TITLE: Asymptotic expansion of $\int_{0}^{\infty} \frac{\sin(\frac{x}{n})}{x(1+x^2)}dx$?
QUESTION [5 upvotes]: What are the first terms of the asymptotic expansion of
$$
I_n=\int_{0}^{\infty} \frac{\sin(\frac{x}{n})}{x(1+x^2)}dx\ ?
$$
Using the dominated convergence theorem to 
$$
n\bigg(I_n-\int_{0}^{\infty} \frac{\frac{x}{n}}{x(1+x^2)}dx\bigg)
$$ 
I obtain $I_n \sim \frac{\pi}{2n}$. But I have no idea for the second term.
ADD-ON: 
Thank you for the beautiful answers.
1) Does there exist an elementary method (without complex analysis) for obtain the second term ? Indeed, the question is provided by a elementary course on real analysis.
2) I guess the first term with the equivalence $\sin(x) \underset{0}{\sim} x$. Why intuitively the approximation $\sin(x) \approx x-\frac{x^3}{6}$ does not give the second term ?

REPLY [2 votes]: $$
\begin{align}
\int_0^\infty\frac{\sin\left(\frac{x}{n}\right)}{x\left(1+x^2\right)}\,\mathrm{d}x
&=\frac12\int_{-\infty}^\infty\frac{\sin(x)}{x\left(1+n^2x^2\right)}\,\mathrm{d}x\tag{1}\\
&=\frac14\int_{-\infty-\frac{i}{2n}}^{\infty-\frac{i}{2n}}\left(\frac2x-\frac1{x-\frac{i}{n}}-\frac1{x+\frac{i}{n}}\right)\sin(x)\,\mathrm{d}x\tag{2}\\
&=\frac1{8i}\int_{\gamma^+}\left(\frac2x-\frac1{x-\frac{i}{n}}\color{#CCC}{-\frac1{x+\frac{i}{n}}}\right)e^{ix}\,\mathrm{d}x\\
&-\frac1{8i}\int_{\gamma^-}\left(\color{#CCC}{\frac2x-\frac1{x-\frac{i}{n}}}-\frac1{x+\frac{i}{n}}\right)e^{-ix}\,\mathrm{d}x\tag{3}\\[6pt]
&=\frac\pi4\left(2-e^{-1/n}\right)+\pi\left(-e^{-1/n}\right)\tag{4}\\[12pt]
&=\frac\pi2\left(1-e^{-1/n}\right)\tag{5}\\[9pt]
&\sim\frac\pi2\left(\frac1n-\frac1{2n^2}+\frac1{6n^3}-\dots\right)\tag{6}
\end{align}
$$
Explanation:
$(1)$: the integrand is even then substitute $x\mapsto nx$
$(2)$: partial fractions and move the contour down $\frac{i}{2n}$
$(3)$: close the contours with arcs whose integrals vanish and $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\phantom{\text{(3):}}$ the greyed out parts have no residue inside the corresponding contours
$(4)$: use residues to compute the contour integrals
$(5)$: algebra
$(6)$: give the asymptotic expansion (actually, the start of the Taylor series)
The contours mentioned above are
$$
\gamma^+=\left[-R-\frac{i}{2n},R-\frac{i}{2n}\right]\cup \left(Re^{i\pi[0,1]}-\frac{i}{2n}\right)
$$
and
$$
\gamma^-=\left[-R-\frac{i}{2n},R-\frac{i}{2n}\right]\cup \left(Re^{-i\pi[0,1]}-\frac{i}{2n}\right)
$$