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Re: Weird trigonometric integral and Simplification question

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

> 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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