$\int_1^\infty \sin(e^x(x-2))dx$ 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.

In general, if $f(x)$ is an increasing function growing faster than $x$, say $\Omega(x^{1+\alpha})$ with $\alpha>0$, from some point on, then
$$ \int_{1}^{+\infty}\sin(f(x))\,dx $$
is a convergent integral (in the improper Riemann sense) by applying the substitution $x=f^{-1}(t)$ and exploiting Dirichlet's test - $\sin(t)$ is a function with mean zero.
