Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
1990
*January
*February
*March
*April
*May
*June
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 1990

[Date Index] [Thread Index] [Author Index]

Search the Archive

Integrate ???

  • To: "mathgroup" <mathgroup at yoda.ncsa.uiuc.edu>
  • Subject: Integrate ???
  • From: "WILLIAM M. GOLDING" <golding at ccf1.nrl.navy.mil>
  • Date: 7 Nov 90 21:48:00 EDT

	Hi there,

		Sorry, to keep bringing up this integrate point but its 
been nagging at me. I agree with everyone that mathematica is doing the 
definite integral by doing the indefinite integral and evaluating at 
0 and 2 Pi. The problem is that it's doing the indefinite integral
incorrectly. It's not totally incorrect, the integral is legit for the range
0 to Pi but negative of the right answer for the range Pi to 2 Pi.
I don't know how important this is in the whole scheme of things but
it sure makes me aware that one has to take with a few grains of 
salt the results obtained from a symbolics package. Obviously the integral
could be done over the two ranges separately and pieced together, but 
when the program comes back and tells you its got the answer I want to be
able to believe it. Anyways here is the original problem an how I think
it should be solved.

	Integrate[Sqrt[2-2 Cos[x]],{x,0,2 Pi}]

simplify the integrand

	2-2 Cos[x] = 2 - Exp[I x] - Exp[-I x]

		   = - [ Exp[I x/2]- Exp[-I x/2] ]^2

		   = - 4 I^2 Sin[x/2]^2

		   =  4 Sin[x/2]^2

We want the square root of this last expression so we get

	Sqrt[ 2-2 Cos[x] ] = Sqrt[ 4 Sin[x/2]^2 ]

			   = Abs [ 2 Sin[x/2] ]

In general,  we have to include the absolute value sign here because 
we might not know the range of the integration variable. That is assuming
x to be real, and there is no reason in this case not to assume x to be
real, we have taken the positve sqrt of number which is definitely >= 0,
and we require the result to be on the positive branch of our square root
function. For example, suppose x = 3 Pi then if we exclude the absolute
value sign we get

	Sqrt[ 4 Sin[3 Pi/2]^2] = 2 Sin[3 Pi/2]

			       = 2 (-1)

			       = - 2

but the Sqrt function must be >= 0 if we're going to stay on the positive 
branch, so the absolute value sign is necessary in the general case. 

Actually, with the integrand in this form and limiting x to the 
range  0 to 2 Pi as for our special case the Abs is not needed since

		2 Sin[x/2] = 2 Abs[ Sin[x/2] ]  0 <= x <= 2 Pi    .

			
Thus we should be able to calculate the indefinte integral from

	Integrate[ Sqrt[2-2 Cos[x]] ,x] = Integral[ 2 Sin[x/2] ,x]

				= -4 Cos[x/2]

			only for   0<= x <= 2 Pi  .

To find the definite integral as in the original problem we get

	Integrate[Sqrt[2-2 Cos[x]],{x,0,2 Pi}] =

                                                 | 2 Pi
				= ( -4 Cos[x/2] )| 
                                                 | 0

				= -4 ( Cos[Pi] - Cos[0] )

				= -4 (   -1    -   1 )

				=  8

And this is the correct result at last. 


The result that Mathematica gives for the indefinite integral is 


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

                  -4
Out[1]= -----------------------
                          2
                    Sin[x]
        Sqrt[1 + -------------]
                             2
                 (1 + Cos[x])


By using a little algebra and trigonometry you can show that this last result 
is the same as ours in the range  0 to Pi , and the negative of ours in 
the range pi to 2 Pi . That is


The indefinite integral that mathematica produces is equivalent to

	-4 Cos[x/2]   0 <= x <= Pi   (Identical to our result except
					our range is 0 to 2 Pi)

	 4 Cos[x/2]   Pi <= x <= 2 Pi

So we evaluate this at 0 and get -4 then we evaluate at 2 Pi and get -4 
take the difference and the result is zero which is what mathematica has
been getting. Notice that this function is continuous at x=Pi, but has
discontinuous slope at x = Pi. Is it unreasonable for the integral of a
continuous function to have discontinuous slope within the integration
range. And if so might this be used as a clue to expect an error in the
integration algorithm.


The thing that bothers me about all this is that it is possible to get
an integral that covers the whole integration range and seems reasonable,
without mathematica once mentioning that there might be a problem with
what we are doing.


Here is an example of the neglect of the range applicability that I thought
interesting.


In[2]:= Sqrt[Sin[x]^2]

Out[2]= Sin[x]    Seems like a reasonable result but what about 



In[3]:= Integrate[Sqrt[Sin[x]^2] , x ]

Out[3]= -Cos[x]

In[4]:= Quit

In[1]:= Sqrt[Sin[x]^2]

Out[1]= Sin[x]       Seems like a reasonable result but what about 
			x = 3 Pi/2


In[2]:= Integrate[Sqrt[Sin[x]^2] , x]

Out[2]= -Cos[x]       It decided that Sin[x] was what I really wanted
			to integrate


In[3]:= Integrate[Sqrt[Sin[x]^2],{x,0,Pi}]

Out[3]= 2                This is right !!!

In[4]:= Integrate[Sqrt[Sin[x]^2],{x,Pi,2 Pi}]

Out[4]= -2		 This is wrong !!

In[5]:=  Integrate[Sqrt[Sin[x]^2],{x,0,2 Pi}]

Out[5]= 0		 This is wrong !

In[6]:= I Quit






  • Prev by Date: Input accuracy
  • Next by Date: Problems converting clipboard to Illustrator format
  • Previous by thread: Re: Input accuracy
  • Next by thread: Problems converting clipboard to Illustrator format