Author | Thread |
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01/20/2008 01:54:58 AM · #1 |
Y=4*integral of[(sinc(x))^2 dx] from 0 to +infinity
Any ideas? |
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01/20/2008 02:02:49 AM · #2 |
Y: 4*pi/2
ETA: I googled this
Message edited by author 2008-01-20 02:03:11. |
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01/20/2008 02:52:04 AM · #3 |
Wouldn't 4*pi/2 just be 2*pi?
~Terry
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01/20/2008 02:53:59 AM · #4 |
Or, we could just round it off to 6.28 |
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01/20/2008 04:25:56 AM · #5 |
From page:
List of integration formulas
int_0_to_inf[ ( (sin(x)/x )^2 ]dx = PI/2
and if (from sinc function
sinc(x) = [sin(PI.x )] / (PI.x)
that means
int_0_to_inf[ ( (sin(PI.x)/(PI.x) )^2 ]d(PI.x) = PI/2
or
PI . int_0_to_inf[ ( (sin(PI.x)/(PI.x) )^2 ]d(x) = PI/2
or
int_0_to_inf[ ( (sin(PI.x)/(PI.x) )^2 ]d(x) = 1/2
or
int_0_to_inf[ ( (sinc(x) )^2 ]d(x) = 1/2
or
4 . int_0_to_inf[ ( (sinc(x) )^2 ]d(x) = 4 . 1/2 = 2
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01/20/2008 07:32:46 AM · #6 |
Originally posted by zxaar: From page:
List of integration formulas
int_0_to_inf[ ( (sin(x)/x )^2 ]dx = PI/2
and if (from sinc function
sinc(x) = [sin(PI.x )] / (PI.x)
that means
int_0_to_inf[ ( (sin(PI.x)/(PI.x) )^2 ]d(PI.x) = PI/2
or
PI . int_0_to_inf[ ( (sin(PI.x)/(PI.x) )^2 ]d(x) = PI/2
or
int_0_to_inf[ ( (sin(PI.x)/(PI.x) )^2 ]d(x) = 1/2
or
int_0_to_inf[ ( (sinc(x) )^2 ]d(x) = 1/2
or
4 . int_0_to_inf[ ( (sinc(x) )^2 ]d(x) = 4 . 1/2 = 2 |
I'm sure this means something to someone, but to your average Bear it's just complete and utter nonsense :-)
R.
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01/20/2008 11:53:21 AM · #7 |
Wow you guys are smarter than the average Bear. :)
How about
integral of[x*dirac_delta(t-4)dt] from -infinity to +infinity= ?
Thanks a ton |
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01/20/2008 12:01:41 PM · #8 |
Originally posted by Niten: Wow you guys are smarter than the average Bear. :)
How about
integral of[x*dirac_delta(t-4)dt] from -infinity to +infinity= ?
Thanks a ton |
x
Message edited by author 2008-01-20 12:02:45.
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01/20/2008 12:06:58 PM · #9 |
the answer is "x"
A Dirac delta function is zero at all points, except for a select point (t=4 for this one) at which it is infinity. The integral from negative infinity to positive infinity of the delta fuction is defined to be one. x in this equation is a constant because the integration is with respect to t, so it can be pulled out in front of the integral, and you are left with 1*x=x
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01/20/2008 12:11:39 PM · #10 |
The Sinc function is interesting.
The integral from zero to infinity of sinc squared is pi/2, so the answer is 2pi
Message edited by author 2008-01-20 12:21:05.
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01/20/2008 08:54:17 PM · #11 |
Originally posted by Bear_Music: Originally posted by zxaar: From page:
List of integration formulas
int_0_to_inf[ ( (sin(x)/x )^2 ]dx = PI/2
and if (from sinc function
sinc(x) = [sin(PI.x )] / (PI.x)
that means
int_0_to_inf[ ( (sin(PI.x)/(PI.x) )^2 ]d(PI.x) = PI/2
or
PI . int_0_to_inf[ ( (sin(PI.x)/(PI.x) )^2 ]d(x) = PI/2
or
int_0_to_inf[ ( (sin(PI.x)/(PI.x) )^2 ]d(x) = 1/2
or
int_0_to_inf[ ( (sinc(x) )^2 ]d(x) = 1/2
or
4 . int_0_to_inf[ ( (sinc(x) )^2 ]d(x) = 4 . 1/2 = 2 |
I'm sure this means something to someone, but to your average Bear it's just complete and utter nonsense :-)
R. |
Yepp, but when it makes sense it is more fun. |
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