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Science Forum Index » Math - Numerical Analysis Forum » Integrating sin x / x
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| Schizoid Man |
 Posted: Fri Mar 16, 2007 1:00 am |
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Hi,
I was wondering whether there's any numerical technique I can use to
integrate sin x / x over [0,+\infty].
The integral converges though it doesn't appear to have a closed form
antiderivative.
Thanks,
Schiz |
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| Carl Barron |
| Posted: Fri Mar 16, 2007 1:45 am |
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In article <etdbqk$dgr$1@geraldo.cc.utexas.edu>, Schizoid Man
<schiz@oid.man> wrote:
Quote: Hi,
I was wondering whether there's any numerical technique I can use to
integrate sin x / x over [0,+\infty].
The integral converges though it doesn't appear to have a closed form
antiderivative.
Thanks,
Schiz
since sin(x)/x is smooth gaus-laguare quadrature should be accurate
since the abcissas are > 0 and x = 0 is a removeable singularity of
sin(x)/x.
the ans is pi/2 if I recall correctly. |
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| user923005 |
| Posted: Fri Mar 16, 2007 2:37 am |
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On Mar 15, 11:00 pm, Schizoid Man <s...@oid.man> wrote:
Quote: Hi,
I was wondering whether there's any numerical technique I can use to
integrate sin x / x over [0,+\infty].
The integral converges though it doesn't appear to have a closed form
antiderivative.
It's pi/2.
http://press.princeton.edu/books/maor/chapter_10.pdf |
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| Badger |
| Posted: Fri Mar 16, 2007 10:17 am |
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On Thu, 15 Mar 2007 23:00:30 -0700, Schizoid Man <schiz@oid.man>
wrote:
Quote: Hi,
I was wondering whether there's any numerical technique I can use to
integrate sin x / x over [0,+\infty].
The integral converges though it doesn't appear to have a closed form
antiderivative.
Thanks,
Schiz
Integral is pi/2. One way of doing the integral is here:
<http://www.math.binghamton.edu/paul/papers/LoyaMathMag.pdf>
Another way is to use the Laplace transform of sin(t) / t.
f(t) := sin(t) / t
F(s) = L{f(t)} = arctan(1/s)
As s go to 0, F goes to pi/2.
HTH |
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| Robert Israel |
| Posted: Fri Mar 16, 2007 4:35 pm |
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Schizoid Man <schiz@oid.man> writes:
Quote: Hi,
I was wondering whether there's any numerical technique I can use to
integrate sin x / x over [0,+\infty].
The integral converges though it doesn't appear to have a closed form
antiderivative.
The antiderivative is called Si(x). That function's definition is,
you guessed it, int_0^x sin(t)/t dt. But it _is_ a "known" function.
If you really want a numerical technique, you might try writing the
integral as
sum_{n=0}^infty int_{2 n pi}^{2 (n+1) pi} sin(x)/x dx
= sum_{n=0}^infty int_0^{2 pi} sin(t)/(2 n pi + t) dt
= sum_{n=0}^infty int_0^pi sin(t) (1/(2 n pi + t) - 1/((2 n + 1) pi + t)) dt
= sum_{n=0}^infty int_0^pi sin(t) pi/((2 n pi + t)((2 n + 1) pi + t))) dt
Note that the summand is O(1/n^2), so the sum converges, and various
acceleration methods can be applied to evaluate it numerically.
--
Robert Israel israel@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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| Schizoid Man |
| Posted: Sat Mar 17, 2007 2:08 pm |
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Carl Barron wrote:
Quote: since sin(x)/x is smooth gaus-laguare quadrature should be accurate
since the abcissas are > 0 and x = 0 is a removeable singularity of
sin(x)/x.
the ans is pi/2 if I recall correctly.
Hi Carl,
You're right - this can be approximated using Gauss-Laguerre quadrature,
though the convergence is quite slow. I tried calculating this using 32
points and it wasn't nearly enough.
I know that the integral is pi/2, but I was wondering whether there is
an easy numerical solution to this, without using techniques like
Laplace Transform or Contour Integration.
Thanks. |
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| widmar |
| Posted: Thu Mar 22, 2007 3:10 pm |
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Schizoid Man wrote:
Quote:
I was wondering whether there's any numerical technique I can use to
integrate sin x / x over [0,+\infty].
The integral converges though it doesn't appear to have a closed form
antiderivative.
There are several, however the most obvious (and reliable) choice is
already implied in your question; "integrating" is the road map to a
quick numerical result. Solve,
d/dt y = sin(t)/t s.t. y(0) = 0
with the initial rate y' = 1. As a side exercise you might try series
evaluation (suggested elsewhere) as an antithesis on how things can go
wrong when engaging (numerically) ill defined methods.
---
sdx - modeling, simulation.
http://www.sdynamix.com |
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| RRogers |
| Posted: Fri Mar 23, 2007 7:41 am |
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On Mar 22, 4:10 pm, widmar <j...@sdynamix.com> wrote:
Quote: Schizoid Man wrote:
I was wondering whether there's any numerical technique I can use to
integrate sin x / x over [0,+\infty].
The integral converges though it doesn't appear to have a closed form
antiderivative.
There are several, however the most obvious (and reliable) choice is
already implied in your question; "integrating" is the road map to a
quick numerical result. Solve,
d/dt y = sin(t)/t s.t. y(0) = 0
with the initial rate y' = 1. As a side exercise you might try series
evaluation (suggested elsewhere) as an antithesis on how things can go
wrong when engaging (numerically) ill defined methods.
---
sdx - modeling, simulation.http://www.sdynamix.com
I can't see the previous posts, but www.mathworld.com has the complete
intergal [-inf,+inf] as pi, so 1/2 would be pi/2.
Ray |
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