Re: Bug in Integrate?

• To: mathgroup at yoda.ncsa.uiuc.edu
• Subject: Re: Bug in Integrate?
• From: uunet!decide!bobk (Bob Korsan)
• Date: Wed, 15 Aug 90 10:32:05 PDT

```You write

----- Begin Included Message -----

>From saxony!uunet!yoda.ncsa.uiuc.edu!mathgroup-adm Tue Aug 14 20:59:55 1990
From: uunet!mathnx.byu.edu!smithw (William V. Smith)
To: mathgroup at yoda.ncsa.uiuc.edu
Subject: Bug in Integrate?

This problem came up in some instructional materials being developed
here. If any of you know what's going on here let me know.  I told
the person who handed it to me, that it might be related to the rather
awkward way Mathematica handles square roots.  So, try doing
this integration problem:

In[1]:=

a Integrate[Sqrt[(1-Cos[theta])^2+Sin[theta]^2],{theta,0,2 Pi}]

Out[1]:=

0

(*this is incorrect*)

In[2]:=

a NIntegrate[Sqrt[(1-Cos[theta])^2+Sin[theta]^2],{theta,0,2 Pi}]

Out[2]:=

8. a

(*this is correct*)

Date: Fri, 10 Aug 90 23:24:01 MDT
From: smithw at mathnx.byu.edu (William V. Smith)
To: mathgroup at yoda.ncsa.uiuc.edu
Subject: Re: Another mathematica bug?

Whoops! I just got a chance to look at this one again and it appears
that NEITHER Integrate nor NIntegrate give the correct answer.  I
get 4aSqrt[2] working it out by hand.  Sorry about that.  Its been
a busy day.  Somebody want to check this?

----- End Included Message -----

Well, you are correct, the trouble is with Sqrt. If you look at the
integrand and simplify

In[1]:=Simplify[Sqrt[(1-Cos[t])^2+Sin[t]^2]/.Sin[t]^2->1-Cos[t]^2]

Out[1]:=Sqrt[2 - 2 Cos[t]]

and then

In[2]:=Integrate[Sqrt[1-Cos[t]],t]

Out[2]:=

-2 Sqrt[2]
-----------------------
2
Sin[t]
Sqrt[1 + -------------]
2
(1 + Cos[t])

If you plot this you will notice that since Mathematica
always (EVEN WHEN IT SHOULDN'T) takes the positive
square root, the result is always negative in the
interval [0,2Pi]. Thus, the integrand evaluated at
both end points is the same and subtracts to zero.
This is a MAJOR BUG in version 1.2.1 and should be
corrected immediately.

Peace/Bob

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```

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