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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1446.0. "Integral of (cos x dx) / x" by LANDO::NATUSCH () Fri May 17 1991 11:42

    I haven't done any calculus for some number of years, the other
    night I tried to solve the following simple intregral and couldn't
    figure out how to do it.  I tried intergrating by parts and got
    the result (I used the $ for the integral sign):
    
    		$ (cos x) dx    =  $ (cos x) dx
    		  ----------       ------------
    			x                x
    
    
    I'm looking for the method to solve this not just the answer, can
    you help?
    
    		thanks
    
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1446.1The indefinite integral has no closed formCOOKIE::PBERGHPeter Bergh, DTN 523-3007Fri May 17 1991 15:055
                      <<< Note 1446.0 by LANDO::NATUSCH >>>
                        -< Integral of (cos x dx) / x >-

I assume you mean integral of cos(x)/x where x is the variable of integration.
This integral can not be expressed in a simple closed form.
1446.2Non-analytic.CADSYS::COOPERTopher CooperFri May 17 1991 16:089
    Maple makes use of several "non-analytic" functions which it reduces
    many integrals without analytic solutions to.  One of these is the
    "cosine integral" which they call Ci.  Ci is defined in the
    documentation to be "gamma + ln(x) + int( (cos(t)-1)/t, t=0..x)" with
    "gamma" being Eulers constant which equals "limit(sum(1/i, i=1..n) -
    ln(n), n=infinity)".  The integral of cos(x)/x w.r.t. x is simply Ci(x)
    according to Maple.

					Topher
1446.3would a polynomial be OK ? Check with expertsEAGLE1::BESTR D Best, sys arch, I/OFri May 17 1991 21:0614
re .0

Playing with sinc functions  ?

Doing something related to filtering or convolution ?

I think .1 is right, but if you want an approximation polynomial you might be
able to get one by taking one of the polynomial infinite series for
cos(x), dividing each term by x and then integrating the series term by
term.

The resulting infinite series MIGHT be a series for the integral
if it converges, but wait until one of the notesfile real analysis experts
responds on whether this is legit.
1446.4just the answer anywayUTROP1::HARTOG_HmoduladaplisprologopsimulalgolSun May 19 1991 11:284
    From some old book:
    
    $ cos(x)/x dx = log(x) - x^2/(2*2!) + x^4/(4*4!) - x^6/(6*6!) ...
    
1446.5I suspect a well-studied problem hereCSSE::NEILSENWally Neilsen-SteinhardtMon May 20 1991 15:2515
.4 looks like the answer to .3 and should converge in a positive interval.

But I suspect that the polynomial will converge very slowly for x>1.

If you are serious about this problem, I'd guess from .3 that what you want
to do is either 

	use MAPLE

		or

	consult that book about approximations to functions that is
	referenced in a couple of these notes, and see what they say 
	about better approximations to Ci.  There are almost always some
	known better approximations than polynomials.
1446.6ALLVAX::JROTHI know he moves along the piersMon May 20 1991 17:009
>    From some old book:
    
>    $ cos(x)/x dx = log(x) - x^2/(2*2!) + x^4/(4*4!) - x^6/(6*6!) ...

     You have to add Euler's constant to the above series, then it is
     correct.  The series part represents an entire function and will
     converge acceptably well, in fact better a little better than exp(x).

    - Jim
1446.7Taylor series OK for "small" xCOOKIE::PBERGHPeter Bergh, DTN 523-3007Mon May 20 1991 19:4619
          <<< Note 1446.3 by EAGLE1::BEST "R D Best, sys arch, I/O" >>>
               -< would a polynomial be OK ? Check with experts >-

>> I think .1 is right, but if you want an approximation polynomial you might be
>> able to get one by taking one of the polynomial infinite series for
>> cos(x), dividing each term by x and then integrating the series term by
>> term.

The cosine series (1 - x^2/2! + x^4/4! - x^6/6! ....) is extremely well behaved
in the mathematical sense, so the series for cos(x)/x is (almost) as well
behaved (it has a pole of order 1 at the origin).  Thus, over any interval that
does not contain zero (but may contain infinity), you may integrate term by
term (for the purposes of integrating term by term, it suffices to show that
the series is absolutely convergent over the interval  --  easy exercise for
the reader).  The practical problem with this approach is that the terms for
large x may get *quite* large before they start to get smaller, i.e., you may
get floating-point overflow before the series has converged.  If you only need
the value of the integral for small x (guess: smaller than 5 or so), you should
not have any problem.