| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1446.1 | The indefinite integral has no closed form | COOKIE::PBERGH | Peter Bergh, DTN 523-3007 | Fri May 17 1991 13:05 | 5 |
| <<< 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.
|
| 1446.2 | Non-analytic. | CADSYS::COOPER | Topher Cooper | Fri May 17 1991 14:08 | 9 |
| 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
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| 1446.3 | would a polynomial be OK ? Check with experts | EAGLE1::BEST | R D Best, sys arch, I/O | Fri May 17 1991 19:06 | 14 |
| 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.
|
| 1446.4 | just the answer anyway | UTROP1::HARTOG_H | moduladaplisprologopsimulalgol | Sun May 19 1991 09:28 | 4 |
| From some old book:
$ cos(x)/x dx = log(x) - x^2/(2*2!) + x^4/(4*4!) - x^6/(6*6!) ...
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| 1446.5 | I suspect a well-studied problem here | CSSE::NEILSEN | Wally Neilsen-Steinhardt | Mon May 20 1991 13:25 | 15 |
| .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.
|
| 1446.6 | | ALLVAX::JROTH | I know he moves along the piers | Mon May 20 1991 15:00 | 9 |
| > 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
|
| 1446.7 | Taylor series OK for "small" x | COOKIE::PBERGH | Peter Bergh, DTN 523-3007 | Mon May 20 1991 17:46 | 19 |
| <<< 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.
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