This is a group challenge:
It's well known that .
I know of a few ways of showing this. My challenge to all of you is to come up with as many ways as you can to prove this result.
Let's see if there's any ways I'm not aware of yet!
This is a group challenge:
It's well known that .
I know of a few ways of showing this. My challenge to all of you is to come up with as many ways as you can to prove this result.
Let's see if there's any ways I'm not aware of yet!
There are different ways to compute the integral...
(1)
... even if not all [in my opinion...] are 'fully convincing'... there is however also way to non compute it: if You consider (1) as a Lebesgue integral it diverges... in my opiniom this 'controversial' has to be, sooner or later, clarified...
Kind regards
I only know of 2 and here they are.
Now switch the order of integration
The other is via the residue theorem
Let be the linear arc from and and
Now notice that is an even function so
Now consider
where is an analytic function.
Using this and the ML bound we get
Now by Jordan's lemma on
This leaves
Since we take the real and immaginary parts to get
This gives
Here is a proof , first note that
By taking log. differentiation , we obtain this equation :
I would like to stop here and back to the integral .
We find that the integrand is even , i write it as
Consider
Sub. and on the first and the second integral respectively ,we obtain :
Therefore ,
the problem here is that you don't have absolute convergence.
in order to justify the interchange, it's better to integrate by parts first by selecting as an antiderivative for from there you introduce the parameter you said and the interchange is justified now by Tonelli's theorem since the function is no negative.
Although diverges, one can show in the following way:
First, note that since is measurable, we have that and so
Now note that using Fubini on gives and changing the order of integration the other way gives, after using integration by parts twice,
Taking the limit as and using DCT finished off the argument.
Here's a way using minimum machinery, but it requires the simple lemma that if then and the fact that exists, although this follows immediately from Dirichlet's test for integrals.
Ok, so note that since exists we have that
and thus clearly . We next claim that for every . But, this follows immediately from the common trigonometric identity associated with the Dirichlet kernel. We then note that
where we've made use of the lemma mentioned at the beginning of the post. Thus, it follows that