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.
Sketch of a proof:
Denote
$$A = (1 - x)^{2(1 - x)}, \quad B = (1 - x)^{2x - 1},
\quad C = x^{2x}.$$
The desired inequality is written as
$$\left(\frac{A}{(1 - x)^2} - \frac{B}{2^4x(1 - x)^2C}\right)(x^{-2}C - 1) \ge 1.$$
Denote $a = \ln 2$. Let
\begin{align*}
A_1 &= \frac{p_1x^2 + p_2x + p_3}{(8a^3 - 24a^2
+ 48a - 24) x - 4a^3 - 12}, \\
B_1 &= \frac{(2a -2)x^2 + (-3a + 2)x + a}{(4a - 2)x^2 + (-4a + 2)x + a}, \\
C_1 &= \frac{(4a - 2)x^2 + (-4a + 2)x + a}{(2a - 2)x - a + 2},
\end{align*}
where
\begin{align*}
p_1 &= -4a^4 + 16a^3 -24a^2 + 24a - 24, \\
p_2 &= 4a^4 - 24a^3 + 48a^2 - 48a + 36, \\
p_3 &= - a^4 + 8a^3 - 24a^2 + 30a - 24.
\end{align*}
Fact 1: $A \ge A_1 > 0$ for all $x \in [1/2, 1)$.
Fact 2: $B \le B_1$ for all $x \in [1/2, 1)$.
Fact 3: $C \ge C_1 > 0$ for all $x \in [1/2, 1)$.
Fact 4: $x^{-2}C_1 - 1 > 0$ for all $x \in [1/2, 1)$.
By Facts 1-4, it suffices to prove that
$$\left(\frac{A_1}{(1 - x)^2} - \frac{B_1}{2^4x(1 - x)^2C_1}\right)(x^{-2}C_1 - 1) \ge 1$$
which is true (simply a polynomial inequality).
We are done.
Best Answer
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 $u > 0$.