$\int_1^\infty \sin(e^x(x-2))dx$ converges or diverges Is the following integral converges or diverges? :
$$\int_1^\infty \sin(e^x(x-2))dx$$
I tried to use the assignment $t=e^x$ and I got:
$$\int_1^\infty \sin(e^x(x-2))dx=\int_e^\infty \frac{\sin(t(\ln\ t-2))}{t}dt$$
I thought of comparing it to $\int_1^\infty \frac{1}{x}dx$ and I am stuck.

Set $e^{x}\left(x-2\right)=t.
 $ We get $$I=\int_{1}^{\infty}\sin\left(e^{x}\left(x-2\right)\right)dx=\int_{-e}^{\infty}\frac{\sin\left(t\right)}{t}\frac{W\left(t/e^{2}\right)}{1+W\left(t/e^{2}\right)}dt
 $$ where $W\left(x\right)
 $ is the Lambert W function. Now since $$\frac{W\left(t/e^{2}\right)}{1+W\left(t/e^{2}\right)}\sim1
 $$ as $t\rightarrow\infty
 $ we have that $$I\sim\int_{-e}^{\infty}\frac{\sin\left(t\right)}{t}dt=\int_{-e}^{0}\frac{\sin\left(t\right)}{t}dt+\int_{0}^{\infty}\frac{\sin\left(t\right)}{t}dt=\textrm{Si}\left(e\right)+\frac{\pi}{2}
 $$ so the integral converges.
