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I have to find the Laplace transform of

$$e^{t^3} u(t)$$

and I know that $u(t)$ will just change the integral from negative infinity to positive infinity to $0$ to positive infinity, but I'm stuck with what to do after that since $e$ is raised to a variable cubed. Help?

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    $\begingroup$ The LT of $e^{t^3}$ does not exist. $\endgroup$
    – Mark Viola
    Feb 9, 2016 at 22:05
  • $\begingroup$ why doesn't it exist? $\endgroup$
    – dms94
    Feb 9, 2016 at 22:05
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    $\begingroup$ $e^{t^3-st}\to \infty$ as $t\to \infty$. $\endgroup$
    – Mark Viola
    Feb 9, 2016 at 22:06
  • $\begingroup$ Formatting tips here. $\endgroup$
    – Em.
    Feb 9, 2016 at 22:15
  • $\begingroup$ try to find the expression for $\int_0^\infty e^{a t^3} e^{-st} dt$ which clearly exists for any $Re(a) < 0$, and see if it makes sense to consider its analytic continuation for $Re(a) \ge 0$ $\endgroup$
    – reuns
    Apr 17, 2016 at 15:18

1 Answer 1

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The Laplace transform existence theorem states that, if $f(t)$ is piecewise continuous on every finite interval in $[0,\infty)$ and there exist constants $M$ and $a$ such that $$|f(t)|\le M\mathrm e^{at} \qquad\text {for sufficiently large }t \in [0,\infty)$$ then $\mathcal L\{f(t)\}=F(s)$ exists for all $s>a$.

So, a function $ f(t)$ has a Laplace transform whenever it is of exponential order.

The function $\mathrm e^{t^3}$ is not of exponential order because $$ \lim_{t\to\infty}\frac{\mathrm e^{t^3}}{M\mathrm e^{at}}=\infty\qquad\forall M,\, a\in\Bbb R $$

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