Fourier transform of a Laplace transform Is there an easy way to find the Fourier transform of a Laplace transform of function? 
$$
F[L[f(t)]_{s}]
$$
Where my $f(t)$ is $\sqrt{t}$. However, Before finding the Fourier transform I do the substitution $s = (x^{2}+1)$. I am doing this because I wish to find out the convolution of $1/(x^{2}+1)$ with itself, using Fourier identities for convolution. I.e.
$$
g(x)\star g(x) = F^{-1}[F[g(x)]\cdot F[g(x)]]
$$
Note that my $g(x) = L[\sqrt{t}]_{s}$.
 A: Since you mention $\sqrt{t}$ I assume that your original function is defined
on $[0,\infty )$. Then the complex Laplace transform is
\begin{equation*}
L(z)=\int_{0}^{\infty }dt\exp [izt]f(t),\;{Im}z>0.
\end{equation*}
With $\theta (t)$ the Heaviside step function and setting $z=\omega +i\delta
$ we can write
\begin{equation*}
L(z)=L(\omega +i\delta )=\int_{-\infty }^{\infty }dt\exp [i\omega t]\theta
(t)\exp [-\delta t]f(t),
\end{equation*}
so the Laplace transform coincides with the Fourier transform of $\theta
(t)\exp [-\delta t]f(t)$.
But why not calculate the convolution directly? We have
\begin{equation*}
\frac{1}{x^{2}+1}\ast \frac{1}{x^{2}+1}=\int_{-\infty }^{+\infty }dy\frac{1}{%
y^{2}+1}\frac{1}{(y-x)^{2}+1}
\end{equation*}
which can be written as
\begin{equation*}
\frac{1}{x^{2}+1}\ast \frac{1}{x^{2}+1}=\int_{-\infty }^{+\infty }dy\frac{1}{%
y+i}\frac{1}{y-i}\frac{1}{y-x+i}\frac{1}{y-x-i}
\end{equation*}
Completing the integration interval with a semi-arc in the upper half-plane
you now pick up the residues in $y=i$ and $y=x+i.$
