Could you help me with this improper integral please? I have tried initially to separate it into the integral from 0 to 1, and 1 to infinity and then use the comparison test, but if i use $ 5x^2 - x +2\over x^7$ it wont always be larger than the original since $ \left | \cos x \right | $ can ocasionally be =0 leaving the dominating factor to be $ x^6$.
$$ \int_0^\infty {5x^2 -x +2\over x^\frac 13 + x + x^6 + x^7\left | \cos x \right |} $$
Thanks a lot!
Let $f(x)$ denote the integrand. On $[0,1],$ $|f(x)| \le 8/x^{1/3}.$ On $[1,\infty),$ $|f(x)|\le (8x^2)/x^6 = 8/x^4.$ Since
$$\int_0^1 (1/x^{1/3})\,dx < \infty,\, \int_1^\infty (1/x^4)\,dx < \infty,$$
$\int_0^\infty |f(x)|\,dx < \infty,$ hence $\int_0^\infty f(x)\,dx$ converges.