Discuss the convergence of $\int_0^\infty x \sin e^x \, dx$

HINT:

Let $x=\log y$. Then, $dx=\frac1y \,dy$ and

$$\int_0^{\infty}x\sin(e^x)\,dx=\int_{1}^{\infty}\frac{\log(y)\sin(y)}{y}\,dy$$

Now note that that the integrand is $\sin y$ times a function that monotonically decreases to $0$ (for $y\ge e$). Given that the integral of the sine function is bounded on any interval, finish by appealing to Abel's Test.

Let $y=e^x$. Then $$ \int_0^\infty x \sin e^x \: dx=\int_1^\infty \frac{\ln y}{y}\sin y\:dy=\int_{1}^{\pi}\frac{\ln y}{y}\sin y\:dy+\sum_{n=1}^{\infty}\int_{n\pi}^{(n+1)\pi}\frac{\ln y}{y}\sin y\:dy $$ Since $\frac{\ln y}{y}\to0$ as $y\to\infty$ and is monotonic decreasing, as well as $|\int_{n\pi}^{(n+1)\pi}\sin y\:dy|=2$ $$ \sum_{n=1}^{\infty}\int_{n\pi}^{(n+1)\pi}\frac{\ln y}{y}\sin y\:dy=\sum_{n=1}^{\infty}(-1)^na_n $$ is an alternating series with $$ 2\frac{\ln (n+1)}{n+1}\leqslant a_n\leqslant 2\frac{\ln n}{n}\quad\text{and }\quad a_n=O(\frac{\ln n}{n})\to0 $$ as $n\to\infty$. So it is converges by Leibniz Criterion. Since $$ \frac{\ln n}{n}=O(\frac1{n^{1-\epsilon}}) $$ It is not absolute convergent.