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!