1 over sqrt_a-bcosx_ Dan

1 Evaluation of an integral PhL 6.23.10 I am tired of doing this integral from scratch over an d over again, so it will now enter the database. I did the integral below, using Maple as a starting point, then I did the second integral using Wolfram for the indefinite integral. Wolfram agreed with Maple on the first integral. I found both these definite integrals in GR7 p 408 with a monstrous huge swap error and I reported this error today to Dan Zwillinger. In any event, here are the correct results for these two related integrals: ( non-AS argument convention) ∫0 π dx(1 / a ± bcosx ) = (2 / a+b ) K( 2b/(a+b) ) a > b > 0 ∫0 π dx a ± bcosx = (2 a+b ) E( 2b/(a+b) ) a > b > 0 and the argument of the complete ellip tic function is in the range (0,1). Doing the first integral The integral in question is this one. I = ∫0 2π dx/ a - bcosx = 2 ∫0 π dx/ a - bcosx a > b 1. Our first adjustment is this I = (2/ a ) ∫0 π dx/ 1 - fcosx ≡ (2/ a ) J f = b/a f < 1 2. According to Maple we have this for integral J which I can simplify to read J = 2 K( 2f/(f-1) ) / 1-f 2 3. GR has useful data on the K function (complete ellip tic of first kind). In that data, symbols k and k' are both used, and they are connected by k2+ k'2= 1 (GR p 904). Suppose we make a local definition k2 = 2f/(1+f) k'2 = (1-f) /(1+f) => k2+ k'2= 1 Then what we really have above is this, where I pick +i arbitrarily J = 2 K(i k/k') / 1-f We can then use GR page 908 which says K(i k/k') = k'K(k) so we then get J = 2 k'K(k) / 1-f = 2 ( 1-f / 1+f ) K(k) / 1-f = 2 K(k)/ 1+f = 2 K ( 2f/(f+1) )/ 1+f So we can now look back at our two results so far: J(f) = 2 K( 2f/(f-1) ) / 1-f = 2 K( 2f/(f+1) )/ 1+f f = b/a (*) Notice this important fact J(-f) = 2 K( 2[-f]/([-f]-1) ) / 1-[-f] = 2 K( 2f/(f+1) )/ 1+f = J(f) which says our result is the same whether we use +b or -b in our integral! For f < 1 which is our interest so far, the second form in (*) has a real number as the K argument, and this real number is in the range 0,1: k 2 ≡ 2f/(1+f) < 1 => 2f < 1+f => f < 1 and we like to have an argument in this range for a K function. This suggests this idea, J = ∫0 π dx/ 1 - fcosx = 2 K( 2|f|/(|f|+1) )/ 1+|f| valid |f| < 1 and we end up with our K argument in the (0,1 ) nice range. Another way to say it is this: J = ∫0 π dx/ 1 ± fcosx = 2 K( 2f/(f+1) )/ 1+f valid f < 1 which requires many fewer symbols and gives the same K arg range (0,1). 3 4. Here then is our result: I = ∫0 2π dx/ a - bcosx = (2/ a ) ∫0 π dx/ 1 - fcosx = (4/ a ) K(k)/ 1+f where f = b/a < 1 and k = 2f/(f+1) = 2b/(a+b) Meanwhile, we have a 1+f = a+b so our final result may be written I = 4K( 2b/(a+b) ) / a+b b/a < 1 arg of K is in range (0,1) So what we claim to have shown is this: ∫0 π dx/ a - bcosx = 2K( 2b/(a+b) ) / a+b a > b Using our trick above, we can extend this result to say ∫0 π dx/ a ± bcosx = 2K( 2b/(a+b) ) / a+b a > b > 0 This integral is not in my GR4, but it was tacked ont o the end of section 3.66 in GR7 where it appears as the second of this pair: WRONG!!! K and E are swapped!!!!! 4 Unfortunately , GR7 has an E instead of a K in its result a nd gives no reference for either of these results, leaving us in some doubt. Is Maple wrong? Wolfram claims this indefinite integral and we know from AS p 589 that F(0|m) = 0 so nothing from the lower endpoint, and then F( π/2|m) is the K function (not the E function), so Wolfram agrees with me. In more detail, this upper endpoint evaluation is: 2 (a+b)/(a-b) F( π/2|-2b/(a-b))/ a+b = 2 F( π/2|-2b/(a-b))/ a-b = 2 F ( π/2, -2b/(a-b) )/ a-b = 2 K( -2b/(a-b) )/ a-b = 2 K ( 2b/(a+b) ) / a+b // using our symmetry rule noted earlier = my result, not the GR7 result Conversely, Wolfram says also This is a very similar situation, notice the simila rity of the factors. Again, only the upper endpoint contributes, and if we again change the sign of b, the answer is = 2 a+b E( 2b/(a+b) ) So that clinches it! GR7 has swapped the E and K in its two equations! I will report this right now. Here is the email I just sent: 5 with copy in the GR7 directory. NOTE ADDED: Marilyn asked me about Wolfram alpha , and it allows you to do definite integrals, unlike their regular online integrator site which says in its FAQ (not set up for definite integrals). Here are some examples: http://www.wolframalpha.com/ The syntax is provided by their Help examples And here I apply this syntax to the current integral of interest , indefinite as in their integrator online, 6 which, by the way, is followed by lots of useful other info, such as alternate forms and power series expansion of the result. And here is the definite integral I want but you see here that they are using the AS convention that k2 is the argument of K. While we're at it, here is the other definite integral I am doing in this doc. But strangely, it cannot do this one! 7 ∫0 π dx a ± bcosx = (2 a+b ) E(2b/(a+b)) a > b > 0