Does $\lim_{M\to\infty}\int_{0}^\infty (1+x^2)^{-s}\frac{\sin Mx}{x}dx$ exist? Does the limit $\displaystyle \lim\limits_{M\to\infty}\int_{0}^\infty (1+x^2)^{-s}\frac{\sin Mx}{x}dx$ exist? Where $s>0$ be a fix real number.
i.e. does the integral $\displaystyle \int_{0}^\infty dy\int_{0}^\infty (1+x^2)^{-s}\cos xydx$ converage in some sense?
 A: By switching to Fourier transforms or through other techniques it is not difficult to prove that
$$ \lim_{M\to +\infty}\frac{\sin(Mx)}{\pi x}=\delta(x) $$
hence:
$$ \lim_{M\to +\infty}\int_{0}^{+\infty}\frac{1}{(1+x^2)^s}\cdot\frac{\sin(Mx)}{x}\,dx = \frac{1}{2}\lim_{M\to +\infty}\int_{-\infty}^{+\infty}\frac{1}{(1+x^2)^s}\cdot\frac{\sin(Mx)}{x}\,dx = \color{red}{\frac{\pi}{2}}.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\color{#f00}{\lim_{M \to \infty}\int_{0}^{\infty}\pars{1 + x^{2}}^{-s}\,
{\sin\pars{Mx} \over x}\,\dd x} =
\half\,\lim_{M \to \infty}
\bracks{M\int_{-\infty}^{\infty}\pars{1 + x^{2}}^{-s}\,
{\sin\pars{Mx} \over Mx}\,\dd x}
\\[3mm] = &\
\half\,\lim_{M \to \infty}
\bracks{M\int_{-\infty}^{\infty}\pars{1 + x^{2}}^{-s}\,\ \overbrace{%
\half\int_{-1}^{1}\expo{\ic kMx}\,\dd k}
^{\ds{=\ {\sin\pars{Mx} \over Mx}}}\,\dd x}
\\[3mm] = &\
{1 \over 4}\,\lim_{M \to \infty}\bracks{M\int_{-1}^{1}
\int_{-\infty}^{\infty}\pars{1 + x^{2}}^{-s}\expo{\ic kMx}\,\dd x\,\dd k}
\\[3mm] = &\
{1 \over 4}\,\lim_{M \to \infty}\bracks{M\int_{-1}^{1}
\int_{-\infty}^{\infty}\exp\pars{\ic kMx - s\ln\pars{1 + x^{2}}}\,\dd x\,\dd k}
\end{align}

For 'large $M$', the asymptotic result for the $x$-integral is given by
$\ds{\root{\pi \over s}\,\exp\pars{-\,{M^{2} \over 4s}\,k^{2}}}$ such that
\begin{align}
&\color{#f00}{\lim_{M \to \infty}\int_{0}^{\infty}\pars{1 + x^{2}}^{-s}\,
{\sin\pars{Mx} \over x}\,\dd x} =
{1 \over 4}\,\root{\pi \over s}\,\lim_{M \to \infty}\bracks{M\int_{-1}^{1}
\exp\pars{-\,{M^{2} \over 4s}\,k^{2}}\,\dd k}
\\[3mm] = &\
{1 \over 4}\,\root{\pi \over s}\,\lim_{M \to \infty}\braces{%
M\bracks{{2\root{\pi s} \over M}\,\mathrm{erf}\pars{M \over 2\root{s}}}}
= {\pi \over 2}\,\
\underbrace{\lim_{M \to\ \infty}\mathrm{erf}\pars{M \over 2\root{s}}}
_{\ds{=\ 1}}\ =\ \color{#f00}{\pi \over 2}
\end{align}


$\ds{\mathrm{erf}}$ is the Error Function.

