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 $g(x) = L[\sqrt{t}]_{s}$.


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 $y=x+i.$

  • $\begingroup$ The relation between Laplace and Fourier transform you have written does not solve my question. And the second part I do not quite understand how that leads to an answer of that integral. $\endgroup$ – jaydeepsb Aug 26 '14 at 13:29
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    $\begingroup$ Calculation of the last integral is trivial. Close the integration path by a semi-circle in the upper half plane and then obtain the residues. $\endgroup$ – Urgje Sep 12 '14 at 10:05

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