How to find the Fourier Transform of $\frac{t}{t^2+9}$ using convolution? I need to find the Fourier Transform of:
$$\frac{t}{t^2+9}$$
I defined $h(t) = t$ and $g(t)=\frac{1}{t^2+9}$.
But, my problem is get the Fourier Transform of $h(t)$. I know that the Fourier Transform of $g(t)$ is $\frac{\pi}{3} e^{-3|w|}$.
Are the functions $h(t),g(t)$ correctly defined?
 A: You want to understand
$$ I(s)=\int \frac t{t^2+9} e^{its}dt$$
Note that the integrand is not in $L^1$ so this needs to be interpreted appropriately, e.g. as an $L^2$ Fourier transform, which is defined as a limit of Fourier transform of $L^1\cap L^2$ functions.
As I mentioned in the comments this formally holds because your $h(t)=t$ Fourier transforms into the derivative of the dirac delta (up to constants).
To make it rigourous one does not need to talk about distributions, you can proceed instead using an approximation to the identity: below, this is the sequence of functions $n \hat\phi(ns)$. Take a smooth cutoff function $\phi_n(t) = \phi(t/n)$ where $\phi$ has compact support and $\phi(0)=1$. Then it makes sense to compute
$$I_n= \int \frac {t\phi_n(t)}{t^2+9} e^{its}dt$$
And we can recover $I$ in the limit $n\to\infty$. By the convolution theorem
\begin{align} 
I_n(s) 
&= \mathcal F(t \phi_n) * \mathcal F\left(\frac1{t^2+9}\right)(s)\\ 
&= -i(\widehat{\phi_n})' * \left(\frac\pi3 e^{-3|\bullet|}\right)(s) \\
&= -in\widehat{\phi}(n\bullet) * \left(\frac\pi3( e^{-3|\bullet|})'\right) (s)
\\
&=-in\widehat{\phi}(n\bullet) * \left(\frac\pi3 e^{-3|\bullet|}\operatorname{sgn}(\bullet)\right)(s)\\
&\to 
-i\frac\pi3 e^{-3|s|}\operatorname{sgn}(s).\end{align}
We also used that $\exp(--3|s|)$ is weakly differentiable, and the scaling property of the Fourier transform. (Your comment copied the answer from W|A wrongly, only $\operatorname{sgn}s$ appears, not $s\operatorname{sgn}s$.)
