How to calculate $\int_{0}^{\infty} x^{-x} \,dx$? I am trying to solve this improper integral: $\int_{0}^{\infty} x^{-x} \,dx$.
First I replace de infinity by a another variable $y$, so:
$\int_{0}^{y} x^{-x} \,dx = \int_{0}^{y} e^{\ln(x^{-x})}\,dx = \int_{0}^{y} e^{-x\ln(x)}\,dx = \int_{0}^{y} \sum_{n=0}^{\infty} \frac{(-x\ln(x))^{n}}{n!}\,dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \int_{0}^{y} (x\ln(x))^n\, dx$.
Solving $\int (x\ln(x))^n\,dx$:

*

*Substitute $u=\ln(x) \implies \int u^ne^{(n+1)n}\,du $

*Substitute $v=u^{n+1} \implies \frac{1}{n+1}\int e^{(n+1)v^{\frac{1}{n+1}}}\,dv$

*Substitute $w=(n+1)^{(n+1)}v  \implies \int e^{(n+1)v^{\frac{1}{n+1}}}\,dv = (n+1)^{-n-1}\int e^{w^{\frac{1}{n+1}}}\,dw$

*$\int e^{w^{\frac{1}{n+1}}}\,dw = -(n+1)(-1)^n \operatorname{\Gamma}(n+1,-w^{\frac{1}{n+1}})$

*Undo substitutions, $\int (x\ln(x))^n\,dx = \dfrac{\left(n+1\right)^{-n-1}\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(x\right)\right)}{\left(-1\right)^n}$

*$\int_{0}^{y} (x\ln(x))^n\, dx  \implies \dfrac{\left(n+1\right)^{-n-1}\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(x\right)\right)}{\left(-1\right)^n}  \Big|_0^y$, when $x=0$, the incomplete gamma function will be evalueate between plus infinity and plus infinity so, $\int_{0}^{y} (x\ln(x))^n\, dx =  \dfrac{\left(n+1\right)^{-n-1}\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(y\right)\right)}{\left(-1\right)^n}$

*$\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \int_{0}^{y} (x\ln(x))^n\, dx = \sum_{n=0}^{\infty} \frac{\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(y\right)\right)}{n!(n+1)^{n+1}} =   \sum_{n=1}^{\infty}\frac{Q\left(n,\ -n\ln\left(y\right)\right)}{n^n}$.

Where $Q$ is the normalized or regularized incomplete gamma function
But what's happen when $y \to \infty$? The incomplete gamma function will be evaluated between zero and minus infinity, is it valid? Is there another way to find this value? Because de the improper integral converges.
 A: See this answer/my question:

Area under $x^{-x}$ over its real domain. What is another non-integral form of $$\int_{\Bbb R^+}x^{-x}dx$$?

In summary, here are the results:
$$\int_{\Bbb R^+}x^{-x}dx =\\ \lim_{x\to \infty}\sum_{n\ge1}\frac{Q(n,-nx)}{n^n}= \\\lim_{b,n\to \infty}\frac bn \\\sum_{k=0}^n\left(\frac{bk}{n}\right)^{-\left(\frac{bk}{n}\right)}=\\ \sum_{n\ge1}n^{-n}-\lim_{x\to\infty}\sum_{n\ge1}(-x)^n\  _1\mathrm{\tilde F}_1(n,n+1,nx) =\\ \sum_{n\ge 1}n^{-n}-\lim_{x\to \infty}\sum_{n\ge1}\sum_{k\ge1}\frac{(-x)^n(nx)^k}{(k+n)k!n!}= \\\lim_{x\to \infty}\sum_{n\ge 1}\sum_{k=0}^{n-1}\frac{(-1)^k e^{nx} n^{k-n}x^k}{k!}$$
Here is proof of the main result and @Nikos Bagis’s representation:
$$\int^{\infty}_{0}\frac{1}{t^t}\textrm{d}t=\sum_{n\geq 1}\frac{1}{n^n}-1+\int^{1}_{0}\frac{1}{t^{t^{t^{ \ldots}}}}\textrm{d}t $$
Where appears the $\,_1
\mathrm {\tilde F}_1(a,b,z) $ Confluent Regularized Hypergeometric function , Regularized Incomplete Gamma function $Q(a,z)$, and infinite tetration
Let’s see if we can use @Rounak Sarkar’s method from here to find a sum only answer.
Let me work on this to actually get some new results.
If you want here is a Mellin Transform representation:
$$\int_0^\infty x^{a-1-x}dx=\int_0^\infty x^{a-1}x^{-x} dx\implies \int_0^\infty x^{-x}dx=\mathcal{M}\{ x^{-x}\}(1)$$
I do not use Ramunajun’s Master Theorem much, but let’s try it:
$$\mathcal{M}\{f(x)\}(s)=\Gamma(s) y(-s), f(x)=\sum_{n=0}^\infty \frac{(-1)^n y(n)x^n}{n!} $$
So we need to find the “alternating” Exponential Generating function for $x^{-x}$
According to this theorem, then:
$$\int_0^\infty x^{-x} dx=\mathcal{M}\{ x^{-x}\}(1) =y(-1),x^{-x}=\sum_{n=0}^\infty\frac{(-1)^n y(n) x^n}{n!}$$
So how do we find $y(n)$?
Here is how to find $y(n)$:
The coefficients are just a Taylor series for $x^{-x}$ at $x=1$ which is just OEIS A176118
Therefore: $y(n)=(-1)^n \text A176118(n)$
And experimentally:
$$\int_0^\infty x^{-x} dx \mathop=^\text{natural}_\text{extension}-\text A176118(-1) $$
But this is stretching the definition. Maybe we can use an nth derivative formula to find a closed form for the nth derivative? Please correct me and give me feedback!
