Can this infinite series be simplified down any further (Trying to find the definite integral of $x^{-x}$ from 0 to infinity)

Question

I was trying to evaluate the integral from 0 to infinity of $x^{-x}$, but I got stuck at a double infinite summation. I know that the integral of said function is non-elementary, but perhaps there is a way to write this double sum in a better way.

Before I begin, I know that $0^0$ is undefined, which is why I, if I can even get that far, will be taking the limit for $\infty$ and $0$.

I've noticed a different post made roughly 5 years ago, in which the same question has been attempted to be answered, and it ended with the use of the Lambert W function (product log). I want to make clear that I will not be heading in the direction that he / she / it went. That question also does not answer the question in this post.

That being said, my first challenge was to rewrite the formula, and then find an indefinite integral. Here is what I tried:

$$d(x)=x^{-x}=$$

$$e^{-x\ln(x)}=$$

$$\sum_{n=0}^\infty\frac{(-x\ln(x))^n}{n!}$$

For clearance, I'll define some functions:

$$d(x)=x^{-x}$$

$$d_{0}(x)=\frac{(-x\ln(x))^0}{0!}=1$$

$$d_{1}(x)=\frac{(-x\ln(x))^1}{1!}=-x\ln(x)$$

$$d_{2}(x)=\frac{(-x\ln(x))^2}{2!}=\frac{x^{2}\ln(x)^2}{2!}$$

$$d_{n}(x)=\frac{(-x\ln(x))^n}{n!}$$

Then, I integrated the infinite summation (making great use of the sum rule), and after some pattern seeking I found a pattern:

$$D_{1}(x)=\int-x\ln(x)dx=\frac{x^2}{2^2}\left(1-\frac{2}{1!}\ln(x)\right)$$

$$D_{2}(x)=\int\frac{(-x\ln(x))^2}{2!}dx=\frac{x^3}{3^3}\left(1-\frac{3}{2!}\ln(x)+\frac{3^2}{2!}\ln(x)^2\right)$$

$$D_{3}(x)=\int\frac{(-x\ln(x))^3}{3!}dx=\frac{x^4}{4^4}\left(1-\frac{4}{3!}\ln(x)+\frac{4^2}{3!}\ln(x)^2-\frac{4^3}{3!}\ln(x)^3\right)$$

$$D_{n}(x)=\int\frac{(-x\ln(x))^n}{n!}dx=\left(\frac{x}{n+1}\right)^{n+1}\sum_{k=1}^{n}\frac{(-(n+1)\ln(x))^k}{k!}$$

I will refer to this generalisation (in this document) as Not A Dolphin's conjecture.

To be sure that this pattern held for all non-negative integers, I tried to differentiate it, but without succes. WolframAlpha however, did manage to differentiate it and confirmed that does hold for all of these values.

Now, I plugged it into an infinte sum and got this as my (temporary) final answer:

$$\int\sum_{n=0}^\infty \frac{(-x\ln(x))^n}{n!}dx=$$

$$\sum_{n=0}^\infty\int\frac{(-x\ln(x))^n}{n!}dx=$$

$$\sum_{n=0}^\infty D_{n}(x)=$$

$$\sum_{n=0}^{\infty}\left(\frac{x}{n+1}\right)^{n+1}\sum_{k=0}^{n}\frac{(-(n+1)\ln(x))^k}{k!}=$$

$$\sum_{n=1}^{\infty}\left(\frac{x}{n}\right)^{n}\sum_{k=0}^{n-1}\frac{(-n\ln(x))^k}{k!}$$

I am currently trying to simplify this expression, but without much succes. I am just a 15-year old with an interest for math. So, if anyone knows a way to simplify this, please let me know!

Answer 1

I could not find a way to simplify the two sums but I thought of 2 ideas you might have missed if you are not necessarily looking for the exact value ,

(1) $$L=\int_0^1x^{-x} dx=\sum_{n=1}^\infty n^{-n}$$ Then $$\int^{\infty}_0 x^{-x} =L+\int_1^{\infty}x^{-x}dx$$

This second integral can be easily approximated like so :

$$\int_1^{\infty}x^{-x}dx \approx \sum ^{\infty}_{n=1}n^{-n}-\frac{1}{2} $$

Thus, we have

$$\int^\infty_0x^{-x}dx \approx2L-\frac{1}{2}\approx2.08$$

(2)

$$\int\limits_0^\infty x^{-x}dx=\int\limits_0^1 x^{-x}dx+\int\limits_0^1 x^{-2+1/x}dx$$

Since it is difficult to develop the second integral into a series we can use the solution $z_1$ for $\int_0^1(x^{zx}-x^{-2+\frac{1}{x}})dx=0$

For $z=z_1$

$$ \int\limits_0^\infty x^{-x}dx\approx\sum\limits_{n=1}^\infty \frac{1+(-z_1)^{n-1}}{n^n}$$

With $z_1\approx 1.45$

I count the second way as an approximation because I'm using an approximation of $z_1$. If you can somehow plug the exact value of $z_1$ then the integral is exactly equal to the sum.