Convergence of $\int_e^\infty \frac{\ln x\cdot \cos x^2}{(x+1)^{\frac{3}{2}}}$ $$\int_e^\infty \frac{\ln x\cdot \cos(x^2)}{(x+1)^{\frac 3 2}} \, dx$$
After an hour of trying to compare this expression to something, I've put it in WolframAlpha which calculated it precisely so basically that means this integral converges(I think?), but I've failed to prove it. I've tried limiting $cos({x^2})$ with $-1$ and $1$, tried looking at the absolute convergence.
Everything that I found bigger than this always seems to diverge, and anything smaller seems to converge or go to $-\infty$ as $x$ goes to $+\infty$.
Any hints would be appreciated,
thank you in advance!
 A: Note that $|\log(x)|<\frac{x^\alpha}{\alpha}$ for all $\alpha>0$.  Hence, we see that 
$$\left|\frac{\log(x)\cos(x^2)}{(x+1)^{3/2}}\right|\le \frac{1}{\alpha x^{3/2-\alpha}}$$
In particular, if we take $0<\alpha<1/2$, then we see by comparison with $\int_e^\infty \frac{1}{\alpha x^{3/2-\alpha}}\,dx$ that the integral $\int_e^\infty \frac{\log(x)\cos(x^2)}{(x+1)^{3/2}}$ is absolutely convergent.
A: $$\int_e^\infty \frac{\ln(x)\cos(x^2)}{(x+1)^{3/2}}dx \stackrel{u=\ln(x)}{=} \int_0^\infty \frac{ue^u\cos(e^{2u})}{(e^u+1)^{3/2}}dx$$
Now, since $-1 \le \cos(x) \le 1$, simply note that
$$ -\color{red}{\int_0^\infty \frac{ue^u}{(e^u+1)^{3/2}}du}\le\int_0^\infty \frac{ue^u\cos(e^{2u})}{(e^u+1)^{3/2}}du \le \color{red}{\int_0^\infty \frac{ue^u}{(e^u+1)^{3/2}}du}$$
And it is clear that
$$ \sqrt{2}=\int_0^\infty \frac{e^u}{(e^u+1)^{3/2}}du\le\color{red}{\int_0^\infty \frac{ue^u}{(e^u+1)^{3/2}}du} \le \int_0^\infty \frac{ue^u}{(e^u+1)^2}du = \ln(2)$$
We therefore conclude that our original integral is convergent
