Compute $\lim\limits_{a \to 0^+} \left(a \int_1^{\infty} e^{-ax}\cos \left(\frac{2\pi}{1+x^{2}} \right)\,\mathrm dx\right)$ How can I compute the following limit?
$$\lim_{a \to 0^+} \left(a \int_{1}^{\infty} e^{-ax}\cos \left(\frac{2\pi}{1+x^{2}} \right)\,\mathrm dx\right)$$
Any hints you can please give?
Cheers
 A: Hint: try a change of the independent variable $y:=ax$.
A: More generally, for every bounded measurable function $u$, consider 
$$
I_a(u)=a \int_0^{\infty} \mathrm e^{-ax}u(x)\mathrm dx,\qquad
J_a(u)=a \int_1^{\infty} \mathrm e^{-ax}u(x)\mathrm dx.
$$
You are interested in $\lim\limits_{a\to0^+}J_a(u)$ for $u(x)=\cos(2\pi/(1+x^2))$.
Since $I_a(u)-J_a(u)$ is $a$ times the integral on $(0,1)$ of a uniformly bounded function,  when $a\to0^+$, $I_a(u)-J_a(u)\to0$. From now on, we study $I_a(u)$.
From here, several methods are available. The one I prefer is to note that $I_a(u)=\mathrm E(u(X_a))$ where $X_a$ is a random variable with exponential distribution of parameter $a$, hence $X_a$ is distributed as $X_1/a$. Since $X_1\gt0$ with full probability, $X_1/a\to+\infty$ with full probability. Thus, if $u$ has a limit $u^*$ at infinity, $u$ is bounded and $I_a(u)=\mathrm E(u(X_1/a))\to u^*$ when $a\to0^+$.
In your case, $u^*=\cos(0)=1$ hence 
$$
\lim_{a\to0^+}a \int_1^{\infty} \mathrm e^{-ax}\cos(2\pi/(1+x^2))\mathrm dx=1.
$$
A: First, what's $\lim_{a \to 0^+} \int_1^\infty ae^{-ax} \: dx$?  Second, can you show that your integral is not so far from $\int_1^\infty ae^{-ax} \: dx$?
A: Another (quick) idea. You can start by using the inequality
$$ 1-\frac{t^2}{2} \leq \cos(t) \leq 1 + \frac{t^2}{2} $$
Then
$$
e^{-ax} - e^{-ax}\frac{4\pi^2}{2(1+x^2)^2} \leq e^{−ax} \cos\left(\frac{2π}{1+x^2}\right) \leq e^{-ax} + e^{-ax}\frac{4\pi^2}{2(1+x^2)^2} 
$$
When you integrate this inequality, you will obtain $a\int_1^{+\infty} e^{-ax} dx$ (which is the limit of your integral) and $a\int_1^{+\infty} e^{-ax}\frac{4\pi^2}{2(1+x^2)^2} dx$ which tends towards $0$ since
$$ 0 \leq a\int_1^{+\infty} e^{-ax}\frac{4\pi^2}{2(1+x^2)^2} dx\leq a\int_1^{+\infty} \frac{4\pi^2}{2(1+x^2)^2}dx$$
A: This is my hint:
Because $a\int_{1}^{\infty} e^{-ax}\cos\left ( \frac{2\pi}{1+x^2} \right )dx=\lim_{x\to \infty}\left ( e^{-a}-e^{-ax} \right )+\int_{1}^{\infty}\frac{4\pi x e^{-ax}}{(1+x^2)^2}\sin\left ( \frac{2\pi}{1+x^2} \right )dx$ 
Hence, the limit does not exist!
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\lim_{a \to 0^+}\bracks{a\int_{1}^{\infty}\expo{-ax}
\cos\pars{2\pi \over 1 + x^{2}}\,\dd x}
\\[3mm]&=
\lim_{a \to 0^+}\braces{a\int_{1}^{\infty}\expo{-ax}\,\dd x
-
a\int_{1}^{\infty}\expo{-ax}
\bracks{1 - \cos\pars{2\pi \over 1 + x^{2}}}\,\dd x}
\\[3mm]&=
\lim_{a \to 0^+}\braces{\int_{a}^{\infty}\expo{-x}\,\dd x
-
a\int_{1}^{\infty}\expo{-ax}
\bracks{2\sin^{2}\pars{\pi \over 1 + x^{2}}}\,\dd x}
\\[3mm]&=
1
-
2\quad\overbrace{%
\lim_{a \to 0^{+}}\braces{a\int_{1}^{\infty}\expo{-ax}
\sin^{2}\pars{\pi \over 1 + x^{2}}}\,\dd x}^{\ds{=\ 0\,,\quad
\pars{~\mbox{see below}~}}}\tag{1}
\end{align}
$$\color{#0000ff}{\large%
\lim_{a \to 0^+}\bracks{a\int_{1}^{\infty}\expo{-ax}
\cos\pars{2\pi \over 1 + x^{2}}\,\dd x} = 1}
$$
Notice that $\verts{\sin\pars{x}} \leq \verts{x}$ and $\expo{-ax} < 1$ when $ax > 0$.
Then,
$$
0 \leq \verts{a\int_{1}^{\infty}\expo{-ax}\sin^{2}\pars{\pi \over 1 + x^{2}}\,\dd x}
\leq
\verts{a}\,\verts{\int_{1}^{\infty}{\pi^{2} \over \pars{x^{2} + 1}^{2}}\,\dd x}
$$
such that the limit in the right hand side of $\pars{1}$ is, indeed,  zero when
$a \to 0^{+}$.
