Calculate the definite integral using approximation methods

Question

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

Answer 1

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 $I(\Omega=0) = -\pi + C_1 = 0$.

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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 $$ \int_\mathbb{R}\frac{\sin(\Omega x)}{x(x^2+1)}\mathrm{d}x = \mathcal{Im}\left(\pi i\cdot1 + 2\pi i\left(-\frac{1}{2}e^{-\Omega}\right)\right) = \pi\left(1-e^{-\Omega}\right) $$

Answer 2

$$ \begin{aligned} \int_{-\infty}^{\infty} \frac{\sin (\Omega x)}{x\left(x^2+1\right)} d x =&\int_{-\infty}^{\infty} \frac{x \sin (\Omega x)}{x^2\left(x^2+1\right)} d x \\ = & \underbrace{\int_{-\infty}^{\infty} \frac{\sin (\Omega x)}{x} d x}_{=\pi}- \underbrace{ \int_{-\infty}^{\infty} \frac{x \sin (\Omega x)}{x^2+1} d x}_{J} \\ \end{aligned} $$ 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 $$ \boxed{I=\pi\left(1-e^{-\Omega}\right)} $$