Calculate the definite integral using approximation methods Calculate the integral  $$\int_{-\infty }^{\infty }  \frac{\sin(\Omega x)}{x\,(x^2+1)} dx$$ given $$\Omega >>1 $$
I tried but couldn't find C1
 A: As you solved a differential equation with respect to $\Omega$, $C_1$ is a constant of integration for this variable, that is why you need an initial/boundary condition such as $I(\Omega=0) = -\pi + C_1 = 0$.

Addendum I don't know how you solved the integral $\int_\mathbb{R}\frac{\cos(\Omega x)}{x^2+1}\mathrm{d}x$ in the middle of your derivation, all this kind of integrals can be tackled pretty easily thanks to complex integration and residues.
In your case, you would start from noticing that $\int_\mathbb{R}\frac{\sin(\Omega x)}{x(x^2+1)}\mathrm{d}x = \mathcal{Im}\left(\int_\mathbb{R}\frac{e^{i\Omega z}}{z(z^2+1)}\mathrm{d}z\right)$, whose integrand has three simple poles at $z = 0,\pm i$. The residues of the poles with a non-negative imaginary part are given by
$$
\begin{array}{l}
   \mathrm{Res}_{z=0}\left(\frac{e^{i\Omega z}}{z(z^2+1)}\right) = \lim_{z\rightarrow0} \frac{e^{i\Omega z}}{(z^2+1)} = 1 \\
   \mathrm{Res}_{z=i}\left(\frac{e^{i\Omega z}}{z(z^2+1)}\right) \,= \lim_{z\rightarrow i} \frac{e^{i\Omega z}}{z(z+i)} \;= -\frac{1}{2}e^{-\Omega}
\end{array}
$$
hence finally
$$
\int_\mathbb{R}\frac{\sin(\Omega x)}{x(x^2+1)}\mathrm{d}x = \mathcal{Im}\left(\pi i\cdot1 + 2\pi i\left(-\frac{1}{2}e^{-\Omega}\right)\right) = \pi\left(1-e^{-\Omega}\right)
$$
A: $$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{\sin (\Omega x)}{x\left(x^2+1\right)} d x =&\int_{-\infty}^{\infty} \frac{x \sin (\Omega x)}{x^2\left(x^2+1\right)} d x \\
= &  \underbrace{\int_{-\infty}^{\infty} \frac{\sin (\Omega x)}{x} d x}_{=\pi}- \underbrace{ \int_{-\infty}^{\infty} \frac{x \sin (\Omega x)}{x^2+1} d x}_{J} \\
\end{aligned}
$$
Using contour integration along anti-clockwise direction of the path $$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where }  \gamma_{1}(t)=t+i 0(-R \leq t \leq R)  \textrm{ and } \gamma_{2}(t)=R e^{i t}  (0<t<\pi) ,$$
we have
$$
\begin{aligned}
J & =\operatorname{Im}\left[\operatorname{Res}\left(\frac{z e^{\Omega z i}}{z^2+1}, z=i\right)\right]\\
& =\operatorname{Im}\left(2 \pi i \cdot \lim _{z \rightarrow i} \frac{z e^{\Omega z i}}{z+i}\right) \\
& =\operatorname{Im}\left(2 \pi i \cdot \frac{1}{2} e^{-\Omega}\right) \\
& =\pi e^{-\Omega}
\end{aligned}
$$
We can now conclude that $$
\boxed{I=\pi\left(1-e^{-\Omega}\right)}
$$
