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 [integral] How to integrate a fourier series ? tommy1729 Ultimate Fellow Posts: 1,654 Threads: 369 Joined: Feb 2009 05/04/2014, 02:11 PM (This post was last modified: 05/04/2014, 03:17 PM by tommy1729.) Let f(z) be a real fourier series with period 1. Let A be a real number. How to find the integral from 0 to sin^2(A) of the function f(z) in closed form ? A closed form here allows an infinite sum or product. (or even an infinite power tower if you wish) Term by term integration of a fourier series fails. And the coefficients provide the values of certain integrals but only taken over its period. Numerical methods and riemann sums can fail ! So, I do not know how to proceed in the general case. --------- Related : when is this integral ... 1) C^2 2) C^oo 3) "tommy-integrable" (if that exists) see thread : ** http://math.eretrandre.org/tetrationforu...hp?tid=861** 4) analytic ( all with respect to the real A ) --------- Also related : f(z) repeats by the rule f(z+1) = f(z). Now assume a " period shift " ; g(z) = f(z) for 0 < z < 0.5 but g(z+0.5) = g(z). Now what is the four. series of g(z) ? Sure I know the formula for the coefficients, but that includes integrals such as above ... Hence why this is related. Lets call going from f to g " period shift -0.5 ". I was fascinated by the idea to " extend " : doing a period shift +0.5. Afterall if we have a method to do period shift -0.5 , then by inverting that we should be able to do other period shifts. (probably +0.5 or +2) ---- Note : I consider also using an averaged continuum sum such as : CS ( f(A) dA ) going from A = 0 to A = +oo and divided by its lenght (A). However just as term by term integration can fail , this probably holds for continuum sums too. Hence probably a failure in most cases, and not a general solution. ---- Your thoughts are appreciated. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,654 Threads: 369 Joined: Feb 2009 05/04/2014, 03:19 PM This question seems somewhat deja-vu ... Did anyone else here post similar ideas ? regards tommy1729 « Next Oldest | Next Newest »

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