Improper integral inequality including the golden ratio and the Sophomore's dream 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 $(I)$ ?
 A: Too long for a comment we have :
$$\int_{6}^{\infty}\sin\left(x^{-x}\right)dx+\int_{2}^{6}\sin\left(e^{\left(-\left(x-1\right)^{\frac{93}{100}}x^{0.5}\cdot2^{0.5}\ln\left(2\right)\right)}\right)dx+\int_{0}^{1}\sin\left(e^{-x\left(x-1\right)\left(\frac{2}{x^{2}+x}\right)^{\frac{1}{3}}}\right)dx+\int_{1}^{2}\sin\left(e^{-\left(\frac{x^{0.48}\left(x-1\right)}{2^{0.48}}\right)\cdot2\ln\left(2\right)}\right)dx-\frac{\left(1+\sqrt{5}\right)}{2}<0$$
A: Here will be my attempt at evaluating the integral. We will use this sine series to evaluate with the help of this graph. The sine expansion actually works which simplfies the problem into a generalized tetration integral. We can interchange the sum and integral term by term:
$$\mathrm{I=\int_{\Bbb R^+}sin(x^{-x})dx=\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\int_0^\infty x^{-(2k+1)x}dx}$$
Using the graph again, one can use any series expansion for $e^y$ here or the classic fact that $e^y= \mathop{\sum}_{x\ge 0} \frac{ y^x}{x!} $ for all real x. Desmos seems to have a hard time computing the following, but each expansion works separately which implies verifiable steps. One again can integrate term by term. Here is info on the regularized gamma function and exponential integral function with which I would evaluate it, but the bound evaluation ends up being tricky:
$$\mathrm{I= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\sum_{n=0}^\infty \frac{(-1)^n(2k+1)^n}{n!}\int_0^\infty x^n ln^n(x)dx=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\sum_{n=0}^\infty \frac{(-1)^n(2k+1)^n}{n!}\frac{Γ(n+1,-(n+1)ln(x)(-(n+1))^{-n}}{n+1}\bigg|_0^\infty=\lim_{x\to -\infty}\sum_{k=0}^\infty\sum_{n=1}^\infty\frac{(-1)^k(2k+1)^{n-2}Q(n,nx)}{n^n (2k)!}=1.612215…<1.6180339…=\phi=\frac{\sqrt 5+1}{2}}$$
As you can see, each step was verified, but seems to be unable to find a nice form for “I” and desmos has trouble calculating the result even though I gave examples as support for the expansion of $x^{-cx}$ and sin($x^{-x}$) which still worked. Please correct me and give me feedback!
