Derive the integration formula

By using the residue theorem, I integrated the function from to , I can only obtain the result , which means my answer is .

I couldnt find my mistake.Thank you.

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- Dec 16th 2011, 05:52 AMfareastmovementImproper integral
Derive the integration formula

By using the residue theorem, I integrated the function from to , I can only obtain the result , which means my answer is .

I couldnt find my mistake.Thank you. - Dec 16th 2011, 06:44 AMgirdavRe: Improper integral
The function you have to integrate is even, so the integral on is two times the integral on .

- Dec 16th 2011, 07:10 AMfareastmovementRe: Improper integral
Thank you for your reply.

Yes i did it (divided by 2)and still get as answer, but not - Dec 16th 2011, 07:23 AMgirdavRe: Improper integral
Did you wrote the exponentials in terms of ?

- Dec 16th 2011, 07:50 AMfareastmovementRe: Improper integral
Yes, by integrating and compare the real part of both sides

- Dec 17th 2011, 08:41 AMFernandoRevillaRe: Improper integral
Another way: denote , then . Using the substitution we get (Dirichlet's integral) . So, . For we get .

Then,

**@fareastmovement**: Show your complete work by the method of residues (of course if you want) and we can check it.

Edited: Method not valid, see the posts below.

- Dec 17th 2011, 09:30 AMfareastmovementRe: Improper integral
Thank you, I got it now. You are really helpful :)

- Dec 18th 2011, 10:25 PMchisigmaRe: Improper integral
There is only a minor detail: writing is a nonsense because the integral diverges...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

- Dec 18th 2011, 11:00 PMchisigmaRe: Improper integral
In order to avoid criticity in the solution of the integral I would suggest to use the trigonometric identity...

(1)

... so that the integral becomes...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

- Dec 18th 2011, 11:01 PMFernandoRevillaRe: Improper integral
- Dec 18th 2011, 11:18 PMchisigmaRe: Improper integral
The real problem is in defining the function ...

(2)

... because the integral diverges for any value of . In my previous post a way to overcame such type of criticity is indicated...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

- Dec 18th 2011, 11:43 PMFernandoRevillaRe: Improper integral
Not for all values of but of course the method I used is not valid. Reviewing the situation I find the explanation: in a foolish way my mind thought about to instead of to and curiously at the end I made the substitution . A complete disaster. :)

Method not valid. - Dec 19th 2011, 01:39 AMAlso sprach ZarathustraRe: Improper integral
- Dec 19th 2011, 03:34 AMFernandoRevillaRe: Improper integral
(a) is convergent for (easily proved using

**chisigma**'s answer #9 )

(b) (differentiation under the integral sign and Dirichlet's Integral).

(c) , so hence . That is, .

(d) Taking limits in as we get . So, - Dec 19th 2011, 11:19 AMRandom VariableRe: Improper integral

Fubini's theorem is not satsified. So to justify changing the order of integration, let's write the double integral more formally as a limit.

Since the Sine Integral is bounded above by 2, we can use the Dominated Convergence Theorem to justify bringing the limit inside of the integral.