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.
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
Some thoughts:
We need to prove that $$\int_0^\infty \left(1 - \Gamma(1 + x^{3/2}\mathrm{e}^{-x^2})\right) \mathrm{d} x > \frac{7}{10} + \frac{\sqrt2 - 1}{2}\zeta(\tfrac12)\sqrt{\pi}.$$
For a function $f(x)$, let $T(f(x),a,n)$ denote the $n$-th order Taylor approximation of $f(x)$ at $x = a$.
Fact 1: $0 \le x^{3/2}\mathrm{e}^{-x^2} < 2/5$ for all $x\ge 0$.
Fact 2: For all $y\in [0, 2/5)$, $$\Gamma(1 + y) \le T(\Gamma(1 + y), 0, 10).$$ (The proof of Fact 2 is given at then end.)
Using Facts 1-2, it suffices to prove that $$\int_0^\infty \left(1 - T(\Gamma(1 + y), 0, 10)\Big\vert_{y = x^{3/2}\mathrm{e}^{-x^2}}\right) \mathrm{d} x > \frac{7}{10} + \frac{\sqrt2 - 1}{2}\zeta(\tfrac12)\sqrt{\pi}$$ which is true.
Note: Here the integral admits a closed form in terms of $\gamma, \zeta(3), \zeta(5), \zeta(7), \zeta(9), \zeta(1/2), \Gamma(3/4)$ etc.
Proof of Fact 2:
We use the Taylor series of $\ln \Gamma(1 + y)$: $$\ln \Gamma(1 + y) = - \gamma y + \sum_{k=2}^\infty \frac{\zeta(k)}{k}(-1)^k y^k, \quad |y| < 1.$$ We have, for all $y\in [0, 2/5)$, $$\ln \Gamma(1 + y) \le - \gamma y + \sum_{k=2}^{10} \frac{\zeta(k)}{k}(-1)^k y^k.$$ It suffices to prove that $$- \gamma y + \sum_{k=2}^{10} \frac{\zeta(k)}{k}(-1)^k y^k \le \ln \Big[T(\Gamma(1 + y), 0, 10)\Big].$$ Let $F(y) = \mathrm{RHS} - \mathrm{LHS}$. We have $F(0) = 0$. One can prove that $F'(y) \ge 0$ for all $y\in [0, 2/5)$.
We are done.