Equivalent of $\int_2^{+ \infty} e^{\Gamma (t) \log x}dt$ when $x \to 1$

Question

I wonder if the equivalent :

$$ \int_2^{+ \infty} e^{\Gamma (t) \log x}dt $$

for $x \to 1^{-}$ (i.e the first term in the asymptotic expansion) had been studied ? Is it tricky to get an equivalent ?

Personnaly I don't get it because the Stirlign formula and its reffinements do not suffice to replace $\Gamma (t)$ by a series development ..

So if you have ideas, please telle me !

Thanks.

Answer 1

We have that $$\left\lfloor{t}\right\rfloor !\geq\Gamma(t)\geq\left\lfloor{t-1}\right\rfloor !\implies e^{-\left\lfloor{t}\right\rfloor !}\leq e^{-\Gamma(t)}\leq e^{-\left\lfloor{t-1}\right\rfloor !}$$ for all $t\in[2,\infty]$. Therefore, if we set $$I(x)=\int_2^{+ \infty} e^{\Gamma (t) \log x}dt$$ we have that, for $x\in(0,1)$, $$\sum_{t=2}^\infty x^{t!}=\int_2^\infty\, e^{\left\lfloor{t}\right\rfloor ! \log x}\,\mathrm{d}t\leq I(x)\leq \int_2^\infty\, e^{\left\lfloor{t-1}\right\rfloor ! \log x}\,\mathrm{d}t = \sum_{t=1}^\infty x^{t!}$$ It is fairly easy to check that $$\lim_{x\to 1^-} \sum_{t=2}^\infty x^{t!}=\infty$$ which proves that $\lim_{x\to 1^-} I(x)=\infty$.

Edit: If you're interested in parametrizing $x$, I would suggest setting $x=1-\frac{1}{y}$. I'm pretty sure that $I(1-\frac{1}{y})\sim y!$, though this should definitely be formalized and made rigorous.