Although, the function e^{t^2} is not exponentially bounded and due to linearity of Laplace transform we may write
L {(e^{t^2}}=L{1+t^2+(t^4)/2!+(t^6)/3!+....}
=1/s+(2!)/s^3+(4!)/(2!s^5)+(6!)/(3!s^7)+.....
By using power series of e^{t^2}, perhaps, L {(e^{t^2}} can be computed.
No:
L(e^(t^2))=\int_0^{infty}(e^(t^2)e^{-st})dt=\int_0^{infty}(e^(t^2-st})dt, which is divergent for all s.
On the other hand you cannot commute (in general) infinite sums (or power expansions) with integration. You need dominated convergence hypothesis.
Try Wolframalpha :)
http://www.wolframalpha.com/input/?i=LaplaceTransform%5BExp%5Bt%5E2%5D%2Ct%2Cs%5D
Exponential order is only a sufficient condition for the existence of the Laplace transform. It is not a necessary condition.
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