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2008-04-28, 07:23   #12
davieddy

"Lucan"
Dec 2006
England

6,451 Posts

Quote:
 Originally Posted by frmky I think so. I suspect the square roots were there to ensure it since, interestingly, the answer does not depend on the power of the logs. Greg
Any function at all would suffice would it not?

 2008-04-29, 01:37 #13 Primeinator     "Kyle" Feb 2005 Somewhere near M50..sshh! 2×3×149 Posts Alright then, thanks for the help. Because I have never heard of an integral being solevd by symmetry, and my text book does not mention, what are the rules behind it or alternatively, where could I look for more information in a format that is easy enough to understand for a second semester calc student? Thanks.
2008-04-29, 09:53   #14
Chris Card

Aug 2004

2·5·13 Posts

Quote:
 Originally Posted by Primeinator Alright then, thanks for the help. Because I have never heard of an integral being solevd by symmetry, and my text book does not mention, what are the rules behind it or alternatively, where could I look for more information in a format that is easy enough to understand for a second semester calc student? Thanks.
I'm not sure there is an exhaustive set of rules, it more a question of looking for symmetries and seeing where they lead. In this case, the integrand looked horrible, so I assumed there was some trick to it.

One obvious rule is that if f is an odd functions, f(-x) = -f(x), then integrating from -a to +a will always give zero, since the integrals from -a to 0 and 0 to a cancel out.

Similarly, for an even function, f(-x) = f(x), integrating from -a to a is the same as twice the integral from 0 to a.

Remember that mathematicians love to prove things "by symmetry", so it's very often worth looking out for.

Chris

 2008-05-01, 06:16 #15 Primeinator     "Kyle" Feb 2005 Somewhere near M50..sshh! 15768 Posts Thanks Chris...one more thing to point out when integrating an odd function over from -a to a...doesn't it have to be a closed interval (i.e. non-infinity) to cancel to zero, otherwise you have to use another technique?
2008-05-01, 06:41   #16
Chris Card

Aug 2004

2028 Posts

Quote:
 Originally Posted by Primeinator Thanks Chris...one more thing to point out when integrating an odd function over from -a to a...doesn't it have to be a closed interval (i.e. non-infinity) to cancel to zero, otherwise you have to use another technique?
well in all these things you have to be careful that the integral exists, but if the integral of -inf to +inf of an odd function exists then I think it would be zero.

Chris

 2008-06-04, 00:56 #17 nibble4bits     Nov 2005 181 Posts Note that I'm using the notation that Int(f(x),x,a,b) is the same as the antiderivative of f(x) on the interval from a to b. If a and b are not specified, of course it's an indefinite integral instead. Well even if at x=0, f(x) is undefined, you can use a limit for something like this: Int(1/x,x,-1,1) is unsolvable directly because 1/0 is undefined. Of course since it's an odd function, we already know there's a symmetry, but we'll try to find another solution anyways. If you break it up, you get: (1-1)(Int(1/x,x,0,1)) but this has a problem since 0/0 is still a hidden possiblity when simplifying this formula. The limit as n goes to z of Int(1/x,x,n,1) is in fact 1/z. Since 0/z=0/0, the final answer is still undefined using this method. Using a table: 0*Int(1/x,x,a,b)=0*(ln(b)-ln(a))=0(ln(1)-ln(0))=0(ln(0)) Again, you're having trouble since the limit as n-> 0 of ln(n) is -infinity. Treating the integral as a summation: Another way to look at that though, is that you're subtracting ln(0) from ln(0) and thus the result is zero so long as all the terms cancle. You normally can NOT just add infinite to negative infinity to get zero, but in this case they're the exact SAME infinity so far as the terms are concerned. You'll see this concept when you have to deal with a series assuming you haven't already. A Riemann sum (limit) definition of the integral would have been really handy right about now. ;) http://www.physicsforums.com/archive.../t-112179.html http://documents.wolfram.com/v4/MainBook/3.5.8.html http://www.msstate.edu/dept/abelc/math/integrals.html http://en.wikipedia.org/wiki/Antiderivative (especially the discontinous functions section) These links are mainly to help provide more an alternate explanation compared to your textbook.
2008-06-04, 04:07   #18

"Richard B. Woods"
Aug 2002
Wisconsin USA

718810 Posts

Quote:
 Originally Posted by Primeinator Because I have never heard of an integral being solevd by symmetry, and my text book does not mention, what are the rules behind it
Think back to your geometry class. Many geometric theorems use symmetry to finish proving something that would be more tedious without considering symmetry.

Symmetry is not limited to explicit geometry. When you consider substituting -x for x, you're folding along the y-axis -- a basically geometric idea, isn't it?

As Chris said,
Quote:
 Originally Posted by Chris Card Remember that mathematicians love to prove things "by symmetry", so it's very often worth looking out for.

Last fiddled with by cheesehead on 2008-06-04 at 04:08

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