7
$\begingroup$

A few years ago, I got interested in an apparently hard integration problem which had me fascinated. This was the integral of a Sophomore Dream like integral except with the bounds over the real positive numbers denoted by $\Bbb R^+$ and not from 0 to 1 over the unit square, or unit line in this case? Nevertheless, this quirky problem caught my attention for having its area be almost 2 with this graph. Here is my solution, although there are a couple of calculations that I could have done differently: enter image description here $$ A\mathit =\mathrm{\int_{\Bbb R^+}x^{-x}dx=\int_0^{\infty}\sum_{n=0}^{\infty}\frac{(-1)^nx^n ln^n(x)}{n!}dx}$$

I naively used Wolfram Alpha to integrate, but this can easily be still done with the substitution of $x=e^{-y}$ which is similar to the other Sophomore Dream integrals, so it will be ignored here for conciseness and should be able to use an interchanging theorem. Here I chose not to do the substitution as the same result appears anyways and this is simpler:

$$\mathrm{\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\int_0^{\infty}x^n ln^n(x)dx= \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}(-x^{-n}x^nln^n(x)ln(x)ln^{-(n+1)}(x)(-1)^{-(n+1)}(n+1)^{-(n+1)}Γ\big(n+1,-(n+1)ln(x)\big)\big|_0^{\infty}= \sum_{n=0}^{\infty}\frac{Γ\big(n+1,-(n+1)ln(\infty)\big)}{Γ(n+1)(n+1)^{n+1}}}$$

This -$\infty$ in the gamma function argument does not seem to exist, but taking a limit helps as well as some other forms with the Regularized Incomplete Gamma Function: $$A=\mathrm{\sum_{n=1}^{\infty}n^{-n}\frac{Γ(n,-\infty n)}{Γ(n)}=\lim_{x\to \infty}\sum_{n\ge1}\frac{Q(n,-nx)}{n^n}=1.99545595…}$$

As seen from the link, the index n has to be a natural number which is true here. Also, there are other forms with the exponential sum function, and other series representations if non-elementary functions are not allowed. Here is proof of this answer from Wolfram Alpha.

Please do not give me any integral expressions of this constant, but rather special functions or other representations are encouraged. I will “check” the best answer. I just need a “nice” non-integral expression of the constant. Please give me feedback and correct me!

An alternate form solution is very needed for this problem also concerning power tower, so please check it out. I have left my context, attempts, and questions, therefore any help with alternate forms for either of these would be appreciated.

Using the definition of the Riemann Sum, we can rewrite this as:

$$\mathrm{\int_{\Bbb R^+}x^{-x}dx=\lim_{b,n\to \infty}\frac bn \sum_{k=0}^n\left(\frac{bk}{n}\right)^{-\left(\frac{bk}{n}\right)},n\gg b}$$

Here is proof of this result: Graph

Here are some unlikely closed forms.

There is also a form of the constant using a summation expansion. Here is proof of a third form of this constant here: $$\mathrm{A= \lim_{x\to \infty}\sum_{n\ge 1}\sum_{k=0}^{n-1}\frac{(-1)^k e^{nx} n^{k-n}x^k}{k!}}$$

Here is another sum and double sum answer here. This has a sophomore dream in the solution plus the same integrand from 1 to $\infty$: $$\mathrm{A=\sum_{n\ge 1}n^{-n}-\lim_{x\to \infty}\sum_{n\ge1}\sum_{k\ge1}\frac{(-x)^n(nx)^k}{(k+n)k!n!}}$$

Let $\bar F$ represent a Regularized Hypergeometric function then, as $\mathrm{Q(n,-nx)= 1-(-nx)^n\ _1\bar F_1(n,n+1,nx)}$, the following can be proven with the Kummer Confluent Hypergeometric Function: $$\mathrm{\sum_{n\ge1}n^{-n}-\lim_{x\to\infty}\sum_{n\ge1}(-x)^n\ _1\bar F_1(n,n+1,nx)}$$

Also, feel free to use these identities to find an alternate form. I could go on, but I have found no form that does not need a limit in the answer.

