Finding a closed form for $\int_0^{\infty} \frac{\sin(x/\epsilon)}{1+x^2}dx$ in terms of $\epsilon$?

Question

$$\int_0^{\infty} \frac{\sin(x/\epsilon)}{1+x^2}dx$$

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.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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 $\ds{\mrm{Ci}\pars{z} \to 0}$.