I'm trying to solve the next problem: Determine if $\intop_{0}^{\infty}\frac{\cos x}{\sqrt{1+x^{3}}}dx$ is absolutetly convergent, conditionally covergent or diverges.

I think that the integral is abosolutely convergent and I tried to do this: For all $x\geq0$ is true that $$ \frac{\cos x}{\sqrt{1+x^{3}}}\leq\frac{\mid\cos x\mid}{\sqrt{1+x^{3}}}\leq\frac{1}{\sqrt{1+x^{3}}}\leq\frac{1}{\sqrt{x^{3}}}=\frac{1}{x^{3/2}}. $$

Then, using the fact that $\int_{1}^{\infty}\frac{1}{x^{\alpha}}dx$ is convergent for $\alpha>1$ and the comparison test we can conclude that the integral $\int_{1}^{\infty}\frac{\cos x}{\sqrt{1+x^{3}}}dx$ is absolutely convergent. Also, since the function $f\left(x\right)=\frac{\mid\cos x\mid}{\sqrt{1+x^{3}}}$ is continuous on $[0,\infty)$ then is Riemann integrable on $\left[0,1\right]$. Therefore, $$\int_{0}^{\infty}\frac{\mid\cos x\mid}{\sqrt{1+x^{3}}}dx=\int_{0}^{1}\frac{\mid\cos x\mid}{\sqrt{1+x^{3}}}dx+\int_{1}^{\infty}\frac{\mid\cos x\mid}{\sqrt{1+x^{3}}}dx.$$

And then, $\int_{0}^{\infty}\frac{\mid\cos x\mid}{\sqrt{1+x^{3}}}dx$ is convergent since in the last equality, the two sumands on the right side are finite. Thus, the integral $\int_{0}^{\infty}\frac{\cos x}{\sqrt{1+x^{3}}}dx$ is absolutely convergent.

I don't know if what I did is right. Could you help me checking or giving me some suggestion?


  • $\begingroup$ Where are you having doubts? $\endgroup$
    – Unit
    Jul 6, 2019 at 17:50
  • $\begingroup$ I have a little doubt where I separated the integral as the sum of those two integrals. $\endgroup$
    – Dendrilo
    Jul 6, 2019 at 18:23
  • $\begingroup$ Here:$$\int_{0}^{\infty}\frac{\mid\cos x\mid}{\sqrt{1+x^{3}}}dx=\int_{0}^{1}\frac{\mid\cos x\mid}{\sqrt{1+x^{3}}}dx+\int_{1}^{\infty}\frac{\mid\cos x\mid}{\sqrt{1+x^{3}}}dx.$$ $\endgroup$
    – Dendrilo
    Jul 6, 2019 at 18:25
  • $\begingroup$ That's perfectly fine. You can convince yourself by writing $\int_0^\infty$ as $\lim_{b \to \infty} \int_0^b$. $\endgroup$
    – Unit
    Jul 6, 2019 at 18:41
  • 1
    $\begingroup$ Yes. Thanks @Unit. $\endgroup$
    – Dendrilo
    Jul 6, 2019 at 19:02

1 Answer 1


This is correct. As you correctly noted, the absolute value of the integrand is continuous on $\Bbb R^+$; and thus in particular Riemann-integrable on any (bounded) interval $[0,I]\subset\Bbb R$.

Also, because you want to prove absolute convergence, your first inequality should be stated as $$\left|\frac{\cos x}{\sqrt{1+x^3}}\right|\le\frac{|\cos x|}{\sqrt{1+x^{3}}}.$$

  • $\begingroup$ Alright, thanks. $\endgroup$
    – Dendrilo
    Jul 6, 2019 at 19:08

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