What about the convergence of $\int_1^\infty \frac{e^{\sin x}\sin 2x}{x}dx$?

Question

What about the convergence of $\int_1^\infty \frac{e^{\sin x}\sin 2x}{x}dx$?

Clearly, $\int_1^\infty \frac{\sin 2x}{x}dx$ is convergence, how to attack $e^{\sin x}$? $e^{\sin x}\cos x\cdot \frac{2\sin x}{x}$, oh.

Answer 1

Hint. Let $M>1$. One may first integrate by parts, \begin{align} \int_1^M \frac{e^{\sin x}\sin 2x}{x}\,dx&=2\int_1^M \frac{\cos x \:e^{\sin x}\cdot\sin x}{x}\,dx \\\\&=2\left[\frac{e^{\sin x}\cdot\sin x}{x}\right]_1^M-2\int_1^M \frac{e^{\sin x}\cdot\cos x}{x}\,dx+2\int_1^M \frac{e^{\sin x}\cdot\sin x}{x^2}\,dx \end{align} As $M \to \infty$, it is not hard to see that the integrated part and the latter integral converge.

One may observe that, by integrating by parts once more, $$ \int_1^M \frac{e^{\sin x}\cdot\cos x}{x}\,dx=\left[\frac{e^{\sin x}}{x}\right]_1^M+\int_1^M \frac{e^{\sin x}}{x^2}\,dx, $$and as $M \to \infty$ the right hand side converges.

The OP integral is then convergent.