Prove or disprove that :
$$\int_{0}^{\infty}\left(\left(x^{\frac{3}{2}}e^{-x^{2}}\right)!-1-\frac{1}{e^{x^{2}}+1}\right)dx+\frac{7}{10}<0$$
It's an experiment using Desmos .
Some facts :
$$\int_{0}^{\infty}\left(\frac{1}{e^{x^{2}}+1}\right)dx=-\frac{1}{2}\left(\sqrt{2}-1\right)\sqrt{\pi}\zeta{(\frac{1}{2})}$$
$$\int_{0}^{\infty}\left(\left(x^{\frac{3}{2}}e^{-x^{2}}\right)!-1\right)dx<-\frac{164}{100}\quad(1)$$
First I have tried to show $(1)$ using elementary method as integrations by parts but it goes nowhere.*To go further I use the Stirling approximation modified (Ramanujan) on $(1,\infty)$ and then use power series but it seems delicate *.On the other hand I don't see a good substitution to tackle it .
*Oups I was working on other integral..This part is not relevant.
Question :
Can we hope to show it by hand or it is illusory ?
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.