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