Discuss the convergence of $\int_0^\infty x \sin e^x \, dx$ $$\int_0^\infty x \sin e^x \, dx$$
I have tried applying the Dirichlet test, Comparison Principle, integration by parts and substitution, but all have failed. None of these prove that the integral is divergent though, so I'm not really sure how to show that this converges/diverges.
My work:
Dirichlet: Fails, because neither $f(x)=x$ nor $g(x)=\sin e^x$ goes to zero
Comparison: Fails. $\sin e^x \le1$, therefore $\int_0^\infty x \, dx\ge \int_0^\infty x \sin e^x \, dx$. However, $\int_0^\infty x \, dx$ does not converge, so this idea is unhelpful
IBP: $\int_a^b FG'=(F(b)G(b)-F(a)G(a))-\int_a^bGF'$ $$F=x,\quad F'= dx,\quad G' = \sin e^x, \quad G =\text{?}$$
Substitution: $u(x)=e^x$ $du=e^x$ therefore: $$\int_0^\infty x \sin e^x \, dx=\int_0^{\infty} \ln {u(x)} \sin u(x) \, dx$$ From here, you can use IBP, resulting in: $$F=\ln(u), \quad F'=\frac 1x, \quad G'= \sin u(x), \quad G = -\cos u(x)\cdot u'(x)$$ $$-\ln u(x) \cos u(x) u'(x)|_0^\infty-\int_0^\infty \frac{-\cos u(x) u'(x)}{u(x)}$$ But I feel like this integral is far too complicated for the scope of the question. Additionally, $\ln\infty$ would go to infinity anyway, so I feel like that is not an acceptable way to solve the problem. Graphically, my calculator says that the integral should be equal to 0.411229, a number which appears to have no numerical significance. Is there any other way to integrate this function?
 A: 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.
A: 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.
A: $$
\int_0^{\infty} x\sin e^x dx = - \int_0^{\infty} xe^{-x}d(\cos e^x) \\
=\left[-xe^{-x}\cos e^x \right]_0^{\infty} +\int_0^{\infty}d(xe^{-x})\cos e^x \\
=\int_0^{\infty}(1-x)e^{-x}\cos e^x dx
$$
and
$$
\left|\int_0^{\infty}(1-x)e^{-x}\cos e^x dx \right| \lt \int_0^{\infty}xe^{-x} dx = 1
$$
A: If divergent means that its absolute value does not integrate, note that it is bounded from below by $1/2$ on the intervals $[\ln(2\pi n +\pi/6),\ln(2\pi n+5\pi/6)]$ ($n=1,2,\ldots$).
