It's an inequality I found nice let me propose it :
Let $0\leq x$ then we have :
$$\int_{0}^{\infty}\sin\left(x^{-x}\right)dx<\phi=\frac{\left(1+\sqrt{5}\right)}{2}\quad (I)$$
My attempt :
First of all I recall two facts :
Fact 1
$$\int_{0}^{1} x^{-x}\ dx = \sum_{n=1}^{\infty} n^{-n}$$
Fact 2:
Let $0\leq x\leq \frac{\pi}{2}$ then we have :
$$\sin\left(x\right)\leq x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}$$
Unfortunately it works on $J=[0,1]$ not on $[0,\infty]$ so see The "natural" Sophomore's Dream integral: $\int_{0}^{\infty} x^{-x}\ dx$ wich is a useful link .
Edit : for $x\in(1,\infty)$ it seems we have :
$$\left(e^{\left(-\frac{e}{\left(e-1\right)^{2}}\left(\left(\ln\left(x^{2}+e-1\right)\right)-e^{-1}\right)^{2}\right)}e^{\left(e^{-\left(1\cdot0.85+0.15\cdot\frac{1}{x}\right)}\right)}\right)\geq x^{-x}\quad(G)$$
Edit 2:
To show $(G)$ we first take the logarihthm on both side and then use the very well know expansion via Taylor's series of logarithm and exponential .Currently I don't see better . Then we have :
$$\int_{0}^{2}\sin\left(x^{-x}\right)dx+\int_{2}^{\infty}\left(e^{\left(-\frac{e}{\left(e-1\right)^{2}}\left(\ln\left(x^{2}+e-1\right)-e^{-1}\right)^{2}\right)}e^{e^{-\left(0.85+\frac{0.15}{x}\right)}}\right)dx<\frac{\left(1+\sqrt{5}\right)}{2}$$
Edit 3 : It seems we have on $(1,2)$ :
$$e^{-\left(\frac{x^{0.48}\left(x-1\right)}{2^{0.48}}\right)\cdot2\ln\left(2\right)}>x^{-x}$$
Edit 4:
It seems we have on $(0,1)$ :
$$e^{\left(1.02-0.02\cdot\frac{2x}{x+1}-x\right)\cdot2^{0.6}x^{0.6}\ln\left(2\right)}> x^{-x}$$
And :
For $2\leq x\leq 6$ :
$$e^{\left(-\left(x-1\right)^{\frac{93}{100}}x^{0.5}\cdot2^{0.5}\ln\left(2\right)\right)}\geq x^{-x}$$
Edit 5:
Following an answer due to user Michael Rozenberg here ( Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ ) we have for $0<x<1$ :
$$e^{-x\left(x-1\right)\left(\frac{2}{x^{2}+x}\right)^{\frac{1}{3}}}> x^{-x}$$
Question :
How to show the inequality $(I)$ ?
Best Answer
We give the following auxiliary results.
Fact 1: $\sin u \le u - \frac{1}{6}u^3 + \frac{1}{120}u^5$ for all $u \ge 0$.
Fact 2: $x^{-x} \le \frac{3 - x}{x^2 - x + 2} \le 1$ for all $x \in [1, 2]$.
Fact 3: $x^{-x} \le a^{-x} \mathrm{e}^{-x + a}$ for all $x, a > 0$.
(Proof: It is equivalent to $\ln u \ge 1 - \frac{1}{u}$ for all $u > 0$.)
