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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?
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.
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(t-4)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(t-4)dt] from -infinity to +infinity= ?

Thanks a ton


x

Message edited by author 2008-01-20 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 2008-01-20 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.
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