2] + constant that does not depend on x In this case, differentiating both functions should give 1/. The sh-1 form also appears in Schaum p 72. One thing I know is, from Schaum p 29, sh-1y = ln(y + ) // valid for all real y Using this, I can obtain a "result to show" that involves only logs. y = (2cx +b)/ y2+1 = (2cx +b)2/(4ac-b2) + 1 = [ (2cx +b)2 + (4ac-b2) ] / (4ac-b2) = [ 4c2x2+ b2 + 4cbx + 4ac-b2 ] / (4ac-b2) = [ 4c2x2 + 4cbx + 4ac] / (4ac-b2) = 4c[ cx2 + bx + a] / (4ac-b2) = 4cR/(4ac-b2) = 2 / . Then sh-1 [ (2cx +b)/] = sh-1y = ln(y + ) = ln [ (2cx +b)/ + 2/] = ln [ 2cx + b + 2] - ln() and there you have it, case closed. Thus I have proven both these results, I = !Syntax Error, Idx/ = (1/) ln [ 2cx + b + 2] + constant I = !Syntax Error, Idx/ = (1/) sh-1 [ (2cx +b)/] + constant This last form assumes that 4ac - b2 > 0 and c > 0. Let's summarize this as I = (1/) sh-1 [ (2cx +b)/] 4ac - b2 > 0 and c > 0 [ a > 0 is forced ] This situation implies that 4ac > b2 and so a > b2/(4c) so it is implied that a > 0. 2. Obtain the arcsin formula Let us now assume that a > 0 before continuing. Then 4ac-b2 is going to go negative for c < 0. Consider this picture At θ = 0, the vector 4ac-b2 lies on the real axis and points to the right assuming 4ac-b2 > 0. But as we swing the vector c over to θ = +π, we end up with 4ac pointing to the left and 4ac-b2 also points even more to the left, so here is the status of things angle(c) = 0 angle(4ac-b2) = 0 // before swing angle(c) = π angle(4ac-b2) = π // after swing We end up with 4ac - b2 = -4a|c| - b2 . Since we have assumed a > 0, we have 4ac - b2 < 0 after doing this swing operation, so this quantity changed sign. After the swing we have c = |c| eiπ = (-c)eiπ (4ac-b2) = |4ac-b2| eiπ = |b2-4ac| eiπ = (b2-4ac)eiπ Therefore (note that the phases are determined here, you don't pick separate ± i for each one) = i = i We can then modify our integral I above using these rules to get I = (1/) sh-1 [ (2cx +b)/] = (1/i ) sh-1 [ (2cx +b)/( i)] = (-i)(1/ ) sh-1 [ -i(2cx +b)/] = (-i)2(1/ ) sin-1 [ (2cx +b)/] = - (1/ ) sin-1 [ (2cx +b)/] Therefore we find I = - (1/) sin-1 [ (2cx +b)/] 4ac - b2 < 0 and c < 0 [ a > 0 assumed ] and this is our first statement of the sin-1 evaluation of the I integral. Now recall from GR7 Recall our two results from above: I = (1/) sh-1 [ (2cx +b)/] 4ac - b2 > 0 and c > 0 [ a > 0 is forced ] I = - (1/) sin-1 [ (2cx +b)/] 4ac - b2 < 0 and c < 0 [ a > 0 assumed ] This agrees exactly with the lines 2 and 4 of the quotes above. I have already derived the other two lines in a separate doc, so all four forms are now derived. Now in the first case a > 0 is forced, while in the second it was only assumed. In the second case we need 4ac < b2 -4a|c| < b2 4a|c| > -b2 a > -b2/(4|c|) This in fact allows a range of negative a without impacting the sin-1 formula shown above. In the first case we have 4ac > b2 a > b2/(4c) So let's restate the above two evaluations this way I = (1/) sh-1 [ (2cx +b)/] 4ac - b2 > 0 and c > 0 [ a > b2/(4c) ] (1) I = - (1/) sin-1 [ (2cx +b)/] 4ac - b2 < 0 and c < 0 [ a > -b2/(4|c|) ] (2) 3. Obtain the arccos formula It is a fact that sin-1(x) + cos-1(x) = π/2 where x is real, this is a very obvious fact when stated in English text. There is no corresponding equation for hyperbolic functions!!!! So we can shift the constant and write the second result above as I = + (1/) cos-1 [ (2cx +b)/] 4ac - b2 < 0 and c < 0 [ a > -b2/(4|c|) ] (3) There is no simple cosh-1 formula. However, I did derive a certain cosh-1 formula with a completely different argument than that shown above. I will gather up my obscure formulas later on. There is another version of the cos-1 which follows from (2), I = (1/) sin-1 [ -(2cx +b)/] 4ac - b2 < 0 and c < 0 [ a > -b2/(4|c|) ] (4) Then I = - (1/) cos-1 [ -(2cx +b)/] 4ac - b2 < 0 and c < 0 [ a > -b2/(4|c|) ] (5) You get this same result going directly from the previous cos result, using Schaum p 18. 4. Summary of formulas derived above I = (1/) sh-1 [ (2cx +b)/] 4ac - b2 > 0 and c > 0 [ a > b2/(4c) ] (1) I = - (1/) sin-1 [ (2cx +b)/] 4ac - b2 < 0 and c < 0 [ a > -b2/(4|c|) ] (2) I = + (1/) cos-1 [ (2cx +b)/] 4ac - b2 < 0 and c < 0 [ a > -b2/(4|c|) ] (3) I = (1/) sin-1 [ -(2cx +b)/] 4ac - b2 < 0 and c < 0 [ a > -b2/(4|c|) ] (4) I = - (1/) cos-1 [ -(2cx +b)/] 4ac - b2 < 0 and c < 0 [ a > -b2/(4|c|) ] (5) 5. Goldstein page 77 Application I claim that Goldstein page (3-45) is item (5) above but with an overall sign error. This was my conclusion a long time ago, though I did not document it at that time (it was marked in pencil). Let us now see if this sign error affects anything of significance. He has a = 2mE/l2 and b = 2mk/l2 and c = -1. Going back to his (3.44) I would say θ = θ' - I = θ' - { - (1/) cos-1 [ -(2cx +b)/] + K} = θ' + (1/) cos-1 [ -(2cx +b)/] + K Now consider b2-4ac = [2mk/l2]2 + 4[ 2mE/l2] = [2mk/l2]2 { 1 +4[ 2mE/l2]/ [2mk/l2]2 } = [2mk/l2]2 { 1 +4[ 2mE/l2]/ [4m2k2/l4]} = [2mk/l2]2 { 1 +[ 2E]/ [mk2/l2]} = [2mk/l2]2 { 1 +2El2/mk2} = q // agrees with page 77 α and also - 2cx - b = +2u - [2mk/l2] = [2mk/l2] { 2u/ [2mk/l2] - 1 } = [2mk/l2] { ul2/ mk - 1 } The above is then θ = θ'+ K + cos-1 [ [2mk/l2] { ul2/ mk - 1 } / [2mk/l2] ] = θ' + K + cos-1 [ (ul2/ mk - 1)/ ] = θ' + cos-1 [ (ul2/ mk - 1)/ ] // set K = 0 So I claim he has a sign error as well in equation β. Continuing. θ-θ' = cos-1 [ (ul2/ mk - 1)/ ] cos(θ-θ') = (ul2/ mk - 1)/ Notice that this last equation is exactly the same with or without the sign error, because cos is even !! Continue along then cos(θ-θ') = (ul2/ mk - 1) u(l2/mk) = 1 + cos(θ-θ') 1/r = (mk/l2) [ 1 + cos(θ-θ') ] // agrees with (3-46) So Goldstein has a sign wrong in (3-45) and in β, but this does not affect (3-46) !! 6. Other forms of the integral These other proven-valid forms I found by doing Euler substitutions, I = GR7(x) = + (1/) ln [ +2cx + b + 2] I = PL(x) = - (1/) ln [ - 2cx - b + 2] I = (2/) tanh-1 [ ] I = (2/) tanh-1 [ ] I = (2/) sech-1 ( ) = (2/) sech-1 ( ) I = (2/)cosh-1( ) In general these are unpleasant formulas because they involve and worse.