Q:

How to prove $\int_{1}^{\infty}\frac{e^{-tx^2}}{x^2} \{x\} dx=1-γ+O(\sqrt t)$ when $t \to 0^+$

Problem:
  Prove that
  $$
I(t)=\int_{1}^{\infty}\frac{e^{-tx^2}}{x^2} \{x\} dx=1-γ+O(\sqrt t), \text{when } t\to 0^+
$$
  where $\{x\}$ denote the Fractional part function.

What I have tried:
Since
$$
J(t)=\large \int_{1}^{\infty}\frac{e^{-tx^2}}{x^2}\, dx =e^{-t} - 2\sqrt{t}\int_{\sqrt{t}}^{\infty}e^{-x^2}\,dx= e^{-t}-\sqrt{\pi t} \,\text{erfc}(\sqrt{t})
$$
thus it is easy to see
$$
J(t) = 1+O(\sqrt t), t\to 0^+
$$
But how can I derive things like this for $I(t)$? Who can give any help?

A:

$$I(t)-I(0)=\int_{\sqrt{t}}^{\infty}\frac{e^{-x^2}}{(x/\sqrt{t})^2} \{x/\sqrt{t}\} d(x/\sqrt{t})=O(\sqrt t)$$
$$I(0) = \lim_{s\to 1}s\int_1^\infty x x^{-s-1}dx-s\int_1^\infty \lfloor x \rfloor x^{-s-1} =\lim_{s\to 1} \frac{s}{s-1}-\sum_{n=1}^\infty s\int_n^\infty x^{-s-1}dx\\=\lim_{s\to 1} 1+\frac{1}{s-1}-\zeta(s)= 1-\gamma$$ 
Where $\gamma$ comes from $$\sum_{n=1}^\infty n^{-s} -\frac1{s-1}= \sum_{n=1}^\infty (\sum_{k=1}^n \frac1k) (n^{1-s}-(n+1)^{1-s}) -\int_1^\infty  x^{-s}dx$$ $$=\sum_{n=1}^\infty (\gamma+\log(n)+O(1/n)) \int_n^{n+1}(s-1)x^{-s}dx-(s-1)\int_1^\infty \log(x) x^{-s}dx$$ $$= \gamma +(s-1) \sum_{n=1}^\infty \int_n^{n+1} (\log(n)+O(1/n)-\log(x))x^{-s}dx$$

