convergence and divergence test on an exponential function I am trying to test the convergence/divergence of this:
$$\int_0^\infty x \sin(e^x) \, dx $$ by any method
I looked through convergence test and integral test
 A: Hint. Integrating by parts, one gets
$$
\int_0^M x \sin(e^x) \, dx =\left[(xe^{-x})(-\cos (e^x)) \vphantom{\frac 1 1} \right]_0^M+\int_0^M (1-x)e^{-x} \cos(e^x) \, dx
$$ and
$$
\left|\int_0^M (1-x)e^{-x} \cos(e^x) \, dx\right|\leq \int_0^M |1-x|e^{-x} \, dx<\infty
$$ giving the convergence of the initial integral.
A: Outline: Use integration by parts, letting $u=\frac{x}{e^x}$ and $dv=e^x\sin(e^x)\,dx$. 
Then $du=(e^{-x}-xe^{-x})\,dx$ and we can take $v=-\cos(e^x)$. 
The rest is straightforward, for $uv$ behaves nicely "at $\infty$", and $\int_0^\infty (e^{-x}-xe^{-x})\cos(e^x)\,dx$ converges, because $|\cos(e^x)|\le 1$.
A: Make the change of variable $x=\log(y)$; then the integral becomes
$$
I=\int_1^\infty \frac{\log(y)\sin(y)}y \, dy
$$
which converges since $\frac{\log(y)}{y}$ tends monotonically (for $y>e$) to $0$. In fact, 
$$
I=\sum_{n=1}^\infty I_n+\int_1^{\pi}f(y)\,dy,\quad I_n:=\int_{n\pi}^{(n+1)\pi}\frac{\log(y)\sin(y)}y \, dy
$$
Then the $I_n$ have alternating signs and the monotonicity $I_{n+1}<I_{n}\leq \pi \log(n)/n$ holds definitively.
