Discussion on answer by Rounak Sarkar: Can $\int\limits_0^\infty e^{ix^ x}dx$ be written without a limit?

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13:39

A: Can $\int\limits_0^\infty e^{ix^ x}dx$ be written without a limit?

I am just going to focus on the indefinite integral, not the definite one. Let, $\textstyle\displaystyle{I=\int e^{ix^x}dx}$ Substitute, $u=x^x$ Then, $x=e^{W(\ln(x))}$ And $\textstyle\displaystyle{dx=\frac{du}{u(W(\ln(u))+1)}}$ This video will help you derive $dx$ Then, $\textstyle\displaystyle{...

Rounak Sarkar

I will request everyone to check my calculations and tell me if there is any mistake.

Tyma Gaidash

Hello @Rounak . It was a bright idea using the Inverse Lagrange Theorem, but could you please verify the result? Finally, if the answer does indeed work, is it possible to “reverse” the substitution only in terms of x?

Rounak Sarkar

@Tyma Gaidash. I have checked all the steps a lot of times. There were many mistakes, I almost edited amd deleted this answer almost a dozen times. It should work now. By the way, you just found out another thing I forgot to put in the answer. Let me edit it, yes you can just reverse substitute by just replacing every $u$ with $x^x$.

Tyma Gaidash

Actually you do not have to verify, but if you like, there is this integration of an inverse function result which may be useful for the future. Is it possible to simplify or list out the first few terms of $t_n$, the nth derivative part of the series? This is a great approach, so take your time.

Rounak Sarkar

@Tyma Gaidash. To be honest, when I first saw the expression of $t_n$, it just scared me off. 🤣🤣. But now I am not scared anymore, so I will try.

Tyma Gaidash

13:39

Is it possible to list out the first few $t_n$ terms please? Thanks for the effort.

Rounak Sarkar

@Tyma Gaidash. I do have a plan to make a list of the first $3$ or $4$ values of $t_n$. I won't have the time today, so I will try tomorrow.

Tyma Gaidash

That is fine, that is why I said these problems are optional. Thanks again.

Gary

@RounakSarkar What is the region of convergence of this expansion?

Rounak Sarkar

@Gary. I didn't really thought about the radius of convergence. But now that you told me, I checked it out using the ratio test. I couldn't figure out the radius of convergence of the final expression. But for the series of $t(u)$ we have, $\textstyle\displaystyle{|u-1|\lt\lim_{n\rightarrow\infty}\left|\frac{nt_n}{t_{n+1}}\right|}$. I am not sure about the boundary.

Tyma Gaidash

@RounakSarkar What does “[Edit removed during grace period]” mean? There is also a limit of w.

Rounak Sarkar

13:39

@Tyma Gaidash. What?? I don't know what does that mean??

Tyma Gaidash

@RounakSarkar I think it is due to the bounty ending. Look at the edits and you will see this made by you. Thanks for the terms of $t_n$.

The OEIS does not have an exact signed match for the $t_n$.

Rounak Sarkar

@Tyma Gaidash. I quickly realized that it is impossible for me to simplify a few terms of $t_n$. Although finding the value of $t_1$ by hand was easy. So for the rest of the terms I just put it in wolfram alpha got $t_2$ and $t_3$. But for some reason wolfram alpha couldn't find the value of $t_4$ or any values $t_n$ where $n>3$. So I am not really sure that the sequence you are showing is really $|t_n|$ or not.

Steven Clark

@TymaGaidash Based on the first 10 values of $t_n$, I believe $t_n$ is actually related to oeis.org/A305981 and is given by $t_n=(-1)^n \sum\limits_{k=1}^n \left| S_n^{(k)}\right| k^k$ where $S_n^{(k)}$ is the Stirling number of the first kind.

Tyma Gaidash

@StevenClark Not how I mentioned that your linked series has 1, but not -1 like in Rounak’s answer. However, the 2 sequences look very similar.

Steven Clark

@TymaGaidash I said it's related, but it's not an exact match, which is why the $(-1)^n$ precedes the sum in the formula I gave where the sum is based on oeis.org/A305981.

Tyma Gaidash

13:39

Here is a graph of $t_4$ and a calculation. The derivative seems to either be 0 or a very large number, like $\infty$. There is also the possibility of the calculation being faulty. Based on desmos, $t_4≈468$. @StevenClark this supports your guess and matches well with your OEIS sequence.

$t_5≈-6876$ which does match well Steven’s OEIS sequence, there may be a typo here though. I did not expect $t_n$ to have such a close OEIS sequence. Note that indeed there is an $(-1)^{n+1}$ or $(-1)^n$ in front of Steven’s sequence (probably). Maybe @RounakSarkar can edit with this alternating series OEIS formula with a disclaimer? The Lagrange Inversion formula has now brought up a new strategy of the coefficients possibly being an OEIS, or other, sequence for integration.

Steven Clark

@TymaGaidash The OEIS sequence starts with $n=0$ instead of $n=1$. Starting at $n=1$, Mathematica gives the first $12$ values of $t_n=\underset{w\to 1}{\text{lim}}\left(\frac{\partial^{n-1}}{\partial w^{n-1}}\left(\frac{w-1}{e^{\left(\frac{1}{w}-1\right) e^{\frac{1}{w}-1}}-1}\right)^n\right)$ as {-1,5,-41,468,-6854,122582,-2589978,63129392,-1743732192,53827681152,-1836453542472,68620052332752} which matches the corresponding values of oeis.org/A305981 except in sign, and which exactly matches the formula $t_n=(-1)^n \sum\limits_{k=1}^n \left| S_n^{(k)}\right| k^k$.

Tyma Gaidash

Let us continue this discussion in chat.

@StevenClark Hopefully it can to into the answer and simplify the final integral a bit.

@RounakSarkar Steven found a likely explicit infinite sum expression for $t_n$.

Steven Clark

@TymaGaidash It's actually a finite sum. I've now verified the Mathematica results for $t_{1}$ to $t_{14}$ are consistent with the OEIS entry and the formula I gave above. Each successive $t_n$ value takes longer and longer for Mathematica to derive the result for the limit of the derivative, so I'm going to abort the evaluations.

Tyma Gaidash

@StevenClark You are right. I am more used to non finite sums, sorry for the error. We probably can safely say that truncated series is the OEIS sequence .

Rounak Sarkar

@Steven Clark. By the way can we have a closed form for $\lim_{n\rightarrow\infty}\left|\frac{nt_n}{t_{n+1}}\right|$?

@Steven Clark. I have edited the answer to add your work.

@Tyma Gaidash. Do you think that the limit diverges?

Tyma Gaidash

13:39

@RounakSarkar It is weird, there is a truncated $k^k$ sum being regularized with the $\frac{1}{n!k!(k-n)Γ(n-k)}$ from the Q function and denominators. Look at the series expansion of W(x) which also seems like it may diverge. There probably is a certain radius of convergence like Gary suggested.

@RounakSarkar This is your best answer yet! $.305≈ \frac1e =.367…$

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