Using Fact 1, we have (cf. Sophomore's dream) \begin{align*} &\int_0^1 \sin (x^{-x})\,\mathrm{d}x \\ \le\,& \int_0^1 \left(x^{-x} - \frac16 x^{-3x} + \frac{1}{120}x^{-5x}\right)\mathrm{d}x\\ =\,& \sum_{n=0}^\infty \frac{1}{(n + 1)^{n + 1}} - \frac16 \sum_{n=0}^\infty \frac{3^n}{(n + 1)^{n + 1}} + \frac{1}{120} \sum_{n=0}^\infty \frac{5^n}{(n + 1)^{n + 1}}\\ =\, & \sum_{n=0}^5 \frac{1 - 6^{-1}3^n + 120^{-1}5^n}{(n + 1)^{n + 1}} + \sum_{n=6}^\infty \frac{1 - 6^{-1}3^n + 120^{-1}5^n}{(n + 1)^{n + 1}}\\ <\, & \sum_{n=0}^5 \frac{1 - 6^{-1}3^n + 120^{-1}5^n}{(n + 1)^{n + 1}} + \sum_{n=6}^\infty \frac{1 - 6^{-1}3^n + 120^{-1}5^n}{7^{n + 1}}\\ =\,& \frac{62861674901693}{65868380928000}. \tag{1} \end{align*}
Using Facts 1-2, we have \begin{align*} &\int_1^2 \sin(x^{-x})\, \mathrm{d} x \\ \le\,& \int_1^2 \sin\left(\frac{3 - x}{x^2 - x + 2}\right)\,\mathrm{d} x\\ \le\,& \int_1^2 \left[\frac{3 - x}{x^2 - x + 2} - \frac{1}{6}\left(\frac{3 - x}{x^2 - x + 2}\right)^3 + \frac{1}{120}\left(\frac{3 - x}{x^2 - x + 2}\right)^5\right]\mathrm{d}x\\ =\,& \frac{625\sqrt7}{1029} \arctan\frac{\sqrt7}{5} + \frac{1687723}{18063360} - \frac12\ln 2. \tag{2} \end{align*}
Using Fact 3, we have $$\int_4^5 \sin (x^{-x})\, \mathrm{d}x \le \int_4^5 x^{-x}\, \mathrm{d} x \le \int_4^5 4^{-x} \mathrm{e}^{-x + 4}\, \mathrm{d} x = \frac{4 - \mathrm{e}^{-1}}{1024 + 2048\ln 2}, \tag{3}$$ and $$\int_3^4 \sin (x^{-x})\, \mathrm{d}x \le \int_3^4 x^{-x}\, \mathrm{d} x \le \int_3^4 3^{-x} \mathrm{e}^{-x + 3}\, \mathrm{d} x = \frac{3 - \mathrm{e}^{-1}}{81 + 81\ln 3}, \tag{4}$$ and $$\int_{5/2}^3 \sin(x^{-x})\,\mathrm{d} x \le \int_{5/2}^3 x^{-x}\,\mathrm{d} x \le \int_{5/2}^3 (5/2)^{-x}\mathrm{e}^{-x + 5/2}\,\mathrm{d} x = \frac{4\sqrt{10} - 8\sqrt{\mathrm{e}^{-1}}}{125\ln \frac{5}{2} + 125}, \tag{5}$$ and $$\int_2^{5/2} \sin(x^{-x})\,\mathrm{d} x \le \int_2^{5/2} x^{-x}\,\mathrm{d} x \le \int_2^{5/2} 2^{-x}\mathrm{e}^{-x + 2}\,\mathrm{d} x = \frac{2 - \sqrt{2\mathrm{e}^{-1}}}{8 + 8\ln 2}. \tag{6}$$
Also, we have $$\int_5^\infty \sin (x^{-x})\, \mathrm{d}x \le \int_5^\infty x^{-x}\, \mathrm{d}x \le \int_5^\infty \mathrm{e}^{-x\ln 5}\, \mathrm{d}x = \frac{1}{3125\ln 5}. \tag{7}$$
With the above results, we obtain the desired result $$\int_0^\infty \sin(x^{-x})\,\mathrm{d}x < \frac{1 + \sqrt5}{2}.$$ Note: $(1) + (2) + (3) + (4) + (5) + (6) + (7)$ gives $\int_0^\infty \sin(x^{-x})\,\mathrm{d}x < 1.617374660$.
We are done.