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

  • $\begingroup$ 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. $\endgroup$ Aug 19, 2020 at 10:37
  • $\begingroup$ And you cannot use the LCT as the integrand isn't positive... $\endgroup$
    – DonAntonio
    Aug 19, 2020 at 10:46

2 Answers 2


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

  • 1
    $\begingroup$ Why can we use the comparison test for the second improper integral? It’s not positive $\endgroup$
    – FreeZe
    Aug 19, 2020 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 $|\sin(1/x)|\lt1/x$. $\endgroup$ Aug 19, 2020 at 11:19

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.

  • $\begingroup$ (+1) Note that $\frac{u+\sqrt{u^2-4}}{2\left(u^2-4\right)}=\frac1{2(u+2)}+\frac1{u^2-4}+\frac1{2\sqrt{u^2-4}}$, which, being the sum of three decreasing functions, is decreasing. $\endgroup$
    – robjohn
    Jan 30 at 13:46

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