The temptation's too great. I gotta do it. This is done within a
distribution-theoretic context.

Let f(k) = integral (sin(kx)/x)^2 dx, from x = -infinity to +infinity.
Then
f(0) = 0
f'(k) = integral 2 sin(kx) cos(kx)/x dx
= integral sin(2kx)/x dx
f'(0) = 0
f''(k) = integral 2 cos(2kx) dx
= integral (exp(2ikx) + exp(-2ikx)) dx
= 2 integral exp(2ikx) dx
= integral exp(iku) du, substituting u for 2x
= 2 pi delta(k).
Thus
f'(k) = pi sgn(k)
and
f(k) = pi |k|.
Thus
integral (sin(kx)/x)^2 dx = pi |k|
and
integral (sin(x)/x)^2 dx = pi.

Since the function is even, the integral of this from 0 to +infinity
should be pi/2.

This is all preliminary. I don't recall ever seeing what the integral
of (sin(x)/x)^2 actually is.

In article , (Mark) writes:
The temptation's too great. I gotta do it. This is done within a
distribution-theoretic context.

Let f(k) = integral (sin(kx)/x)^2 dx, from x = -infinity to +infinity.
Then
f(0) = 0
f'(k) = integral 2 sin(kx) cos(kx)/x dx
= integral sin(2kx)/x dx
f'(0) = 0
f''(k) = integral 2 cos(2kx) dx
= integral (exp(2ikx) + exp(-2ikx)) dx
= 2 integral exp(2ikx) dx
= integral exp(iku) du, substituting u for 2x
= 2 pi delta(k).
Thus
f'(k) = pi sgn(k)
and
f(k) = pi |k|.
Thus
integral (sin(kx)/x)^2 dx = pi |k|
and
integral (sin(x)/x)^2 dx = pi.

Neat.

Since the function is even, the integral of this from 0 to +infinity
should be pi/2.

Yep.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

(Mark) wrote in message ...
The temptation's too great. I gotta do it. This is done within a
distribution-theoretic context.


.....
integral (sin(x)/x)^2 dx = pi.

Since the function is even, the integral of this from 0 to +infinity
should be pi/2.



Very beautiful, congratulations

Sergio