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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 ?

Miss and Mister cassoulet char
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    The integral $\int_0^1 x^{-x}\text{d}x$ known as sophomore's dream was computed by Bernouilli and numerically is $1.291285997\dots$. The full integral is quite close to 2 (numerical value $1.995455958\dots$). – Davius Jul 10 '22 at 11:58
  • @Davius Thanks for the comment . – Miss and Mister cassoulet char Jul 10 '22 at 14:33
  • A rough upper bound is given by $x^{-x}\le e^{1-x}$ giving $\int_0^\infty x^{-x}dx \le e$ – Joshua Tilley Jul 10 '22 at 15:58
  • For the exact upper bound of 2, why not just numerically compute the integral $\int_0^bx^{-x}dx$ up to some finite bound $b$, and then use this together with $\int_b^\infty x^{-x}dx \le \int_b^\infty e^{1-x} dx=e^{1-b}$ – Joshua Tilley Jul 10 '22 at 16:00

1 Answers1

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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 $a\geq 100$ :

$$h\left(x\right)=f\left(x\right)-x^{-x}\geq 0$$

Miss and Mister cassoulet char
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