I'm trying to work through an exercise on Fourier sine transform. The function given is $$ f(x) = \frac{\sin(\omega x)}{x^2 + 1}\quad \omega > 0 $$ with Fourier sine transform pair defined as $$ f_s(x) = \sqrt{2}f(x),$$ $$\hat{F}_s(k) = \int_{0}^{\infty}f_s(x) \sin(kx)\mathrm{d}x.$$
I'm told I need to consider two cases: $\omega > k$ and $\omega < k$ and I'm not sure why, I think it is something to do with Jordan's lemma when performing the contour integration.
You have $$ \hat{F}_s(k) = \int_{0}^{\infty}f_s(x) \sin(kx)\mathrm{d}x = \int_{0}^{\infty} \sqrt{2} f(x) \sin(kx)\mathrm{d}x = \sqrt{2} \int_{0}^{\infty} \frac{\sin(\omega x)}{x^2+1} \sin(kx)\mathrm{d}x \\ = \sqrt{2} \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin(\omega x)}{x^2+1} \sin(kx)\mathrm{d}x = \sqrt{2} \frac{1}{4} \int_{-\infty}^{\infty} \frac{\cos(\omega x-k x) - \cos(\omega x+k x)}{x^2+1} \mathrm{d}x \\ = \operatorname{Re} \sqrt{2} \frac{1}{4} \int_{-\infty}^{\infty} \frac{e^{i(\omega-k)x} - e^{i(\omega+k)}}{x^2+1} \mathrm{d}x $$
Now calculate $$ I_\pm := \int_{-\infty}^{\infty} \frac{e^{i(\omega\pm k)x}}{x^2+1} \mathrm{d}x $$ by a standard contour integration along $[-R,R]$ on the real axis and a semicircle. Be careful with whether the semicircle should be in the upper or lower half-plane to make the integral over it vanish when $R\to\infty.$ For that you will have to look at the sign of $\omega\pm k.$