Evaluate the area of the moustache This is more a soft question to calculate :
Let $x\geq 0$ then we have
$$\int_{0}^{\infty}-\left(e^{-x^{2}}-x^{-x}\right)dx-\frac{10}{9}<0$$
My attempt :
We have the following inequalities :
For $1\leq x\leq 2$ :
$$e^{-\left(\frac{x^{0.5}\left(x-1\right)}{2^{0.5}}\right)\cdot2\ln\left(2\right)}\geq x^{-x}$$
For $2\leq x\leq 6$ :
$$e^{\left(-\left(x-1\right)^{\frac{92}{100}}x^{0.5}\cdot2^{0.5}\ln\left(2\right)\right)}\geq x^{-x}$$
And :
$$\int_{0}^{1} x^{-x}\ dx = \sum_{n=1}^{\infty} n^{-n}$$
I let you a picture :

Question :
How to show the first inequality ?
Thanks and have fun!
 A: We need to prove that
$$\int_0^\infty x^{-x}\, \mathrm{d} x
< \frac{1}{2}\sqrt{\pi} + \frac{10}{9}. \tag{1}$$
It is known that
$$\int_0^1 x^{-x}\, \mathrm{d} x = \sum_{n=1}^\infty \frac{1}{n^n}
\le \sum_{n = 1}^4 \frac{1}{n^n}
+ \sum_{n=5}^\infty \frac{1}{5^n} = \frac{5578603}{4320000}.$$
Also, we have
$$\int_5^\infty x^{-x}\, \mathrm{d} x = \int_5^\infty \mathrm{e}^{-x\ln x}\, \mathrm{d} x \le \int_5^\infty \mathrm{e}^{-x\ln 5}\, \mathrm{d} x = \frac{1}{3125\ln 5}.$$
By Facts 1-4 (given later), we have
\begin{align*}
 &\int_1^2 x^{-x}\, \mathrm{d} x
 \le \int_0^1 \frac{3 - x}{x^2 - x + 2}\, \mathrm{d} x = \frac{5}{\sqrt{7}}\arctan \frac{\sqrt{7}}{5} - \frac{1}{2}\ln 2,\\
 &\int_2^3 x^{-x}\, \mathrm{d} x \le
 \frac{21}{2\mathrm{e}} + \frac{1001}{72}
 - \frac{1903 + 1518\mathrm{e}}{6\mathrm{e}}\ln 3
 + \frac{2927 + 2286\mathrm{e}}{6\mathrm{e}}\ln 2, \\
 &\int_3^4 x^{-x}\, \mathrm{d} x \le \frac{3\mathrm{e} - 1}{81\mathrm{e}\ln 3 + 81 \mathrm{e}}, \\
 &\int_4^5 x^{-x}\, \mathrm{d} x \le
 \frac{4\mathrm{e} - 1}{2048\mathrm{e}\ln 2 + 1024\mathrm{e}}.
\end{align*}
With these facts, it is easy to verify (1).
We are done.

Fact 1: $x^{-x} \le \frac{3 - x}{x^2 - x + 2}$ for all $x \in [1, 2]$.
Fact 2: For all $x \in [2, 3]$,
$$x^{-x} \le \frac{x}{6\mathrm{e}} -  \frac{19 + \mathrm{e}}{6\mathrm{e}} - \frac{293 + 250\mathrm{e}}{2\mathrm{e}x}
+ \frac{213\mathrm{e} + 159}{2\mathrm{e}x^2} - \frac{53}{x^3} + \frac{384\mathrm{e} + 512}{3\mathrm{e}(1 + x)}.$$
Fact 3: For all $x \in [3, 4]$,
$$x^{-x} \le 3^{-x} \mathrm{e}^{-x + 3}.$$
Fact 4: For all $x \in [4, 5]$,
$$x^{-x} \le 4^{-x} \mathrm{e}^{-x + 4}.$$
Remarks: Actually, $x^{-x} \le a^{-x} \mathrm{e}^{-x + a}$ for all $x, a > 0$
which is equivalent to $\ln u \ge 1 - \frac{1}{u}$ for all $u > 0$.
