Show $\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{y}{1+y^2}e^{-ay} dy =0 $ How to use LDC and MCT to prove?
 A: 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 $I+J \to 0$.

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.
A: 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,
$$
\lim_{a\to\infty}\int_0^\infty\frac{y}{1+y^2}e^{-ay}\,\mathrm{d}y=0
$$

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.
A: $\newcommand{\+}{^{\dagger}}
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$\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}

