how to find this integral ?

Forum for the GRE subject test in mathematics.
hadimotamedi
Posts: 34
Joined: Thu Dec 30, 2010 4:36 am

how to find this integral ?

Postby hadimotamedi » Tue Mar 05, 2013 1:05 am

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

devgen
Posts: 1
Joined: Tue Mar 05, 2013 10:13 am

Re: how to find this integral ?

Postby devgen » Tue Mar 05, 2013 10:25 am

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.

waiting512
Posts: 61
Joined: Sat Dec 10, 2011 10:41 pm

Re: how to find this integral ?

Postby waiting512 » Tue Mar 05, 2013 11:14 am

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+2/3t|from -pi to pi.

I don't want to evaluate this lol.

Rikimaru
Posts: 25
Joined: Thu Feb 07, 2013 3:11 pm

Re: how to find this integral ?

Postby Rikimaru » Tue Mar 05, 2013 11:20 am

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.

hadimotamedi
Posts: 34
Joined: Thu Dec 30, 2010 4:36 am

Re: how to find this integral ?

Postby hadimotamedi » Sat Mar 09, 2013 12:50 am

Thank you for your reply. Can you please let me know how can I find upper and lower bounds for this integral ?

longtm1989
Posts: 54
Joined: Sat Mar 24, 2012 5:01 pm

Re: how to find this integral ?

Postby longtm1989 » Sat Mar 09, 2013 1:00 am

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.

Rikimaru
Posts: 25
Joined: Thu Feb 07, 2013 3:11 pm

Re: how to find this integral ?

Postby Rikimaru » Sun Mar 10, 2013 8:41 pm

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

hadimotamedi
Posts: 34
Joined: Thu Dec 30, 2010 4:36 am

Re: how to find this integral ?

Postby hadimotamedi » Tue Mar 12, 2013 6:09 am

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.




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