how to find this integral ?

 Posts: 34
 Joined: Thu Dec 30, 2010 4:36 am
how to find this integral ?
Dear All
The function is f(t) = ((1  cos2t)^2)/(t^2)
I want to find its integral from t= pi to t= +pi
Can you please let me know how can I do that ?
Thank you
The function is f(t) = ((1  cos2t)^2)/(t^2)
I want to find its integral from t= pi to t= +pi
Can you please let me know how can I do that ?
Thank you
Re: how to find this integral ?
In my oponion, it's impossible to find an analytical solution, but numerically, it's clear that integral greater than five and less than six.

 Posts: 61
 Joined: Sat Dec 10, 2011 10:41 pm
Re: how to find this integral ?
I get a really long answer. It involved to substitutions:
(1/t)(2cos 2t^2/t)8tsin2t^2+(32/3)t^3cos2t^2+(sqrt(2)/2)sin 2(sqrt(2))t+2/3tfrom pi to pi.
I don't want to evaluate this lol.
(1/t)(2cos 2t^2/t)8tsin2t^2+(32/3)t^3cos2t^2+(sqrt(2)/2)sin 2(sqrt(2))t+2/3tfrom pi to pi.
I don't want to evaluate this lol.
Re: how to find this integral ?
Are you sure you need an answer or a bound? If an answer, mathematica has your back. Otherwise, you can easily find some bounds for this.

 Posts: 34
 Joined: Thu Dec 30, 2010 4:36 am
Re: how to find this integral ?
Thank you for your reply. Can you please let me know how can I find upper and lower bounds for this integral ?

 Posts: 54
 Joined: Sat Mar 24, 2012 5:01 pm
Re: how to find this integral ?
I think you can find the value of the integral using complex analysis.
Edit: Sorry, I was wrong, I was thinking of the integral from inf to inf.
In that case you can use complex analysis, ie :
You can square it, and make the numerator into (3/2  2 cos 2x + 1/2 * cos 4x)/x^2, and you need to find the integral of this function from inf/2 to inf/2.
The integral of (3/2  2 e^(2x)+ 1/2* e^(4x))/x^2 from R to R, and half of the circle = integral from R to R of the function you have + the integral of half of the circle (this one goes to 0 as R > infty) = integral you have (if it's from inf to inf). and the integral from R to R and half of the circle is easy to calculate using residue.
I don't know how to do with the integral from pi to pi.
Edit: Sorry, I was wrong, I was thinking of the integral from inf to inf.
In that case you can use complex analysis, ie :
You can square it, and make the numerator into (3/2  2 cos 2x + 1/2 * cos 4x)/x^2, and you need to find the integral of this function from inf/2 to inf/2.
The integral of (3/2  2 e^(2x)+ 1/2* e^(4x))/x^2 from R to R, and half of the circle = integral from R to R of the function you have + the integral of half of the circle (this one goes to 0 as R > infty) = integral you have (if it's from inf to inf). and the integral from R to R and half of the circle is easy to calculate using residue.
I don't know how to do with the integral from pi to pi.
Re: how to find this integral ?
Well, you can use Complex Analysis, and the fact that since the integrand is positive, the desired integral is bounded by the improper one.
Perhaps more simple is to note the following :
(1 cos 2t)^2 / t^2 = 4 sin^2 t / t^2 < 4 min(1, 1/t^2) ( by using sin^2 t <= t^2, and 0 <= sin^2 t <= 1).
So you can bound the integrand by some function f, such that:
1. f = 4, on some interval around 0,
2. f(t) = 4/t^2 outside of said interval.
Now, compute the integral of f over pi to pi
Perhaps more simple is to note the following :
(1 cos 2t)^2 / t^2 = 4 sin^2 t / t^2 < 4 min(1, 1/t^2) ( by using sin^2 t <= t^2, and 0 <= sin^2 t <= 1).
So you can bound the integrand by some function f, such that:
1. f = 4, on some interval around 0,
2. f(t) = 4/t^2 outside of said interval.
Now, compute the integral of f over pi to pi

 Posts: 34
 Joined: Thu Dec 30, 2010 4:36 am
Re: how to find this integral ?
Thank you very much for your help. So you mean if the function "f" is bounded by the functions "g" & "h" so its integral as well. So I need to find its maximum and minimum values in the mentioned interval and thus find the upper and lower bounds of its integral.