How to calculate $\int_{0}^{\infty} x^{-x} \,dx$?

Question

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 $\int (x\ln(x))^n\,dx$:

1. Substitute $u=\ln(x) \implies \int u^ne^{(n+1)n}\,du $
2. Substitute $v=u^{n+1} \implies \frac{1}{n+1}\int e^{(n+1)v^{\frac{1}{n+1}}}\,dv$
3. Substitute $w=(n+1)^{(n+1)}v \implies \int e^{(n+1)v^{\frac{1}{n+1}}}\,dv = (n+1)^{-n-1}\int e^{w^{\frac{1}{n+1}}}\,dw$
4. $\int e^{w^{\frac{1}{n+1}}}\,dw = -(n+1)(-1)^n \operatorname{\Gamma}(n+1,-w^{\frac{1}{n+1}})$
5. Undo substitutions, $\int (x\ln(x))^n\,dx = \dfrac{\left(n+1\right)^{-n-1}\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(x\right)\right)}{\left(-1\right)^n}$
6. $\int_{0}^{y} (x\ln(x))^n\, dx \implies \dfrac{\left(n+1\right)^{-n-1}\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(x\right)\right)}{\left(-1\right)^n} \Big|_0^y$, when $x=0$, the incomplete gamma function will be evalueate between plus infinity and plus infinity so, $\int_{0}^{y} (x\ln(x))^n\, dx = \dfrac{\left(n+1\right)^{-n-1}\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(y\right)\right)}{\left(-1\right)^n}$
7. $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \int_{0}^{y} (x\ln(x))^n\, dx = \sum_{n=0}^{\infty} \frac{\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(y\right)\right)}{n!(n+1)^{n+1}} = \sum_{n=1}^{\infty}\frac{Q\left(n,\ -n\ln\left(y\right)\right)}{n^n}$.

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.

Answer 1

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!

Answer 2

Using the known identities $$\int\limits_0^1\ln^n z\,\text dz = (-1)^n n!,\tag1$$ $$\sum\limits_{j=0}^k \dfrac{a^m}{m!}=e^a\, \dfrac{\Gamma(k+1,a)}{\Gamma(k+1)},\tag2$$ one can get $$I_1(a)=\int\limits_0^a x^{-x}\text dx = a \int\limits_0^1 (ay)^{-ay} \text dy = a\int\limits_0^1 e^{-ay\ln a} e^{-ay\ln y}\text dy$$ $$= \sum\limits_{m=0}^\infty \dfrac{\ln^m a}{m!}\sum\limits_{n=0}^\infty (-1)^{m+n} \dfrac {a^{m+n+1}}{n!}\int\limits_0^1 y^{m+n}\ln^n y\,\text dy$$ $$= \sum\limits_{m=0}^\infty \dfrac{\ln^m a}{m!}\sum\limits_{n=0}^\infty (-1)^{m+n} \dfrac {a^{m+n+1}}{n!(m+n+1)^{n+1}}\int\limits_0^1 \ln^n (y^{m+n+1})\,\text dy^{m+n+1}$$ $$= \sum\limits_{m=0}^\infty \dfrac{\ln^m a}{m!}\sum\limits_{n=0}^\infty (-1)^{m+n} \dfrac {a^{m+n+1}}{n!(m+n+1)^{n+1}}\cdot(-1)^n n!$$ $$= \sum\limits_{m=0}^\infty (-1)^m\dfrac{\ln^m a}{m!}\sum\limits_{n=1}^\infty \dfrac {a^{m+n}}{(m+n)^n} = \sum\limits_{s=1}^\infty \dfrac{a^s}{s^s}\sum\limits_{m=0}^{s-1} \dfrac{(-s\ln a)^m}{m!}$$ $$I_1(a)= \sum\limits_{s=1}^\infty \dfrac{\Gamma(s,-s \ln a)}{s^s\Gamma(s)}.\tag3$$

On the other hand,

$$I_2(a)=\int\limits_a^\infty x^{-x}\text dx = a \int\limits_0^\infty (a(1+y))^{-a-ay}\text dy = a^{1-a}\int\limits_0^\infty \dfrac{a^{-ay}}{(1+y)^a} (1+y)^{-ay}\,\text dy.$$ Let $\;z=a^{-ay} = e^{-(a\ln a)y},\;$ then $\;y=-\dfrac{\ln z}{a\ln a},\;$ $$(1+y)^{-ay}=\left(1-\dfrac{\log_a z}a\right)^{\log_a z} =1 - \dfrac{(z-1)^2}{a\ln^2a}+\dfrac{(z-1)^3(2a\ln a-1)}{2a^2\ln^3a} +\dots =\dfrac{(2a^2 \ln^3a-4a\ln a+1) + z(10a\ln a-3) + z^2 (3-8a\ln a) + z^3(2a\ln a-1)}{2a^2\ln^3 a}+\dots,$$

Taking in account the integral $$\int\limits_0^\infty\dfrac{a^{-kay}}{(1+y)^a} \,\text dy =a^{ka}\operatorname E_a(ka\ln a),\tag4$$ one can get $$\begin{align} &I_2(a)\approx \dfrac1{2a\ln^3 a}\big((2a^2\ln^3a-4a\ln a+1) E_a(a\ln a) + a^a(10a\ln a-3) E_a(2 a\ln a)\\[4pt] & - a^{2a}(8\ln a-3) E_a(3a\ln a) + a^{3a}(2a\ln a-1) E_a(4a\ln a)\big). \end{align}\tag5$$ Approximation $(5)$ is not too accurate (accuracy $\;1.5\%\;$ for $\;a=e^2\;$), because only three terms of the primary series was used. However, it shows the effectiveness of the used approach.

If $\;a=e^3,\;$ then $I_1(a)\approx 1.99545\,59575\,00138\,00041\,87246\,96764\,22,$

$I_2(a)\approx 1.69475\cdot10^{-27},$

$\;I_1(a)+I_2(a)\approx 1.99545\,59575\,00138\,00041\,87246\,98458\,97.$

Numeric value is $I=1.99545\,59575\,00138\,00041\,87246\,98452\,72.$

Easily to see that the second integral increases the result accuracy.

$\color{green}{\mathbf{Edition\ of\ 23.12.21.}}$

The alternative way is

$$I_2(a)=\int\limits_a^\infty x^{-x}\text dx = a \int\limits_0^\infty (a(1+y))^{-a-ay}\text dy = a^{1-a}\int\limits_0^\infty a^{-ay}(1+y)^{-a-ay}\,\text dy.$$ Let $\;z=a^{-ay} = e^{-(a\ln a)y},\;$ then $\;y=-\dfrac{\ln z}{a\ln a},\;$ $$I_2(a)= \dfrac1{a^a\ln a}\int\limits_0^1 \left(1-\dfrac{\log_a z}a\right)^{-a+\log_a z}\text dz =\sum\limits_{n=0}^\infty c_n\ln^n z,$$ $$I_2(a) = \sum\limits_{n=0}^\infty (-1)^n n! c_n,\tag6$$ where $$\sum\limits_{n=0}^\infty c_n y^n = \dfrac 1{a^a\ln a} \left(1-\dfrac y{a \ln a}\right)^{\large-a+\frac y{\ln a}}.\tag7$$

By the direct numeric calculations, $$I_2(e^3)\approx 1.68850\,13010\,68575\,08677\,67860\,60236\,26323\,88582\cdot10^{-27}.$$ Calculations by $(6)-(7)$ by $20$-term model give $$I_2(e^3)\approx 1.68850\,13010\,38888\,76584\cdot 10^{-27},$$ i.e. ten additional correct digits of the result.