 # Existence of Laplace Transform Do the Laplace transforms for the following functions exist? Explain your answers. (You do not need to find the transforms , just show if they exist or not) a) f(t)=t^2sin(omega t) b) f(t)=e^{t^2}sin(omega t) Reeves 2020-12-16 Answered
Existence of Laplace Transform
Do the Laplace transforms for the following functions exist? Explain your answers. (You do not need to find the transforms , just show if they exist or not)
a) $f\left(t\right)={t}^{2}\mathrm{sin}\left(\omega t\right)$
b) $f\left(t\right)={e}^{{t}^{2}}\mathrm{sin}\left(\omega t\right)$
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Step 1
To determine:
The existence of Laplace transforms of the functions:
a) $f\left(t\right)={t}^{2}\mathrm{sin}\left(\omega t\right)$
b) $f\left(t\right)={e}^{{t}^{2}}\mathrm{sin}\left(\omega t\right)$
Step 2
Definition of existence of Laplace transforms :
Let f be a piece – wise continuous function in $\left[0,\mathrm{\infty }\right)$ and is of exponential order. The Laplace transformation F (s) of f exists for some $s>c$, where c is a real number depends on f.
Step 3
a)Since, the function $f\left(t\right)={t}^{2}\mathrm{sin}\left(\omega t\right)$ can be expressed as an exponential order. Hence the Laplace transform exists for the function $f\left(t\right)={t}^{2}\mathrm{sin}\left(\omega t\right)$
Step 4
b)The $f\left(t\right)={e}^{{t}^{2}}\mathrm{sin}\left(\omega t\right)$ cannot be expressed as an exponential order, since it grows too fast than the exponential function and bounded by ${e}^{-t}$ and hence, the integral doesn’t converge
it would at some point exceed the ${e}^{-st}$ term.
This is the reason for the Laplace transformation of this function is undefined.

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