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! (Wink)

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- August 3rd 2010, 05:23 PMchiph588@Number of ways to evaluate an integral
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! (Wink) - August 3rd 2010, 06:01 PMchisigma
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

- August 3rd 2010, 06:03 PMTheEmptySet
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

- August 3rd 2010, 06:12 PMchiph588@
- August 3rd 2010, 06:23 PMTheEmptySet
On the arc

We have

So we get

The 2nd integral is bounded by it's maximum on the arc multiplied by its length so we get

but as we get - August 3rd 2010, 06:27 PMchiph588@
- August 6th 2010, 05:45 PMsimplependulum
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 ,

- January 26th 2011, 12:49 PMKrizalid
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. - January 26th 2011, 07:18 PMGeneral
You could use the following laplace transform's property:

- January 26th 2011, 08:12 PMRandom Variable
- January 27th 2011, 05:41 AMKrizalid
yes, that approach works fine and is justified.

- January 27th 2011, 07:32 PMRandom Variable
I asked a professor, and he said something about that because the issue is at a single point (the origin), that what the TheEmptySet did was OK. I think the absolute convergence requirement is too restrictive. I don't know.

- January 27th 2011, 07:35 PMKrizalid
perhaps is restrictive, but it works, but you'd need a good justification when reversing the integration order by not having absolute convergence.

- January 28th 2011, 12:17 AMDefunkt
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. - January 31st 2011, 12:59 PMDrexel28
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