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
Best Answer
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 ${|\sin(x+1/x)|\over\sqrt x}\le{1\over\sqrt x}$.