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Apparently the laplace transform of $sin(8t^3)$ doesn't exist. No program lets me calculate it. I was asked if it existed in an exam and I said yes because it happens to meet all the criteria. Can anyone tell me if it exists or not?

Thanks.

DLV
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$\displaystyle\int_0^\infty e^{-st}\sin(8t^3)\;dt$ isn't an elementary function.

Vladimir Vargas
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  • Hmm. To be honest I don't know what an elementary function is (I'm looking it up), but why can't you say it exists if all criteria are met? The function is continuous, its of exponential order etc. – DLV Oct 31 '14 at 22:10
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    Read through Liouville's Theorem – Vladimir Vargas Oct 31 '14 at 22:12
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    Basically there are functions that aren't elementary functions such as $\displaystyle \int e^{-x^2} ;dx$ – Vladimir Vargas Oct 31 '14 at 22:14
  • Crap :(. I hope my teacher didn't expect me to know this. Thanks though. – DLV Oct 31 '14 at 22:15
  • So the typical conditions for the existence of the laplace transform should also refer to the integral existence? – DLV Oct 31 '14 at 22:17
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    Laplace transform may exist but you there isn't an elementary function for that transform. If the integral converges, the function is of exponential order, piecewise continuous, etc.. then Laplace transform exist. However the integral may be an ugly one. Just like this one. – Vladimir Vargas Oct 31 '14 at 22:25
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    In fact there is a theorem that states that if the function is of exponential order and piecewise continuous on $[0,\infty)$ then the Laplace transform of the function exists. Again, let me stress that this doesn't imply the existence of a elementary function to express the antiderivative. – Vladimir Vargas Oct 31 '14 at 22:26
  • Ok, so so would you say it exists then? It does meet the criteria, doesn't it? – DLV Oct 31 '14 at 23:34
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    It does, indeed. – Vladimir Vargas Oct 31 '14 at 23:43
  • Not sure it is better looking than the Laplace transform definition itself, but according to this, this specific transform should be equal to $\frac{8\sqrt{s}}{3\pi} \int_{0}^{\infty} \frac{1}{(t^2+64).\sqrt{t}} . K_{\frac{1}{3}} \Big( \frac{2}{3} \frac{s^{\frac{3}{2}}} {\sqrt{3t}} \Big) dt$ – edrezen Nov 19 '23 at 17:02