Asymptotic expansion of $\int_{0}^{\infty} \frac{\sin(\frac{x}{n})}{x(1+x^2)}dx$? 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 ? 
 A: Note that $$I_{n}=\int_{0}^{\infty}\frac{\sin\left(x/n\right)}{x\left(1+x^{2}\right)}dx=\int_{0}^{\infty}\frac{\sin\left(x/n\right)}{x}dx-\int_{0}^{\infty}\frac{x\sin\left(x/n\right)}{1+x^{2}}dx$$ and the first integral is easy to evaluate $$\int_{0}^{\infty}\frac{\sin\left(x/n\right)}{x}dx=\int_{0}^{\infty}\frac{\sin\left(u\right)}{u}du=\color{blue}{\frac{\pi}{2}}$$ for the secon integral we may observe that $$-\int_{0}^{\infty}\frac{x\sin\left(x/n\right)}{1+x^{2}}dx=-\frac{1}{2}\textrm{Im}\left(\int_{-\infty}^{\infty}\frac{xe^{xi/n}}{1+x^{2}}dx\right)$$ then taking as path the upper semicircumference and noting that the integral over the semicircle vanish as the radius $R\rightarrow\infty$ we get by residue theorem$$\int_{-\infty}^{\infty}\frac{xe^{xi/n}}{1+x^{2}}dx=2\pi i\underset{x=i}{\textrm{Res}}\left(\frac{xe^{xi/n}}{1+x^{2}}\right)=\pi ie^{-1/n}$$ so $$-\int_{0}^{\infty}\frac{x\sin\left(x/n\right)}{1+x^{2}}dx=\color{red}{-\frac{\pi e^{-1/n}}{2}}$$ as wanted. Now the expansion should be very simple. 
A: This is not an answer;
We can get the antiderivative first using $$\frac 1{x(1+x^2)}=\frac{1}{x}-\frac{1}{2 (x-i)}-\frac{1}{2 (x+i)}$$ and so, we face the problem of $$\int \frac {\sin(\frac xn)}{x-a}dx=\sin \left(\frac{a}{n}\right) \text{Ci}\left(\frac{x-a}{n}\right)+\cos
   \left(\frac{a}{n}\right) \text{Si}\left(\frac{x-a}{n}\right)$$ where appears the sine and cosine integrals. Then, assuming $n>0$, 
 $$\int_0^\infty \frac {\sin(\frac xn)}{x-a}dx=\frac{1}{2} \left(2 \text{Si}\left(\frac{a}{n}\right)+\pi \right) \cos
   \left(\frac{a}{n}\right)-\text{Ci}\left(-\frac{a}{n}\right) \sin
   \left(\frac{a}{n}\right)$$ and I am stuck with the limits when $x \to \infty$ and $a=\pm i$.
Cheating, that is to say using a CAS, I have been more than surprised to learn that
$$\int_{0}^{\infty} \frac{\sin(\frac{x}{n})}{x(1+x^2)}dx=\frac{\pi}{2}   \left(1-e^{-1/n}\right)$$ as spaceisdarkgreen reported in his/her comment while I was typing. Frm here, the expansion to any order becomes simple.
As spaceisdarkgreen also commented, I suppose that a contour integral would be a nice and elegant solution.
A: $$
\begin{align}
\int_0^\infty\frac{\sin\left(\frac{x}{n}\right)}{x\left(1+x^2\right)}\,\mathrm{d}x
&=\frac12\int_{-\infty}^\infty\frac{\sin(x)}{x\left(1+n^2x^2\right)}\,\mathrm{d}x\tag{1}\\
&=\frac14\int_{-\infty-\frac{i}{2n}}^{\infty-\frac{i}{2n}}\left(\frac2x-\frac1{x-\frac{i}{n}}-\frac1{x+\frac{i}{n}}\right)\sin(x)\,\mathrm{d}x\tag{2}\\
&=\frac1{8i}\int_{\gamma^+}\left(\frac2x-\frac1{x-\frac{i}{n}}\color{#CCC}{-\frac1{x+\frac{i}{n}}}\right)e^{ix}\,\mathrm{d}x\\
&-\frac1{8i}\int_{\gamma^-}\left(\color{#CCC}{\frac2x-\frac1{x-\frac{i}{n}}}-\frac1{x+\frac{i}{n}}\right)e^{-ix}\,\mathrm{d}x\tag{3}\\[6pt]
&=\frac\pi4\left(2-e^{-1/n}\right)+\pi\left(-e^{-1/n}\right)\tag{4}\\[12pt]
&=\frac\pi2\left(1-e^{-1/n}\right)\tag{5}\\[9pt]
&\sim\frac\pi2\left(\frac1n-\frac1{2n^2}+\frac1{6n^3}-\dots\right)\tag{6}
\end{align}
$$
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
$$
\gamma^-=\left[-R-\frac{i}{2n},R-\frac{i}{2n}\right]\cup \left(Re^{-i\pi[0,1]}-\frac{i}{2n}\right)
$$
A: As an alternative to Marco's answer, we have
$$\mathcal{L}\left(\sin\frac{x}{n}\right)=\frac{n}{1+n^2 s^2},\qquad \mathcal{L}^{-1}\left(\frac{1}{x(1+x^2)}\right) = 1-\cos(s) \tag{1}$$
hence
$$ I_n = \int_{0}^{+\infty}\frac{n(1-\cos s)}{1+n^2 s^2}\,ds = \int_{0}^{+\infty}\frac{1-\cos\frac{s}{n}}{1+s^2}\,ds = \frac{\pi}{2}-\int_{0}^{+\infty}\frac{\cos\frac{s}{n}}{1+s^2}\,ds \tag{2} $$
where $\int_{0}^{+\infty}\frac{\cos\frac{s}{n}}{1+s^2}\,ds = \frac{\pi}{2} e^{-1/n}$ is a straightforward consequence of the residue theorem.
The full asymptotic expansion of $I_n$ is so given by
$$\boxed{\ I_n = \color{red}{\frac{\pi-1}{2}}+\color{blue}{\frac{\pi}{2}\sum_{m\geq 1}\frac{(-1)^{m+1}}{m!\,n^m}}.\,}\tag{3} $$
