As my instructor mentioned more times than could ever seem necessary, arguments of the Laplace transform have to have growth of exponential or less.
I typed this "nonexample" into W|A and noticed that it returned a reasonable answer in terms of Dawson functions and an imaginary term.
Assuming this is not an error, what version of the Laplace transform produces this result? Does the inverse Laplace transform produce the original function? I am curious as to what formula they use to calculate this because I cannot find any literature relating to an extension of the Laplace transform.
This integral definition of the laplace transform does not converge.
Consider the Laplace transform of $e^{x^{2}}$. \begin{align} e^{t^{2}} &\doteqdot \int_{0}^{\infty} e^{-st + t^{2}} dt \\ &\doteqdot \int_{0}^{\infty} e^{(t-s/2)^{2}- s^{2}/4} \ dt \\ &\doteqdot e^{- s^{2}/4} \ \int_{0}^{\infty} e^{(t-s/2)^{2}} \ dt \\ &\doteqdot e^{- s^{2}/4} \ \int_{-s/2}^{\infty} e^{u^{2}} du = e^{-s^{2}/4} \left[ \int_{-s/2}^{0} + \int_{0}^{\infty} \right] e^{u^{2}} du \\ &\doteqdot e^{- s^{2}/4} \ \int_{0}^{s/2} e^{u^{2}} du + i e^{-s^{2}/4} \int_{0}^{\infty} e^{-t^{2}} dt \\ &\doteqdot F\left( \frac{s}{2} \right) + \frac{i \sqrt{\pi}}{2} \ e^{- s^{2}/4} \end{align} where $F(z)$ is Dawson's Integral given by \begin{align} F(z) = e^{- z^{2}} \int_{0}^{z} e^{u^{2}} du. \end{align}