Which is entity is greater and by how much ?

$\large \displaystyle A = \int_0^1 x^x dx$

$\large \displaystyle A = \int_0^1 e^{x \ln(x) } dx$

$\large \displaystyle A = \int_0^1 \sum_{k=0}^{\infty}\frac{ [x \ln(x)]^k}{k!} dx$

$\large \displaystyle A = \sum_{k=0}^{\infty} \int_0^1 \frac{ [x \ln(x)]^k}{k!} dx$

Make $u = -\ln(x)$ , $du = -\frac1x dx \rightarrow dx = -e^{-u} du$

$\large \displaystyle A = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \int_0^{\infty} e^{-u(k+1)} u^k du$

$\large \displaystyle A = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! \cdot (k+1)^k} \int_0^{\infty} e^{-u(k+1)} [u(k+1)]^k du$

Make $t = u(k+1)$ , $dt = du(k+1)$

$\large \displaystyle A = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! \cdot (k+1) \cdot (k+1)^k} \int_0^{\infty} e^{-t} t^k dt$

$\large \displaystyle A = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! \cdot (k+1)^{k+1}} \Gamma(k+1)$

$\large \displaystyle A = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! \cdot (k+1)^{k+1}} k!$

Make $n = k+1$

$\color{#20A900} \boxed{ \large \displaystyle A = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^n} }$

Likewise:

$\large \displaystyle B = \int_0^1 x^{-x} dx$

$\large \displaystyle B = \int_0^1 e^{-x \ln(x) } dx$

$\large \displaystyle B = \int_0^1 \sum_{k=0}^{\infty} \frac{[-x \ln(x)]^k}{k!} dx$

$\large \displaystyle B = \sum_{k=0}^{\infty} \int_0^1 \frac{ [-x \ln(x)]^k}{k!} dx$

$\large \displaystyle B = \sum_{k=0}^{\infty} (-1)^k \int_0^1 \frac{ [x \ln(x)]^k}{k!} dx$

We already know this integral from $A$ :

$\large \displaystyle B = \sum_{k=0}^{\infty} (-1)^k \frac{(-1)^k}{(k+1)^{k+1}}$

$\large \displaystyle B = \sum_{k=0}^{\infty} \frac{1}{(k+1)^{k+1}}$

Make $n = k+1$

$\color{#20A900} \boxed{ \large \displaystyle B = \sum_{n=1}^{\infty} \frac{1}{n^n} }$

So, $A-B$ will cancel out each term of odd $n$ , remaining only the terms of even $n$ :

$\large \displaystyle |A-B| = \left | \sum_{n=1}^{\infty} \frac{-2}{(2n)^{2n}} \right |$

$\large \displaystyle |A-B| = 2 \sum_{n=1}^{\infty} \frac{1}{(2n)^{2n}}$

This sum quickly converges to:

$\color{#20A900} \boxed{ \large \displaystyle |A-B| = 0.507855486 }$

Correct to $9$ decimal places. Thus:

$\color{#3D99F6} \boxed{ \large \displaystyle \left \lceil 10000 |A-B| \right \rceil = 5079 }$

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Proof of convergence of the series above:

$\large \displaystyle (2n)^{2n} \geq 2^{2n}$

$\large \displaystyle \frac{1}{(2n)^{2n}} \leq \frac{1}{2^{2n}}$

$\large \displaystyle \sum_{n=1}^{\infty} \frac{1}{(2n)^{2n}} < \sum_{n=1}^{\infty} \left ( \frac14 \right) ^n$

$\color{#20A900} \boxed{\large \displaystyle \sum_{n=1}^{\infty} \frac{1}{(2n)^{2n}} < \frac13 } \color{#333333} \ \ q.e.d.$