Asymptotic behavior of integral $\int_1^\infty \frac{e^{-xt}}{\sqrt{1+t^2}}dt$ as $x \to 0$ I wish to prove that:
$$ \int_1^\infty \frac{e^{-xt}}{\sqrt{1+t^2}}dt \sim - \ln x \quad  \mathrm{as} \quad x \to 0^+$$  
using the fact that:
$$ f \underset{b}{\sim} g \Rightarrow \int_a^x f \underset{x \to b}{\sim} \int_a^x g$$ 
if $\int_a^x g \to \infty$ as $x \to b$ and $f$ and $g$ are integrable on every interval $[a,c]$ with $c < b$.
Does anyone have an idea?
Thank you!
 A: You may prove that
$$ \lim_{x\to 0^+}\frac{\int_{1}^{+\infty}\frac{e^{-xt}}{\sqrt{1+t^2}}\,dt}{-\log x}=1 \tag{1}$$
by proving that:
$$ \lim_{x\to 0^+}\frac{\frac{d}{dx}\int_{1}^{+\infty}\frac{e^{-xt}}{\sqrt{1+t^2}}\,dt}{-1/x}=\lim_{x\to 0^+}\int_{1}^{+\infty}\frac{xt e^{-xt}}{\sqrt{1+t^2}}\,dt = 1.\tag{2} $$
The last integral equals:
$$ \int_{x}^{+\infty}\frac{t}{\sqrt{x^2+t^2}}\cdot e^{-t}\,dt\tag{3}$$
hence $(2)$ (then $(1)$) follows from the dominated convergence theorem.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\left.\int_{1}^{\infty}{\expo{-xt} \over \root{1 + t^{2}}}\,\dd t\,
\right\vert_{\ x\ >\ 0} =
\int_{1}^{\infty}{\expo{-xt} \over t}\,\dd t +
\int_{1}^{\infty}\expo{-xt}\pars{{1 \over \root{1 + t^{2}}} - {1 \over t}}
\,\dd t
\\[5mm] & =
x\int_{1}^{\infty}\ln\pars{t}\expo{-xt}\,\dd t
-
\int_{1}^{\infty}{\expo{-xt} \over t\root{1 + t^{2}}\pars{\root{1 + t^{2}} + t}}
\,\dd t
\\[1cm] & =
-\ln\pars{x}\expo{-x}
\\[5mm] & + \int_{x}^{\infty}\ln\pars{t}\expo{-t}\,\dd t + \int_{1}^{\infty}\ln\pars{t}\expo{-t}\,\dd t -
\int_{1}^{\infty}{\expo{-xt} \over t\root{1 + t^{2}}\pars{\root{1 + t^{2}} + t}}
\,\dd t
\end{align}
The 'remaining' integrals are finite as $\ds{x \to 0^{+}}$ such that
$$\bbx{\ds{%
\int_{1}^{\infty}{\expo{-xt} \over \root{1 + t^{2}}}\,\dd t \sim -\ln\pars{x}
\quad\mbox{as}\ x \to 0^{+}}}
$$
