Fourier Transform of $f(x)=\frac{1}{1+(x-1)^2}$ and convolution

Question

I was given the function $f(x)=\frac{1}{1+(x-1)^2}$ and asked to calculate its Fourier Transform. I thought about substituting the (x-1) for a variable and calculating it but I couldn't get anywhere. Is there a way to do it that I haven't figured it out? Also, I have to determine the function $h$ knowing that $f(x)*h(x)=\frac{2}{4+x^2}$ and I am having dificulties on it as well. I hope someone would bring me some direction on how solve this.

Answer 1

\begin{align} &\int_{-\infty}^{\infty}e^{-isx}\frac{1}{1+(x-1)^2}dx \\ &=\int_{-\infty}^{\infty}e^{-is(x+1)}\frac{1}{1+x^2}dx \\ &= e^{-is}\int_{-\infty}^{\infty}e^{-isx}\frac{1}{1+x^2}dx \\ &= e^{-is}\int_{-\infty}^{\infty}e^{-isx}\frac{1}{2i}\left[\frac{1}{x-i}-\frac{1}{x+i}\right]dx \\ &= e^{-is}\int_{-\infty}^{\infty}e^{-isx}\frac{1}{2}\left[\int_{0}^{\infty}e^{is(x+i)}ds-\int_{-\infty}^{0}e^{is(x-i)}ds\right]dx \\ &= \frac{e^{-is}}{2}\int_{-\infty}^{\infty}e^{-isx}\left[\int_{-\infty}^{\infty}\chi_{(0,\infty)}(s)e^{isx}e^{-s}ds-\int_{-\infty}^{\infty}\chi_{(-\infty,0)}(s)e^{isx}e^sds\right]dx \\ &= 2\pi\frac{e^{-is}}{2}\left[\chi_{[0,\infty]}(s)e^{-s}-\chi_{(-\infty,0]}(s)e^{s}\right] \\ &= \pi e^{-is}e^{-|s|}\mbox{sgn}(s) \end{align}