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.
Remark: Although we get a closed-form bound (in terms of elementary functions), the solution is still quite complicated.
We have
$$\int_{-\infty}^0 (\cosh x - 1)^x\, \mathrm{d}x
= \int_0^\infty 2^{-x}\left(\sinh\frac{x}{2}\right)^{-2x}\,\mathrm{d} x = \int_0^\infty 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x.$$
Fact 1: $\int (\sinh x)^{-5}\, \mathrm{d}x =
-\frac{\cosh x}{4\sinh^4 x} + \frac{3\cosh x}{8\sinh^2 x} - \frac{3}{8}\ln\frac{\mathrm{e}^x + 1}{\mathrm{e}^x - 1} + C$.
Fact 2: $\int (\sinh x)^{-4}\, \mathrm{d}x =
-\frac{\cosh x}{3\sinh^3 x} + \frac{2\cosh x}{3\sinh x} + C$.
Fact 3: $\sinh x \ge x + \frac16 x^3$ for all $x\in [0, 1]$.
Let $T(f(x),a,n)$ denote the $n$-th order Taylor approximation of $f(x)$ at $x = a$.
Fact 4: For all $x\in [0, 1]$,
$(1 + x^2/6)^{-4x} \le T((1 + x^2/6)^{-4x}, 0, 12)$.
(Proof: Take logarithm, then take derivative and the 2nd derivative.)
Fact 5: For all $x\in [0, 1]$,
$2^{-2x} \le T(2^{-2x}, 0, 6)$.
(Proof: Take logarithm, then take derivative.)
Fact 6: For all $u \in [-1, 0]$,
$\mathrm{e}^{-4u} \le T(\mathrm{e}^{-4u}, 0, 8) + \frac{512}{315}u^8$.
(Proof: Take logarithm, then take derivative.)
First, using Fact 1, we have
\begin{align*}
\int_{5/4}^\infty 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x
&\le \int_{5/4}^\infty 2^{1 - 2\cdot \frac{5}{4}}(\sinh x)^{-4\cdot \frac{5}{4}}\, \mathrm{d}x \\
&= -\frac{\sqrt2}{4}\left(-\frac{\cosh (5/4)}{4\sinh^4 (5/4)} + \frac{3\cosh (5/4)}{8\sinh^2 (5/4)} - \frac{3}{8}\ln\frac{\mathrm{e}^{5/4} + 1}{\mathrm{e}^{5/4} - 1}\right)\\
&< \frac{3}{500}.
\end{align*}
Second, using Fact 2, we have
\begin{align*}
\int_1^{5/4} 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x
&\le \int_1^{5/4} 2^{1 - 2\cdot 1}(\sinh x)^{-4\cdot 1}\, \mathrm{d}x\\
&= -\frac{\cosh (5/4)}{6\sinh^3 (5/4)} + \frac{\cosh (5/4)}{3\sinh (5/4)} - \left(-\frac{\cosh 1}{6\sinh^3 1} + \frac{\cosh 1}{3\sinh 1}\right)\\
&< \frac{3}{80}.
\end{align*}
Third, using Facts 3-6, we have
\begin{align*}
&\int_0^1 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x\\
\le\,& \int_0^1 2^{1 - 2x}(x + x^3/6)^{-4x}\, \mathrm{d}x\\
=\, & \int_0^1 2^{1 - 2x}(1 + x^2/6)^{-4x}x^{-4x}\, \mathrm{d}x\\
\le\, & 2\int_0^1 T(2^{-2x}, 0, 6) \, T((1 + x^2/6)^{-4x}, 0, 12)
\, \left[T(\mathrm{e}^{-4u}, 0, 8) + \frac{512}{315}u^8\right]_{u = x\ln x}\, \mathrm{d} x \tag{1}\\
<\, & 1549/500.
\end{align*}
Note: The integral in (1) admits a closed form
$a_0 + a_1\ln 2 + a_2\ln^2 2 + a_3\ln^3 2 + a_4\ln^4 2 + a_5\ln^5 2 + a_6\ln^6 2$
where $a_i$'s are all rational numbers.
Thus, we have $\int_0^\infty 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x
< \frac{3}{500} + \frac{3}{80} + 1549/500 < \pi$.
We are done.
Best Answer
We need to prove that $$\int_0^1 \frac{1}{x^2 + \ln^2 x}\, \mathrm{d} x > 1.$$
Fact 1: For all $y \in (0, \mathrm{e}]$, $$\frac{-y^4 + 64y^3 - 318y^2 + 400y + 35}{132y + 48} \ge y(1 - \ln y)^2.$$ (Note: LHS is the (4, 1)-Pade approximant of $y(1 - \ln y)^2$ at $y = 1$.)
With the substitution $y = \mathrm{e}x$, using Fact 1, we have \begin{align*} & \int_0^1 \frac{1}{x^2 + \ln^2 x}\, \mathrm{d} x\\ =\, & \int_0^\mathrm{e} \frac{\mathrm{e}}{y^2 + \mathrm{e}^2 (1 - \ln y)^2}\, \mathrm{d} y\\ \ge\, & \int_0^\mathrm{e} \frac{132\mathrm{e}y^2 + 48\mathrm{e}y}{(132 - \mathrm{e}^2)y^4 + (64\mathrm{e}^2 + 48)y^3 - 318\mathrm{e}^2 y^2 + 400\mathrm{e}^2y + 35\mathrm{e}^2}\, \mathrm{d} y\\ \ge\, & \int_0^\mathrm{e} \frac{132\mathrm{e}y^2 + 48\mathrm{e}y}{(132 - \mathrm{e}^2)(y + 1/12)(y + 463/64)[(y - 69/44)^2 + (55/54)^2]}\, \mathrm{d} y \tag{1} \\[8pt] =\, & \frac{1253273472\mathrm{e}}{911936989\mathrm{e}^2 - 120375682548} \int_0^\mathrm{e} \frac{1}{12y + 1}\, \mathrm{d} y\\[8pt] &\quad + \frac{29129459463143424\mathrm{e}}{38952637956653\mathrm{e}^2 - 5141748210278196} \int_0^\mathrm{e} \frac{1}{64y + 463}\, \mathrm{d} y \\[8pt] &\quad + \frac{372594816\mathrm{e}}{2487336612548656836 - 18843459185974673\mathrm{e}^2}\\[8pt] &\quad\quad \times \int_0^\mathrm{e} \frac{842186377401732y + 746090344086823}{1411344y^2 - 4426488y + 4934869}\, \mathrm{d} y \tag{2}\\ >\, & 1 \end{align*} where in (1) we have used (easy): for all $y\in [0, \mathrm{e}]$, \begin{align*} &(132 - \mathrm{e}^2)y^4 + (64\mathrm{e}^2 + 48)y^3 - 318\mathrm{e}^2 y^2 + 400\mathrm{e}^2y + 35\mathrm{e}^2\\ \le\, & (132 - \mathrm{e}^2)(y + 1/12)(y + 463/64)[(y - 69/44)^2 + (55/54)^2]. \end{align*} Note: The three integrals in (2) admit closed form expressions in terms of $\ln $ and $\arctan$ only.
We are done.