Improving the result $\int_0^\infty\frac1{x^x}\, dx<2.$ Problem/Conjecture :
Let $a\geq 1$ be a real number then it seems we have :
$$\int_{0}^{\infty}x^{-x}dx<\int_{0}^{\infty}\left(h^{-\frac{1}{2}h^{\int_{0}^{1}y^{y^{e^{ah}}}dy}}\cdot h^{-\frac{1}{2}h^{\int_{0}^{1}y^{-y^{e^{ah}}}dy}}\right)dh<2$$
If my conjecture is true it refine the problem Prove that $\int_0^\infty\frac1{x^x}\, dx<2$ . Where I gives some materials to show it .
Some remarks :
It seems for $a=1$ is the maximum and as $a\to \infty$ we get the lower bound :
Warning :
This problem is not so trivial because we doesn't have the inequality for $x>0$ :
$$x^{-\frac{1}{2}x^{\int_{0}^{1}y^{y^{e^{ax}}}dy}}\cdot x^{-\frac{1}{2}x^{\int_{0}^{1}y^{-y^{e^{ax}}}dy}}\ge x^{-x}$$
We don't have the inequality above but it seems we have for $x>0$
$$f\left(x\right)-x^{-x}+e^{-x}-e^{-1.01x}>0$$
Where :
$$f\left(x\right)=x^{-\frac{1}{2}x^{\int_{0}^{1}y^{y^{e^{x}}}dy}}x^{-\frac{1}{2}x^{\int_{0}^{1}y^{-y^{e^{x}}}dy}}$$
This problem seems very distinct .
How to (dis)prove it ?
 A: Too long for a comment :
We have the inequality $a=1$:
$$\int_{0}^{\infty}\left(f\left(x\right)-g\left(x\right)\right)dx<0$$
Where :
$$f\left(x\right)=x^{-\frac{1}{2}x^{1.1\int_{0}^{1}y^{y^{e^{ax}}}dy}}\cdot x^{-\frac{1}{2}x^{0.9\int_{0}^{1}y^{-y^{e^{ax}}}dy}}$$
And :
$$g\left(x\right)=x^{-\frac{1}{2}x^{\int_{0}^{1}y^{y^{e^{x}}}dy}}\cdot x^{-\frac{1}{2}x^{\int_{0}^{1}y^{-y^{e^{x}}}dy}}$$
On the other hand  it seems we have for $0<x\leq \sqrt{3}$ and $a=1$:
$$x^{-x}\leq x^{-\frac{1}{2}x^{1.1\int_{0}^{1}y^{y^{e^{ax}}}dy}}\cdot x^{-\frac{1}{2}x^{0.9\int_{0}^{1}y^{-y^{e^{ax}}}dy}}$$
Some tought for the general case :
Conjecture :
It seems we have, $a\ge 100$ :
$$\int_{0}^{1}\left(f\left(x\right)-g\left(x\right)\right)dx<0$$
where :
$$f\left(x\right)=x^{-\frac{1}{2}x^{\left(1+\frac{1}{a^{2}}\right)\int_{0}^{1}y^{y^{e^{ax}}}dy}}\cdot x^{-\frac{1}{2}x^{\left(1-\frac{1}{a^{2}}\right)\int_{0}^{1}y^{-y^{e^{ax}}}dy}},g(x)=x^{-\frac{1}{2}x^{\int_{0}^{1}y^{y^{e^{ax}}}dy}}\cdot x^{-\frac{1}{2}x^{\int_{0}^{1}y^{-y^{e^{ax}}}dy}}$$
And for $0.1<x\leq 1$ and $a\geq 100$ :
$$h\left(x\right)=f\left(x\right)-x^{-x}\geq 0$$


