Show $\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{y}{1+y^2}e^{-ay} dy =0 $

Question

How to use LDC and MCT to prove?

Answer 1

The limit $$ \lim_{a\to\infty}\int_0^\infty\frac{y}{1+y^2}e^{-ay}\,\mathrm{d}y $$ can be handled by Dominated Convergence.

For $a\ge1$, $\frac{y}{1+y^2}e^{-ay}$ is dominated by $\frac12e^{-y}$ and $$ \int_0^\infty\frac12e^{-y}\,\mathrm{d}y=\frac12 $$ Pointwise, $$ \lim_{a\to\infty}\frac{y}{1+y^2}e^{-ay}=0 $$ therefore, $$ \lim_{a\to\infty}\int_0^\infty\frac{y}{1+y^2}e^{-ay}\,\mathrm{d}y=0 $$

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Note that you cannot dominate this integrand by $\frac{y}{1+y^2}$ since that function is not in $L^1$. However, $\frac{y}{1+y^2}\le\frac12$, so we can use $\frac12e^{-y}$ to dominate the integrand.

Answer 2

Off the top of my head (as usual), I would split the integral into two parts divided, somewhat arbitrarily, at $1$.

Let $I = \int_0^1$ and $J = \int_1^{\infty}$.

$J < \int_1^{\infty} y e^{-ay} dy = e^{-ay}(\frac{y}{-a}-\frac1{a^2})\big|_1^\infty = -e^{-a}(\frac1{-a}-\frac1{a^2}) \to 0 $.

$I < \int_0^{1} y e^{-ay} dy = e^{-ay}(\frac{y}{-a}-\frac1{a^2})\big|_0^1 = e^{-a}(\frac1{-a}-\frac1{a^2})-\frac1{a^2} \to 0 $.

So $I+J \to 0$.

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Looking at this, you could just show that $\int_0^{\infty} y e^{-ay} dy \to 0$ and that would be enough.

Oh well. I'll leave the whole thing there.

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\lim_{a \to \infty}\int_{0}^{\infty}{y \over 1 + y^{2}}\expo{-ay}\,\dd y = 0 :\ {\large ?}}$

With $\ds{a > 0}$: \begin{align} &\color{#66f}{\large% 0<\verts{\int_{0}^{\infty}{y \over 1 + y^{2}}\expo{-ay}\,\dd y}} <\verts{\int_{0}^{\infty}y\expo{-ay}\,\dd y} \\[5mm]&={1 \over a^{2}}\verts{\int_{0}^{\infty}y\expo{-y}\,\dd y}={1 \over a^{2}} \stackrel{a\ \to\ \infty}{\Huge\to} \color{#66f}{\large 0} \end{align}