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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!

2 Answers 2

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $$\bbx{\ds{% \int_{1}^{\infty}{\expo{-xt} \over \root{1 + t^{2}}}\,\dd t \sim -\ln\pars{x} \quad\mbox{as}\ x \to 0^{+}}} $$

  • 0
    I'm not sure about the integration of the term $x\int_{1}^{\infty}\ln(t)e^{-xt}\, dt$. Isn't it: $$ x\int_{1}^{\infty}\ln(t)e^{-xt}\, dt = \int_{x}^{\infty}\ln(\frac{v}{x})e^{-v}\, dv \quad\mathrm{by \ doing} \; v= xt $$ $$ = -\ln(x)\int_x^\infty e^{-v}dv + \int_x^\infty \ln(v) e^{-v}dv$$ $$ = -\ln(x) e^{-x} + \int_x^\infty \ln(v) e^{-v}dv \quad \sim - \ln(x) \quad \mathrm{as} \; x \to 0^+ $$ – 2017-02-05
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    @omardrwch You're right. I just fixed it. Thanks a lot. – 2017-02-05
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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.