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I tried to directly compute the integral, but I was unable to and wolfram alpha says it cannot find the answer in terms of elementary integrals. How can I prove the existence of the Laplace transform without directly computing it? Helpful hints leading me to the answer will be accepted as the answer(and I actually prefer this to outright stating it. Any help is appreciated!

I messed up and put the wrong function in the title initially- it should be the derivative of the function I gave. I am sorry for wasting people's time.

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  • $\begingroup$ It'd probably be best if you provided us with the exact integral that you're trying to show exists. We can make a reasonable guess as to what you're talking about when you ask about the Laplace transform if we assume this is at an introductory level, but clarity is always preferred over guessing, no matter how minimal the guessing is. e.g. I'm assuming the definition you're using, with possible different variable names, is $\mathcal{L}\{f\}(\alpha) = \int_0^{+\infty} f(t) e^{-\alpha t} \, dt$? $\endgroup$ – tilper Jun 6 '17 at 17:46
  • $\begingroup$ @tilper Yes. That is the definition I'm using $\endgroup$ – Aperson123 Jun 6 '17 at 17:49
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    $\begingroup$ $\sin(e^{t^2})$ is continuous and bounded. And $\int_0^\infty e^{-st}\,dt<\infty$ for $\text{Re}(s)>0$. $\endgroup$ – Mark Viola Jun 6 '17 at 17:49
  • $\begingroup$ @Mark Viola I put the wrong function in-it's changed now. $\endgroup$ – Aperson123 Jun 6 '17 at 17:54
  • $\begingroup$ See my solution for an explanation. $\endgroup$ – Mark Viola Jun 6 '17 at 18:06
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Note that $f(t)=\frac{d}{dt}\sin(e^{t^2})=2te^{t^2}\cos(e^{t^2})$. Then, we have

$$\begin{align} \int_0^R f(t)e^{-st}\,dt&=\int_0^R e^{-st}\frac{d}{dt}\sin(e^{t^2})\,dt\\\\ &e^{-sR}\sin(e^{R^2})-\sin(1)+s\int_0^R \sin(e^{t^2})e^{-st}\,dt \end{align}$$

The function $\sin(e^{t^2})e^{-st}\in C[0,R]$ and is therefore integrable on $[0,R]$ for $\text{Re}(s)>0$. Moreover, we can write for $\text{Re}(s)>0$

$$\begin{align} \left|\int_0^R \sin(e^{t^2})e^{-st}\,dt\right|&\le \int_0^R e^{-st}\,dt\\\\ &=\frac{1-e^{-sR}}{r}\to \frac1s \end{align}$$

Hence for $\text{Re}(s)>0$, $\int_0^\infty \sin(e^{t^2})e^{-st}\,dt$ exists as an improper Riemann integral and is finite and $\int_0^\infty 2te^{t^2}\cos(e^{t^2})e^{-st}\,dt$ also exists and is finite as an improper Riemann integral.

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  • $\begingroup$ One might argue that for any $s>0$ the function $2t e^{t^2}\cos(e^{t^2})e^{-st}\,dt$ does not belong to $L^1(\mathbb{R}^+)$, so it is best to state that we are talking about improper Riemann integrability. $\endgroup$ – Jack D'Aurizio Jun 6 '17 at 19:45
  • $\begingroup$ Jack, is that not clear from the development inasmuch as the limit as $R\to \infty$ is explictly taken? $\endgroup$ – Mark Viola Jun 6 '17 at 19:50
  • $\begingroup$ It probably is, you're right, sorry for the useless nitpicking. $\endgroup$ – Jack D'Aurizio Jun 6 '17 at 19:55
  • $\begingroup$ Jack, I added some language to help clarify. You weren't uselessly nitpicking. Thanks. $\endgroup$ – Mark Viola Jun 6 '17 at 19:56
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Notice that

$$2te^{t^2}\cos(e^{t^2}) = \dfrac{d}{dt} \sin(e^{t^2})$$

And that if $$\mathscr{L}\{f(t)\} = F(s)$$ then $$\mathscr{L}\{f'(t)\} = sF(s) -f(0)$$

and also maybe apply @Mark_Viola's rationale for why F(s) must exist.

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    $\begingroup$ Does this mean that the laplace transform of the derivative of any function with a laplace transform must always exist? Thank you for taking the time to answer my question. $\endgroup$ – Aperson123 Jun 6 '17 at 19:25
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    $\begingroup$ To the best of my understanding, as long as f(t) was differentiable on $[0, \infty) $, then yes. (I'm intentionally avoiding the unit step and dirac delta generalized function here, but it should work for them too). But I'm just an EE. I'm sure there are math professors here that might know of more nuances. $\endgroup$ – Andy Walls Jun 6 '17 at 19:42
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This is an answer to the original question about the Laplace transform of $\sin(e^{t^2})$.
I recommend the OP to try avoiding chamaleon questions in the future and simply ask a new question.


For any $s>0$ the integral

$$ (\mathcal{L}f)(s)=\int_{0}^{+\infty}\sin(e^{t^2})e^{-st}\,dt $$ is finite since $\left|\sin(r)\right|\leq 1$ for any $r\in\mathbb{R}$ and $\int_{0}^{+\infty}e^{-st}\,dt = \frac{1}{s}$ is finite.
We have $$ (\mathcal{L}f)(s)=\frac{1}{2}\int_{0}^{+\infty}\frac{\sin(e^t)}{\sqrt{t}}e^{-s\sqrt{t}}\,dt = \frac{1}{2}\int_{1}^{+\infty}\frac{\sin(u)\,du}{ u\,e^{s\sqrt{\log u}}\sqrt{\log u}}$$ that is convergent by Dirichlet's test too, since $\sin u$ has a bounded primitive while $u\,e^{s\sqrt{\log u}}\sqrt{\log u}$ is an increasing function on $(1,+\infty)$.

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  • $\begingroup$ My question had the wrong function for less than 90 seconds. Making a new question would have been excessive. Couldn't we conclude the existence of the Laplace transform simply from the fact that the function is of exponential order? I think that would be a simpler way to do it, as it appeals to a standard result in undergrad D.E. textbooks. Thank you for answering my question. $\endgroup$ – Aperson123 Jun 7 '17 at 19:23
  • $\begingroup$ @Aperson123: it would not be excessive, since in those 90 seconds many users worked for nothing or almost nothing, and that is not fine. $\sin(e^{t^2})$ and its derivative are not entire functions of order $1$, they have order $2$, hence there is no shortcut, the "standard result" does not apply here. $\endgroup$ – Jack D'Aurizio Jun 7 '17 at 19:42

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