# Is the integral $\intop_{0}^{\infty}\frac{\cos x}{\sqrt{1+x^{3}}}dx$ absolutely convergent, conditionally convergent or divergent?

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?

Thanks.

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

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 $$\left|\frac{\cos x}{\sqrt{1+x^3}}\right|\le\frac{|\cos x|}{\sqrt{1+x^{3}}}.$$