$\endgroup$
12
  • 7
    $\begingroup$ It's not a competition to see who can use the most fonts in one post. $\endgroup$ May 20, 2021 at 13:05
  • 1
    $\begingroup$ @GerryMyerson it is to remind people that the links are relevant. There is just 2 words in one “frakture” font. Feel free to click the link! $\endgroup$ May 20, 2021 at 14:18
  • 3
    $\begingroup$ There is a funny-looking A. There is a display that's about ten feet high. There is boldface. There is Fraktur in blue. It's too busy (for my taste). $\endgroup$ May 20, 2021 at 23:05
  • 2
    $\begingroup$ @GerryMyerson Do you have any alternate forms for this constant? You seem very talented and I would love to see some possible closed forms of this constant, although a closed form is optional as this does not seem to have one. You can also click the fraktur link, that is the default link color. Please no integral defined forms as we are trying to evaluate this integral, and find the area more importantly! $\endgroup$ May 20, 2021 at 23:10
  • 2
    $\begingroup$ Sorry, I don't know anything more than what's at the oeis. I encourage you to follow the links there. $\endgroup$ May 21, 2021 at 2:48

2 Answers 2

3
$\begingroup$

From : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function $$\text{Closed form : }\quad\int_0^\infty x^{-x}dx=\text{Sphd}(-1;\infty)$$ $$\text{Numerical value : Eq.(8:3).}$$

Before downvoting, please read "A schoolboy joke" pp.2-3 in the above referenced paper.

$\endgroup$
11
  • 2
    $\begingroup$ If you can prove this function is as popular as the normal gamma function or exponential sum function, ie it is used and accepted in the math community as a well known function, then I will consider more than an upvote. I like the answer, but this seems to be another case of rename the integral as the answer. Maybe rewrite the function as the sum definition and see if a new interesting result pops up? $\endgroup$ Jun 21, 2021 at 11:48
  • 2
    $\begingroup$ Of course my "answer" is more a joke that an academic answer since the Sphd function is not a standardized special function and it's clear this will not happen soon. $\endgroup$
    – JJacquelin
    Jun 21, 2021 at 13:05
  • 1
    $\begingroup$ I see you and Claude Leibovici both love jokes. Any other ideas on how to find a form not as an integral? It does not have to be a closed form, just evaluation of the antiderivative. $\endgroup$ Jun 21, 2021 at 13:08
  • $\begingroup$ Wait, you wrote this article on Scribd? $\endgroup$ Jun 21, 2021 at 13:14
  • $\begingroup$ Right. On Scribd : fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . $\endgroup$
    – JJacquelin
    Jun 21, 2021 at 15:22
2
+50
$\begingroup$

This is a note. I will try to find more in the near future. $$ I=\int^{\infty}_{0}x^{-x}dx=\int^{\infty}_{1}e^{-x\log x}dx+\int^{1}_{0}x^{-x}dx. $$ Using $\sum_{n\geq 1}n^{-n}=\int^{1}_{0}x^{-x}dx$ and making the change of variable $x\rightarrow e^{W(x)}$ where $W(x)$ is the well known Lambert function, we get $$ I=\sum_{n\geq 1}n^{-n}+\int^{\infty}_{0}\exp\left(-e^{W(x)}W(x)\right)e^{W(x)}W'(x)dx= $$ $$ =\sum_{n\geq 1}n^{-n}+\int^{\infty}_{0}e^{-x}e^{W(x)}W'(x)dx=\sum_{n\geq 1}n^{-n}+\int^{\infty}_{0}e^{-x}d\left(e^{W(x)}\right)= $$ $$ \textrm{ integration by parts } $$ $$ =\sum_{n\geq 1}n^{-n}-1+\int^{\infty}_{0}e^{-x}e^{W(x)}dx=\sum_{n\geq 1}n^{-n}-1+\int^{\infty}_{0}e^{-x}\frac{x}{W(x)}dx $$ If we make the change of variable $x\rightarrow -\log t$ we arrive easily to $$ I=\sum_{n\geq 1}n^{-n}-1+\int^{1}_{0}t\frac{-\log t}{W(-\log t)}\frac{-dt}{t}=\sum_{n\geq 1}n^{-n}-1+\int^{1}_{0}\frac{-\log t}{W(-\log t)}dt. $$ But the function $h(z)=\frac{W(-\log z)}{-\log z}$, is the analytic continuation of the Euler's iterated exponential function $z^{z^{z^{\ldots}}}$. Hence finaly we can write $$ \int^{\infty}_{0}\frac{1}{x^x}dx=\sum_{n\geq 1}\frac{1}{n^n}-1+\int^{1}_{0}\frac{1}{t^{t^{t^{ \ldots}}}}dt. $$ In the sense $z^{z^{z^{\ldots}}}=h(z)$.

