Hi,
I need to compute the integral between minus infinity and + infinity of
[sin(x)/x]^2, or {[1 - cos(x)]/x }^2. It should be pi. It has to be done
in the complex domain. Do you have any suggestions?
Thanks in advance


Sergio

In article , (Sirjo Lee) writes:
Hi,
I need to compute the integral between minus infinity and + infinity of
[sin(x)/x]^2, or {[1 - cos(x)]/x }^2. It should be pi. It has to be done
in the complex domain. Do you have any suggestions?
Thanks in advance

residua

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

wrote in message ...
In article , (Sirjo Lee) writes:
Hi,
I need to compute the integral between minus infinity and + infinity of
[sin(x)/x]^2, or {[1 - cos(x)]/x }^2. It should be pi. It has to be done
in the complex domain. Do you have any suggestions?
Thanks in advance

residua

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


Show please, contour and singular points

Sergio

Sirjo Lee wrote:

wrote in message ...
In article , (Sirjo Lee) writes:
Hi,
I need to compute the integral between minus infinity and + infinity of
[sin(x)/x]^2, or {[1 - cos(x)]/x }^2. It should be pi. It has to be done
in the complex domain. Do you have any suggestions?
Thanks in advance

residua

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

Show please, contour and singular points

Sergio

Write sin^2(x) as [1-cos(2x)]/2 and then evaluate the integral from -R to
-\epsilon
and +\epsilon to +R of [1-exp(2ix)]/2x^2 . You will, of course, want the real
part.
(The imaginary part vanishes identically---can you figure out why?) Close the
contour
with a big semicircle (radius R) in the upper half z-plane, and with a small
semi-
circle (radius \epsilon ) that avoids the singularity at z=0. You can pick the
little
semicircle to be convex upward also so there is no singularity inside the
contour.
Cauchy's Theorem then tells you that the contour integral is some value K. Set
the
contributions from real axis, and the two semicircles equal to K and you are
done.

I think I have told you enough to be getting on with, while at the same time
leaving
something for you to do.

--
Julian V. Noble
Professor Emeritus of Physics

^^^^^^^^
http://galileo.phys.virginia.edu/~jvn/

"God is not willing to do everything and thereby take away that share of
glory that rightfully belongs to ourselves."

-- N. Machiavelli, "The Prince".

In article , (Sirjo Lee) writes:
wrote in message ...
In article , (Sirjo Lee) writes:
Hi,
I need to compute the integral between minus infinity and + infinity of
[sin(x)/x]^2, or {[1 - cos(x)]/x }^2. It should be pi. It has to be done
in the complex domain. Do you have any suggestions?
Thanks in advance

residua


Show please, contour and singular points

Write the expression explicitly (expressing the trig functions as sums
of exponents) Deform the integration path to bypass the origin (a
legitimate deformation, since x=0 *is not* a singularity of the whole
expression), either above or below, doesn't matter. Now, integrate
the separate pieces of the expression you got (x = 0 is a singularity
for the separate pieces) along this path, closing the countour above
or below, depending on the piece. I'll leave it to you to find when
to close above and when below.

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

wrote in message
Write the expression explicitly (expressing the trig functions as sums
of exponents) Deform the integration path to bypass the origin (a
legitimate deformation, since x=0 *is not* a singularity of the whole
expression), either above or below, doesn't matter. Now, integrate
the separate pieces of the expression you got (x = 0 is a singularity

Write sin^2(x) as [1-cos(2x)]/2 and then evaluate the integral from -R to
-\epsilon
and +\epsilon to +R of [1-exp(2ix)]/2x^2 . You will, of course, want the real
part....

--
Julian V. Noble

Gentlemen, thanks for answering but I am puzzled by your answers:
1) The function [sin(x)/x]^2 is analytic everywhere, so the integral around
x axis plus semicircle is 0, i.e. the integral on the semicircle is -pi
(because we know the result on the x-axis as pi), but this is not easy
(at least for me) to show
2) 1/x^2 diverges in the origin and it is NOT a simple pole, so I can not
compute the principal value since according to all the books I have the
p.v. can be computed only when the pole is simple (as i*pi*res).
Actually 1/x^2 is not integrable and it is only when combined with cosx,
so I can not separate it from cos

I am probably missing something, I would be grateful if you could tell me what.
Thanks again

Sergio

Sirjo Lee wrote:

wrote in message
Write the expression explicitly (expressing the trig functions as sums
of exponents) Deform the integration path to bypass the origin (a
legitimate deformation, since x=0 *is not* a singularity of the whole
expression), either above or below, doesn't matter. Now, integrate
the separate pieces of the expression you got (x = 0 is a singularity

Write sin^2(x) as [1-cos(2x)]/2 and then evaluate the integral from -R to
-\epsilon
and +\epsilon to +R of [1-exp(2ix)]/2x^2 . You will, of course, want the real
part....

--
Julian V. Noble

Gentlemen, thanks for answering but I am puzzled by your answers:
1) The function [sin(x)/x]^2 is analytic everywhere, so the integral around
x axis plus semicircle is 0, i.e. the integral on the semicircle is -pi
(because we know the result on the x-axis as pi), but this is not easy
(at least for me) to show
2) 1/x^2 diverges in the origin and it is NOT a simple pole, so I can not
compute the principal value since according to all the books I have the
p.v. can be computed only when the pole is simple (as i*pi*res).
Actually 1/x^2 is not integrable and it is only when combined with cosx,
so I can not separate it from cos

I am probably missing something, I would be grateful if you could tell me what.
Thanks again

Sergio

Since this is surely a homework problem, I don't want to work it for you.
If you follow precisely the steps I laid out you will see how to get the
answer. Part of the exercise is to justify those steps!


--
Julian V. Noble
Professor Emeritus of Physics

^^^^^^^^^^^^^^^^^^
http://galileo.phys.virginia.edu/~jvn/

"Science knows only one commandment: contribute to science."
-- Bertolt Brecht, "Galileo".

(Sirjo Lee) writes:

2) 1/x^2 diverges in the origin and it is NOT a simple pole, so I can not
compute the principal value since according to all the books I have the
p.v. can be computed only when the pole is simple (as i*pi*res).
Actually 1/x^2 is not integrable and it is only when combined with cosx,
so I can not separate it from cos

Sin(x)/x does have a limit of 1 as x-0.


--
-Mike

Apologies for joining the thread late but the integral from 0 to infinity is
pi/2.


"Michael Moroney" wrote in message
...
(Sirjo Lee) writes:

2) 1/x^2 diverges in the origin and it is NOT a simple pole, so I can not
compute the principal value since according to all the books I have
the
p.v. can be computed only when the pole is simple (as i*pi*res).
Actually 1/x^2 is not integrable and it is only when combined with
cosx,
so I can not separate it from cos

Sin(x)/x does have a limit of 1 as x-0.


--
-Mike

"Julian V. Noble" wrote in message ...

Since this is surely a homework problem, I don't want to work it for you.
If you follow precisely the steps I laid out you will see how to get the
answer. Part of the exercise is to justify those steps!


--
Julian V. Noble
Professor Emeritus of Physics

^^^^^^^^^^^^^^^^^^

Finally I solved the problem using the lim for a- 0 of
[1-exp(iz)]/[z^2+a^2]. This trick is very powerful.

I am not sure any other method works, at least I did not succeed.
What is interesting is that a similar easier integral sin x/x
has P.V. of pi, and the integral on the vanishing semicircle around
the origin is -pi. If this was a result of a physics problem what
would the sense be? Is the overall result pi or zero?

Sergio