1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ The problem is that the use of LCT doesn't address the other singularity, at infinity. But that should work. My advice: try to expand the numerator using the sine addition law. At infinity, estimate $\sin\frac{1}{x}$ and $\cos\frac{1}{x}$. The right estimates should give you the convergence you desire. $\endgroup$ – Paco Adajar Aug 19 at 10:37
  • $\begingroup$ And you cannot use the LCT as the integrand isn't positive... $\endgroup$ – DonAntonio Aug 19 at 10:46
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why can we use the comparison test for the second improper integral? It’s not positive $\endgroup$ – FreeZe Aug 19 at 11:12
  • $\begingroup$ @Waizman, the "best" way for an improper integral to be convergent is for the integral of the absolute value of the integrand to converge. In this case $|(1/\sqrt x)\cos x\sin(1/x)|\lt1/x^{3/2}$, since $|\cos x|\le1$ and $|\sin(1/x)|\lt1/x$. $\endgroup$ – Barry Cipra Aug 19 at 11:19
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.