REVISED.

Assume that $W(t)$ is the well known Lambert function. Then for $k=0,1,2,\ldots$, we have $$ \int^{M}_{0}\frac{t^k}{1+W(t)}dt=\frac{(-1)^{k+1}k!}{(k+1)^{k+1}}+ $$ $$ +\frac{M^{k+1} k!}{W(M)^{k+1}(k+1)^{k+1}}\sum_{0\leq j\leq k}(-1)^{j+k}\frac{(k+1)^jW(M)^j}{j!}. $$ Hence $$ \int^{M}_{0}\frac{e^{-t}}{1+W(t)}dt=-\sum^{\infty}_{k=0}\frac{1}{(k+1)^{k+1}}+ $$ $$ +\sum^{\infty}_{k=0}\frac{e^{-(k+1)W(M)} M^{k+1}W(M)^{-k-1}\Gamma(k+1,-(k+1)W(M))}{(k+1)^{k}(k+1)!}= $$ $$ =-\sum_{k\geq 1}k^{-k}+\sum_{k\geq 1}\frac{\Gamma(k,-kW(M))}{k^{k-1}k!},\tag 1 $$ where $$ \Gamma(a,z)=\int^{\infty}_{z}e^{-t}t^{a-1}dt $$ is the incomplete Gamma function. But one can see with the help of (1) easily that $$ \int^{M}_{1}x^{-x}dx=\int^{M\log M}_{0}\frac{e^{-t}}{1+W(t)}dt=-\sum^{\infty}_{k=1}k^{-k}+\sum^{\infty}_{k=1}\frac{\Gamma(k,-kW(M\log M))}{(k!) k^{k-1}}. $$ Hence using $\int^{1}_{0}x^{-x}dx=\sum^{\infty}_{k=1}k^{-k}$, we get $$ \int^{M}_{0}x^{-x}dx=\sum^{\infty}_{k=1}\frac{\Gamma(k,-k\log M)}{(k!) k^{k-1}}\textrm{, }\forall M>0.\tag 2 $$

$\endgroup$
7
  • $\begingroup$ I have been trying to find a summation expansion for the integral, but the Lagrange Inversion Theorem gave multiple summation results for the sums. One of them was an expansion for exp(W(-lnx))=$\frac1{h(z)}$ from exp(W(y)), but this came up with a bad result. Finding the reciprocal W(x) expansion left me with part of the answer as this for $g_n$ as in the Wikipedia link. I forgot to write it as $\frac 1{h(x)}$ in the integral. $\endgroup$ Jun 22, 2021 at 20:13
  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Lambert_W_function#Identities at the end of the section and deffinition of the W-Lambert function. $\endgroup$ Jun 22, 2021 at 20:20
  • $\begingroup$ The function $h(x)=x^{x^{x^{\ldots}}}$ converges in $\textbf{R}$ only for $e^{-e}\leq x\leq e^{1/e}$. That is why I use analytic continuation. Actualy $g(x)=W(-\log x)/(-\log x)$ coinsides with $h(x)$ at $[e^{-e},e^{1/e}]$, but $g(x)$ is defined in all $x\neq 1$. $\endgroup$ Jun 22, 2021 at 20:45
  • $\begingroup$ I did something similar here trying to find the area under h(x) instead from 0 to $e^{1/e}$. I am very close to finding the answer, but I kind of had a similar problem as you with as the interval of convergence of the series I used failed to find the full area. Problem here $\endgroup$ Jun 22, 2021 at 20:51
  • $\begingroup$ Nice! I will check this soon enough. You can simplify by using the regularized gamma function. Please evaluate the final answer with M=$\infty$. Thanks. You will see you get the same result I did with a limit unfortunately as you will have to make the second argument of the incomplete gamma function be: -$\infty$k. $\endgroup$ Jul 12, 2021 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.