Determine if $ \intop_{1}^{\infty}\frac{1}{\sqrt{x}}\sin\left(x+\frac{1}{x}\right) \mathrm{d}x$ converge I have to determine if $\displaystyle \int_{1}^{\infty}\frac{1}{\sqrt{x}}\sin\left(x+\frac{1}{x}\right) \mathrm{d}x$ converge/diverge.
My intuition is that the integral converge, because $\displaystyle\int_{1}^{\infty}\frac{1}{\sqrt{x}}\sin\left(x\right)\mathrm{d}x$ converge from Dirichlet's test, therefore the addition of $ \frac{1}{x} $ shouldnt be much of a difference for $ x\to\infty $.
I guess the right way to prove it is to show that $\displaystyle \int_{1}^{u}\sin\left(x+\frac{1}{x}\right)\mathrm{d}x $ is bounded for any $ u $, and then I could use Dirichlet's test. I tried and couldn't prove it.
Also, I'd like to hear what you think about my proof that the integral  $ \displaystyle \int_{0}^{1}\frac{\sin\left(x+\frac{1}{x}\right)}{\sqrt{x}}\mathrm{d}x $ converge.
I used the limit comparison test in the following way:
$ \displaystyle \lim_{x\to0}\frac{\frac{|\sin\left(x+\frac{1}{x}\right)|}{x^{0.5}}}{\frac{1}{x^{0.8}}}=0 $
and since $ 0.8 <1 $ the integral $ \displaystyle \int_{0}^{1}\frac{1}{x^{0.8}}\mathrm{d}x $ converge, thus the integral $\displaystyle \int_{0}^{1}\frac{\sin\left(x+\frac{1}{x}\right)}{x^{0.5}} \mathrm{d}x$ absolutly converge.
I'll appreciate some help here. Thanks in advance
 A: Start with the angle addition formula:
$$\int_1^\infty{1\over\sqrt x}\sin\left(x+{1\over x}\right)\,dx=\int_1^\infty{1\over\sqrt x}\sin x\cos(1/x)\,dx+\int_1^\infty{1\over\sqrt x}\cos x\sin(1/x)\,dx$$
and note that the second improper integral is convergent since $\sin(1/x)\lt1/x$ (for $x\gt0$) and $\int_1^\infty{1\over x^{3/2}}\,dx$ converges. So it remains to show that the first improper integral is also convergent.
To do this, use integration by parts with $u=\cos(1/x)/\sqrt x$ and $dv=\sin x\,dx$, so that $du=(\sin(1/x)/x^{5/2}-\cos(1/x)/(2x^{3/2}))dx$ and $v=-\cos x$:
$$\begin{align}
\int_1^\infty{1\over\sqrt x}\sin x\cos(1/x)\,dx
&={-\cos(1/x)\cos x\over\sqrt x}\Big|_1^\infty+\int_1^\infty{\sin(1/x)\cos x\over x^{5/2}}\,dx+\int_1^\infty{\cos(1/x)\cos x\over2x^{3/2}}\,dx\\
&=\cos^21+\int_1^\infty{\sin(1/x)\cos x\over x^{5/2}}\,dx+\int_1^\infty{\cos(1/x)\cos x\over2x^{3/2}}\,dx
\end{align}$$
where the final two improper integrals are again convergent.
As for the improper integral from $0$ to $1$, the OP's proof is OK but more complicated that necessary; it suffices to note that ${|\sin(x+1/x)|\over\sqrt x}\le{1\over\sqrt x}$.
A: You may just let $x+\frac{1}{x}=z$ and get
$$ \int_{1}^{+\infty}\sin\left(x+\frac{1}{x}\right)\frac{dx}{\sqrt{x}}=\int_{2}^{+\infty}\sin(z)\underbrace{\frac{\sqrt{z+\sqrt{z^2-4}}}{\sqrt{2}\sqrt{z^2-4}}}_{g(z)}\,dz $$
where $g(z)$ behaves like $\frac{C}{\sqrt{z-2}}$ in a right neighbourhood of $z=2$ and it is decreasing over $z>2$, since
$$ g(2\cosh t) = \frac{e^{t/2}}{e^t-e^{-t}}=\sum_{n\geq 0}\exp\left(-\left(2n+\frac{1}{2}\right)t\right)$$
is clearly decreasing on $\mathbb{R}^+$. It follows that you may apply Dirichlet's lemma here as well.
