Finding a closed form for $\int_0^{\infty} \frac{\sin(x/\epsilon)}{1+x^2}dx$ in terms of $\epsilon$? $$\int_0^{\infty} \frac{\sin(x/\epsilon)}{1+x^2}dx$$
We can use complex analysis to show that $\int_0^{\infty} \frac{\cos(x/\epsilon)}{1+x^2}dx = \frac{\pi}{2}e^{-1/\epsilon}$ but this sin version is causing trouble. Does anyone know a closed form solution like this for it or is it a wasted effort? Thanks.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Note that

  $\ds{\int_{0}^{\infty}{\sin\pars{x/\epsilon} \over 1 + x^{2}}\,\dd x =
\mrm{sgn}\pars{\epsilon}\int_{0}^{\infty}
{\sin\pars{x/\verts{\epsilon}} \over 1 + x^{2}}\,\dd x =
\epsilon\int_{0}^{\infty}
{\sin\pars{x} \over 1 + \epsilon^{2}x^{2}}\,\dd x =
{1 \over \epsilon}\int_0^{\infty}
{\sin\pars{x} \over x^{2} + \epsilon^{-2}}\,\dd x}$.

Then, with $\ds{\Lambda > 0}$:
\begin{align}
\int_{0}^{\Lambda}{\sin\pars{x/\epsilon} \over 1 + x^{2}}\,\dd x & =
{1 \over \epsilon}\int_{0}^{\Lambda}
{\sin\pars{x} \over x^{2} + \epsilon^{-2}}\,\dd x =
\epsilon\,\Im\int_{0}^{\Lambda}
{\sin\pars{x} \over x - \epsilon^{-2}\ic}\,\dd x =
\epsilon\,\Im\int_{-\epsilon^{-2}\ic}^{\Lambda - \epsilon^{-2}\ic}
{\sin\pars{x + \epsilon^{-2}\ic} \over x}\,\dd x
\\[5mm] & =
\epsilon\,\Im\int_{-\epsilon^{-2}\ic}^{\Lambda- \epsilon^{-2}\ic}
{\sin\pars{x}\cosh\pars{\epsilon^{-2}} +
\cos\pars{x}\,\ic\sinh\pars{\epsilon^{-2}} \over x}\,\dd x
\\[5mm] & =
\epsilon\cosh\pars{1 \over \epsilon^{2}}
\,\Im\int_{-\epsilon^{-2}\ic}^{\Lambda- \epsilon^{-2}\ic}
{\sin\pars{x} \over x}\,\dd x +
\epsilon\sinh\pars{1 \over \epsilon^{2}}
\,\Re\int_{-\epsilon^{-2}\ic}^{\Lambda- \epsilon^{-2}\ic}
{\cos\pars{x} \over x}\,\dd x
\\[1cm] & =
\epsilon\cosh\pars{1 \over \epsilon^{2}}\,\Im\bracks{%
\mrm{Si}\pars{\Lambda - {\ic \over \epsilon^{2}}} -
\mrm{Si}\pars{-\,{\ic \over \epsilon^{2}}}}
\\[3mm] & +
\epsilon\sinh\pars{1 \over \epsilon^{2}}\,\Re\bracks{%
\mrm{Ci}\pars{\Lambda - {\ic \over \epsilon^{2}}} -
\mrm{Ci}\pars{-\,{\ic \over \epsilon^{2}}}}
\end{align}

$\ds{\mrm{Si}}$ and $\ds{\mrm{Ci}}$ are the Sine Integral and Cosine Integral Functions, respectively.

As $\ds{\Lambda \to \infty}$:
$$
\bbx{\int_{0}^{\infty}{\sin\pars{x/\epsilon} \over 1 + x^{2}}\,\dd x =
-\epsilon\cosh\pars{1 \over \epsilon^{2}}\,\Im
\mrm{Si}\pars{-\,{\ic \over \epsilon^{2}}} -
\epsilon\sinh\pars{1 \over \epsilon^{2}}\,\Re
\mrm{Ci}\pars{-\,{\ic \over \epsilon^{2}}}}
$$

Note that ( see this page ), as $\ds{\verts{z} \to \infty}$,
  $\ds{\mrm{Si}\pars{z} \to {\pi \over 2}}$ and
  $\ds{\mrm{Ci}\pars{z} \to 0}$.

