Re: Weird trigonometric integral and Simplification question

*To*: mathgroup at smc.vnet.net*Subject*: [mg31814] Re: [mg31777] Weird trigonometric integral and Simplification question*From*: David Withoff <withoff at wolfram.com>*Date*: Mon, 3 Dec 2001 01:45:04 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

> Message 1: Weird trigonometric integral > > The function Sqrt[1-Cos[t]] is continuous for all real t. Hence its > integral must be continuous for all real t. In fact there is a general > solution that is continuous for all real t. However Mathematica gives > only a "particular" solution that is only true on the interval (0, 2 > Pi). Is there any way to know in advance when to expect these subtle and > difficult problems? I presume this is referring to the indefinite integral In[1]:= Integrate[Sqrt[1-Cos[t]],t] Out[1]= -2 Sqrt[1 - Cos[t]] Cot[t/2] The behavior of this result is primarily a mathematics issue rather than a Mathematica issue. If you look closely at a table of integrals, for example, you will find that all but the most elementary entries show this behavior. Mathematically, unless the integrand is free of branch cuts or other singularities everywhere in the complex plane, indefinite integrals that are claimed to be correct everywhere in the complex plane are undetermined up to a piecewise constant function. The result from Integrate is correct for all t, in the sense that the derivative of this result is Sqrt[1-Cos[t]] everywhere except at steps in the aforementioned piecewise constant function. In response to your closing question, you might try studying a book on calculus in the complex plane, after which these "subtle and difficult problems" should become obvious and straightforward. The conclusion that the integral must be continuous if the integrand is continuous is based on the implicit assumptions that the integral is along a particular contour of integration (in this case, along the real line) and that derivatives are understood as directional derivatives along that contour (in this case, directional derivatives along the real line). For most integrals there isn't an antiderivative that is continuous and differentiable for all contours. Dave Withoff Wolfram Research

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