Fourier transform and Limit of complex functions multiplied trigo functions

Question

I would like to calculate the Fourier Transform for $$ \frac{1}{1+x^2}. $$

By definition \begin{align}\DeclareMathOperator{atan}{atan} \int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-i\omega x}dx &= e^{-i\omega x}\atan(x) + i\omega\int_{-\infty}^\infty e^{-i\omega x}\atan(x)dx \\ &= e^{-i\omega x}\atan(x) + i\omega\frac{\atan(x)e^{-i\omega x}}{-i\omega} - i\omega\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-i\omega x}. \end{align}

Thus $$ \int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-i\omega x}dx = \left.\frac{e^{-i\omega x}\atan(x) + i\omega\frac{atan(x)e^{-i\omega x}}{-i\omega}}{1+i\omega}\right |_{-\infty}^{\infty} $$

and now I'm stuck with how I should carry on.

Original question : How can I calculate the following limit $$ \lim_{x\rightarrow\infty}\atan(x)e^{-i\omega x} $$

where $\omega\in \mathbb{R}$.

Answer 1

Except for $\omega=0$ the limit doesn't exist, since $\lim_{x\to\infty}\arctan(x)=\pi/2$ and $$\lim_{x\to\infty}e^{-i\omega x}\overset{\omega\neq0}{=}\lim_{x\to\infty}e^{\pm ix}$$ ($\pm$ dependent on the sign of $\omega$) which doesn't exist as $$e^{\pm ix}=\cos(x)\pm i\sin(x)$$ is a oscillating function.

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Regarding your Edit:

I think you are following a wrong path. Because of the reasons above, you can't do integration by parts. Instead you could try to use $$\frac{1}{1+x^2}=\frac{1}{x+i}\cdot\frac{1}{x-i}$$ and the formular for the Fourier transform of products. With the substitution $u:=x\pm i$ you just have to calculate the Fourier transform of the function $f(u)=1/u$, which has been described here.

Answer 2

$\arctan(x)\to \frac{\pi}{2}+n\pi$ and $e^{-i\omega x}$ has no limit, so the product has no limit.