# Analyzing the convergence of an improper integral

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!

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(1) Don't use the symbol of norm when you actually mean absolute value; (2) $\,x>0\Longrightarrow x^{5/2}>x^{1/2}\Longleftrightarrow x^5>x\Longleftrightarrow x>1\,$ , what makes your last inequality in the fourth row of equations wrong. –  DonAntonio Dec 12 '12 at 3:03
I can see your point... so how would you reduce the last expression, in order to have $p < 1$, if such $p$ exists? –  pmartelletti Dec 12 '12 at 3:16
because if i cannot find a function g that converge, i cannot say anything for f (by these, i mean that if g diverge, it does not mean the f also does... ). I'm ok? –  pmartelletti Dec 12 '12 at 3:33