TITLE: Analyzing the convergence of an improper integral
QUESTION [5 upvotes]: I have to analyze the convergence of 
$$\int _{0}^{+\infty} \frac{\cos \left( x\right) -1} {x^{5 / 2}+5x^{3}}\,dx$$
I've rewritten the integral as
$$
\int _{0}^{+\infty} \frac{\cos \left( x\right) -1} {x^{5 / 2}+5x^{3}}\,dx = \int _{0}^{1} \frac{\cos \left( x\right) -1} {x^{5 / 2}+5x^{3}}\,dx + \int _{1}^{+\infty} \frac{\cos \left( x\right) -1} {x^{5 / 2}+5x^{3}}\,dx
$$
For the first part, I can write:
$$
\int _{0}^{1} \frac{\cos \left( x\right) -1} {x^{5 / 2}+5x^{3}}\,dx = \lim_{a \to 0^{+}} \int _{a}^{1} \frac{\cos(x) - 1}{x^{5/2} + 5x^3}\, dx
$$
Then, I search for function $g(x)$ so that $ 0 \leq f(x) \leq g(x)$ for $(0,1]$ in order to use convergence criteria. Then, I narrow the expresion in the integral to find something bigger, but that it converges. I make the following steps:
$$
\frac{||\cos(x) - 1||}{||x^{5/2} + 5x^3||} \leq \frac{2}{||x^{5/2} + 5x^x||} \leq \frac{2}{||x^{5/2}||} \leq \frac{2}{\sqrt{x}}
$$
knowing that $0 < x \leq 1$ .
So, $g(x)$ converge because $\int_0^1\frac{1}{x^p}$, with $p < 1$ converges. Then, by the comparison criteria, 
$$
\int _{0}^{1} \frac{\cos \left( x\right) -1} {x^{5 / 2}+5x^{3}}\,dx
$$
also converges.
Is the reasoning actually correct?
I know I still have to check the second part of the integral, but is more of the same procedure. I just want to check if I'm in the good way!
Thanks a lot for your help!

REPLY [3 votes]: Unfortunately, we have $x^{-5/2}\gt x^{-1/2}$ if $0\lt x\lt 1$, so the approach to treat the first integral does not work. However, the bound 
$$\left|\frac{\cos \left( x\right) -1} {x^{5 / 2}+5x^{3}}\right|\leqslant \frac 2{x^{5/2} }  $$
is sufficient to deduce the convergence of the second integral. 
Back to the first one. Since $$\lim_{x\to 0}\frac{\cos x-1}{x^2}=l$$ for some positive and finite constant $l$, the convergence of 
$\int_0^ 1 (\cos x-1)/ (x^{5/2}+5x^3)\mathrm dx$ is equivalent to that of 
$\int_0^1\mathrm dx/ (x^{1/2}+5x )$, which is convergent since for $x\in(0,1)$, 
$$0\leqslant \frac 1{x^{1/2}+5x }\leqslant \frac 1{x^{1/2}}.   $$