I have been summoned?

Hello @XanderHenderson !

(I am a little short on time, FYI.)

@XanderHenderson Okay! I saw your comments under the fourier post and it seems you are have quite some knowledge in this area. So I'd like to ask you a question about a related topic :).

My knowledge is actually pretty shallow---Katznelson's book would likely be a better resource, but ask away.

I defined this thing here:

Yeah, that looks about right... though I am not sure about "modified".

Where is this from?

This integral is strongly related to the Inverse Mellin-Transform. The one thing which is different, is that the exponent in the kernel is $s$ and not $-s$.

I defined it. Is there already a name for this? How would you call it? (I also have two more questions.)

@XanderHenderson I forgot to ping you :).

@vitamind Yeah, that, and the logarithmic term out front.

But... why?

That also looks more like an inverse Mellin transform to me... I think.

Given $f : \mathbb{R}_{> 0} \to \mathbb{C}$, isn't the Mellin transform usually given by $\int_{0}^{\infty} f(x) x^{s-1} \,\mathrm{d} x$?

@XanderHenderson Ah right but you can forget about it. It is an Inverse mellin transform but with a different kernel. A simple sub won't help.

@XanderHenderson Maybe a better name would be : Modified inverse mellin tranform (but it's kind of long)

Why are you considering this object?

I have some functions of which I have to take the transform. I want a more compact notation since I'm using it more than a few times.

I also don't think that the change from $s$ to $-s$ is as important as you make it out to be. The substitution $s \mapsto -s$ gives you something like $$\int_{-\gamma+i\infty}^{-\gamma-i\infty} \Phi(-s) x^{-s}\, \mathrm{d}s, $$ which seems to be the inverse Mellin transform of $s \mapsto \Phi(-s)$.

Correct but what about the limits/bounds?

Assuming that you can integrate along the line $-\gamma + i\tau$ ($\tau \in \mathbb{R}$).

It kind of depends on where $\Phi$ is analytic.

In any event, I'm about done for the day.

I originally had Phi(s)=Log(s-1).

@XanderHenderson Okay then see you. I'll ask my other two questions tomorrow, if it's okay. Thanks for helping so far.

@XanderHenderson Do you have time right now?

Not really, but chat is asynchronous. Feel free to say anything you need to say, and I am sure that I will get to it when I have time.

@XanderHenderson All right, thanks. I was wondering if my understanding of the "extension" of the domain for $\Phi$ is correct. After reading this, is it correct that we can safely say that: $\varphi(s):L_{\nu,q}(R^+)\to\mathbb{C}$, where $\nu,q\in\mathbb{R}$ and $q\ge2$.

(The choice of q comes from q=p/(p-1), where 1<p\le2.)

...for the Inverse mellin transform to exist (and that it bijects).

I think that the thing to understand there is that there is not just one Mellin Inversion Theorem, but a number of closely related theorems which all give meaning to inverting the Mellin transform.

The hypotheses are all a little different, and give rise to slightly different animals.

I'm don't really have the energy to track down a better statement of the theorem, but the version you have cited on Wikipedia seems to assert that if a function is in a particular space (basically, $x^\nu f$ is in $L^p(\mathbb{R}, \mathrm{d}x/x)$), then the Mellin transform of $f$ exists, and is invertible.

And yes, the Melin transform lives in the "Hölder dual" of $L^p$.

@XanderHenderson Thanks for your comment. You gave conditions for the Mellin Transform to exist but I need something that says if the Inverse mellin transform exists whith conditions on $\varphi$ and not $f$. So does this condition "$\varphi(s):L_{\nu,q}(R^+)\to\mathbb{C}$, where $\nu,q\in\mathbb{R}$ and $q\ge2$" suffice to verify the existence of an Inverse mellin transform?

I would suggest that you need to look beyond a wikipedia article for better theorems. A good place to start might be Titchmarsh's book on Fourier Integrals. There is a pretty lengthy discussion of the topic starting on page 46 of my edition of the text (second edition; I have a printing from the late 60s).

