Author  Thread 

01/20/2008 01:54:58 AM · #1 
Y=4*integral of[(sinc(x))^2 dx] from 0 to +infinity
Any ideas? 


01/20/2008 02:02:49 AM · #2 
Y: 4*pi/2
ETA: I googled this
Message edited by author 20080120 02:03:11. 


01/20/2008 02:52:04 AM · #3 
Wouldn't 4*pi/2 just be 2*pi?
~Terry



01/20/2008 02:53:59 AM · #4 
Or, we could just round it off to 6.28 


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



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.



01/20/2008 11:53:21 AM · #7 
Wow you guys are smarter than the average Bear. :)
How about
integral of[x*dirac_delta(t4)dt] from infinity to +infinity= ?
Thanks a ton 


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(t4)dt] from infinity to +infinity= ?
Thanks a ton 
x
Message edited by author 20080120 12:02:45.



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



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 20080120 12:21:05.



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. 
