Prove the existence of Laplace transform of $2te^{t^2}\cos(e^{t^2})$

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.

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 $\text{Re}(s)>0$

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

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.

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.