@XanderHenderson All right, thanks for the suggestion. Sorry to annoy you with the same question again: Does this mean there is no simple yes or now to the condition for the theorem I wrote?

I am less familiar with the Mellin transform than I am with the Fourier transform---the answer to your question is "I don't know". I can suggest places to look, but I don't know of a simple answer.

I will definetely look into the book. Is the page available for free?

@XanderHenderson I understand. If the page is not for free - is there any possibility where you could send the few pages in this chat?

I really don't have the ability to do that. Do you not have access to an institutional library which can get books for you?

It isn't instant, but you get the books at the end of the day.

@XanderHenderson Complicated in my case. I'll find a solution.

@XanderHenderson Was my trick for the conditional convergent series useful in some way or was it written too confusing? (if that's the case, it was for 4am sorry)

It was far to vague and without context, but I don't think that it applies. It seems like you were able to take advantage of the way way in which certain terms died off, and I don't have any such uniformity.

@XanderHenderson Yes I already thougth so. I'm very interested in your sums (not with the intention to solve it, but to simply look at it).

@XanderHenderson Do I interpret your silence as not willing to share the summation correctly :)? If yes, please say so!

Oh, sorry. Did not notice.

Hang on...

ucr.primo.exlibrisgroup.com/discovery/…

Start on page 110.

The ugliness happens on page 116.

Thank you!

@XanderHenderson What are $\ell$ and $s$? (Not sure where they're introduced) And the series in 6.5.4 is absolute convergent -> you're converting a summand in a conditionally convergent series -> exchange not justified. Right?

$s$ is a complex variable.

I don't remember what $\ell$ was...

Uh...

It helps to count rectangles...

I wrote that section five or six years ago, and don't have all of the details handy in my head right now.

$\ell$ first shows up in Table 6.1 (which references Figure 6.2).

@vitamind Yeah, that is basically the nut of it. The original series is absolutely convergent, but I need more details about how it behaves on the boundary where it may or may not converge. So I replace the summand with a Fourier series representation of the summand.

However, the Fourier series does not converge uniformly (the summand is a step function), and so the exchange of summations is not obviously justified.

step function. uff

What is frustrating about this is that if the sums can be swapped, you get a line of singularities which precisely agree with what you would expect them to be.

@XanderHenderson Frustrating. I assume you asked some experts, who couldn't help you too?

Like, the "wrong" answer is exactly what you expect it to be, so... ARG!

And yeah, I've had a lot of conversations with folk. I'm not sure what qualifies one as an "expert", but people who know trig series pretty well have looked at it.

@XanderHenderson Yup this is what I meant by expert. Have you considered posting it on MO?

@vitamind I have, but I have zero interest in dealing with MO.

@XanderHenderson Why is that, if I may ask?

@vitamind I'm not a big fan of the culture there.

@XanderHenderson People, comments or answers (or all together)?

Yes.

Okay. I can see what you mean, but at least for me most of the answers are helpful. (I am also planning on answering my first question.)

My gut feeling about MO is that questions in category theory are often given a free pass, whereas questions in hard analysis (e.g. very fine estimates; improving bounds; etc) are often shot down very quickly.

Thus MO is typically poorly suited to the kinds of things that I find interesting.

And, since I am not really in a position where I have to publish or perish, I can continue to bang my head against the problem for as long as I like.

And if I figure it out, it is good for one MPU.

@XanderHenderson I cannot say anything to that since I'm not an experienced MO user. (Your statement makes absolutely sense though. I have the feeling that category theory is the godlike branch of mathematics for many people.) But the question I want to answer, contains some hard analysis.

Category theory is dumb. :P

@XanderHenderson I had a nice conversion with my mentor about category theory today. He said it's nice to know one or two basic concepts but well

that's it.

@XanderHenderson I am so sorry for my overcautiousness: Since you are a moderator you are able to delete this and the previous comments, right?

I can delete anything!

@vitamind BWAHAHAHAHAHAAHAH!

(And off to dinner.)

@XanderHenderson thanks :`D