Improper integral comparison test of $\int _0^1\frac{\cos\left(x\right)}{x^{1/3}}dx\:$

Question

I'm trying to find the convergence/divergence of this integral but I can't seem to find an integral to compare to

$$\int _0^1\frac{\cos\left(x\right)}{x^{1/3}}dx$$

I tried to compare to integral of $\cos(x)$ but I don't get anywhere

Answer 1

The only impropriety is near $0$ and since $\displaystyle \lim_{x\to 0}\frac{\frac{\cos x}{x^{1/3}}}{\frac{1}{x^{1/3}}}=1$. So $\displaystyle \frac{\cos x}{x^{1/3}}$ and $\displaystyle \frac{1}{x^{1/3}} $ they're asymptotically equivalent near $0$. Therefore the convergence is follows since $\displaystyle \int_{0}^{1}\frac{1}{x^{1/3}}\, {\rm d}x$ converges.

Moreover, with change of variables $\displaystyle u=x^{1/3}$ we have $\displaystyle u'(x)=\frac{1}{3}x^{-2/3}$ the integral is the same that $\displaystyle 3\int_{0}^{1}u\cos u^{3}\, {\rm d}u$ then IBP with $\displaystyle U=u$ and $\displaystyle V'=\cos u^{3}$ and power series for $\displaystyle \cos u^{3}$ we get a numerical approximation: $\displaystyle \int_{0}^{1}\frac{\cos x}{x^{1/3}}\, {\rm d}x\approx 1.32122$.