We showed in lecture: The fourier transform of $f(t):=\frac{1}{t^2+a^2}$ is $\hat f(x)=\frac{\sqrt{\frac{\pi}{2}}e^{-a|x|}}{a}$.
I want to determine the Fourier transform of $t \mapsto \frac{4}{t^2+2t+3}$ and $t \mapsto \frac{4t}{t^2+2t+3}$
Let $g(t)=\frac{4}{t^2+2t+3}=4\frac{1}{(t+1)^2+2}$. Therefore $g(t)=4f(t+1)$ and $a=\sqrt 2$. This means $\hat g(x)=4e^{ix}\frac{\sqrt{\frac{\pi}{2}}e^{-\sqrt{2}|x|}}{\sqrt{2}}=2\sqrt{\pi}e^{ix-\sqrt{2}|x|}$.
Let Let $h(t)=\frac{4t}{t^2+2t+3}=4i(-it)\frac{1}{(t+1)^2+2}=4i(-it)f(t+1)$. Therfore $\hat h(x)=4ie^{ix}\hat f'(x)=4ie^{ix}(\frac{\sqrt{\frac{\pi}{2}}e^{-\sqrt 2|x|}}{\sqrt 2})'$
Is that correct? I am specifically unsure if $\hat h(x)$ is correct.
I will be detailed.
Assume that $$ \widehat{f}(x)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}f(t)e^{-itx}dt\tag 1 $$ is the Fourier transform of $f(t)$ . Then if $g(t)=4f(t+1)$, the Fourier transform of $g$ is by definition (we make the change of variable $t\rightarrow 1-t$ in the integral): $$ \widehat{g}(x)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}g(t)e^{-itx}dt=4\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}f(t+1)e^{-itx}dt= $$ $$ =4\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}f(t)e^{-i(t-1)x}dt=4e^{ix}\widehat{f}(x).\tag 2 $$ Hence from the well known Fourier transform pair ($a$ real positive): $$ f(t)=\frac{1}{t^2+a^2}\leftrightarrow\widehat{f}(x)=\sqrt{\frac{\pi}{2}}\frac{e^{-a|x|}}{a} $$ and relation (2) we get with $a=\sqrt{2}$ $$ \frac{4}{t^2+2 t+3}=4\frac{1}{(t+1)^2+(\sqrt{2})^2} \leftrightarrow 4e^{ix}\frac{\sqrt{\frac{\pi}{2}}e^{-\sqrt{2}|x|}}{\sqrt{2}}=2\sqrt{\pi}e^{ix}e^{-\sqrt{2}|x|}.\tag 3 $$ For the function $h(x)=tg(t)$, we have
$$ \frac{1}{\sqrt{2\pi}}\int^{N}_{-N}h(t)e^{-itx}dt=\frac{1}{\sqrt{2\pi}}\int^{N}_{-N}tg(t)e^{-itx}dt=\frac{1}{-i\sqrt{2\pi}}\int^{N}_{-N}(-it)g(t)e^{-itx}dt= $$ $$ =\frac{1}{-i\sqrt{2\pi}}\int^{N}_{-N}g(t)\frac{d}{dx}\left(e^{-itx}\right)dt=\frac{1}{-i\sqrt{2\pi}}\frac{d}{dx}\int^{N}_{-N}g(t)e^{-itx}dt.\tag 4 $$ If $x\neq 0$, we have $$ \lim_{N\rightarrow\infty}\left|\int^{N}_{-N}h(t)e^{-it x}dt\right|<\infty\tag 5 $$ and $$ \int^{\infty}_{-\infty}\left|g(t)\right|dt<\infty.\tag 6 $$ Since the limit when $N\rightarrow \infty$, for $x\neq0$ exists for both sides of (4), we can interchange the order of limit and differentaition and get $$ \widehat{h}(x)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}h(t)e^{-itx}dt=i\frac{d}{dx}\widehat{g}(x)=i\widehat{g}'(x)=2i\sqrt{\pi}\frac{d}{dx}\left(e^{ix}e^{-\sqrt{2}|x|}\right)\textrm{, }x\neq 0.\tag 7 $$ I will prove next (5).
The function $h(t)=\frac{4t}{t^2+2t+3}=\frac{4t}{(t+1)^2+2}$ is clearly an $L_2\left(\textbf{R}\right)$ function i.e. $\int^{\infty}_{-\infty}|h(t)|^2dt<\infty$. Because $$ \int^{\infty}_{-\infty}\left|h(t)\right|^2dt=\int^{\infty}_{-\infty}\frac{16|t|^2}{|(t+1)^2+2|^2}dt=16\int^{\infty}_{-\infty}\left(\frac{|t+1-1|}{|t+1|^2+2}\right)^2dt\leq $$ $$ \leq 16 \int^{\infty}_{-\infty}\left(\frac{|t+1|+1}{|t+1|^2+2}\right)^2dt\leq 16\int^{\infty}_{-\infty} \left(\frac{|t|+\sqrt{2}}{|t|^2+(\sqrt{2})^2}\right)^2dt\leq $$ $$ \leq 16 \int_{\textbf{R}}\frac{2}{|t|^2+2}dt<\infty.\tag 8 $$ Where we have used the inequality $$ \left(\frac{|x|+|y|}{|x|^2+|y|^2}\right)^2\leq \frac{2}{|x|^2+|y|^2}\textrm{, }x,y\in\textbf{R}\textrm{ such that }|x|+|y|\neq0.\tag 9 $$ Hence (from the Plancherel theory of $L_2(\textbf{R})$): Every function that belongs to $L_2(\textbf{R})$ have unique Fourier transform on $L_2(\textbf{R})$.
Since the Placherel theory is valid, $\widehat{h}(x)$ is defined almost everywere. The only discontinuity here is in $x=0$.