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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.

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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.

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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.

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    By $x=f^{-1}(t)$ you mean the inverse function? because in this integral it's quite hard, I think, to find the inverse function, without using the Lambert function. – 2017-02-06
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    @TrueTopologist: you do not need to find an explicit inverse, that is the point. It is enough to have suitable grow conditions for $f$ to deduce how fast (or, better, how *slow*) $f^{-1}$ goes to $+\infty$, and that settles the question through Dirichlet's test. – 2017-02-06
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    Oh now I understand, thank you very much. This idea can help me with many integrals like so in the future. – 2017-02-06