I'm new treating with power-law probability distributions and I need to remember the conditions under certain integrals converge. So, let $I_n(a,b)=\int_a^b \frac{1}{x^n}dx$.
Questions.
What are the conditions that should satisfy $n$ in order to $I_n(a,b)$ be a convergent integral when $a=0$ and $b=\infty$?
Are this conditions flexible when $a>0$ and/or $b<\infty$?
Any books references or links will be useful. Preferable not too formal, since I'm a physicist.
If $a$ is a positive real number, then $\int_a^{\infty}\frac{dx}{x^p}$ and $\int_{0}^a\frac{dx}{x^p}$ can be evaluated using the limit definition of the improper (Riemann) integral by finding an antiderivative of $x^{-p}$ and evaluating the resulting limit.
It turns out that $$\int_a^{\infty}\frac{dx}{x^p}<\infty$$ if and only if $p>1$, and $$\int_{0}^a\frac{dx}{x^p}<\infty$$ if and only if $p<1$. Since $$ \int_0^{\infty}\frac{dx}{x^p}=\int_{0}^1\frac{dx}{x^p}+\int_1^{\infty}\frac{dx}{x^p}$$ it follows that there are no values of $p$ for which this integral converges.
In general, for $0\le a < b$, the power rule gives $$ \int_a^b\frac{1}{x^n} = \frac{1}{n-1}\left(\frac{1}{a^{n-1}}-\frac{1}{b^{n-1}}\right)$$ for $n\ne1$ and $\log(b/a) $ for $n=1.$
There are three cases of interest:
If $n>1$ then when $a\rightarrow 0,$ the first term diverges. The second term remains finite as $b\rightarrow\infty$
If $n<1$ then when $b\rightarrow \infty,$ the second term diverges. The first term remains finite as $a\rightarrow 0.$
When $n=1$ then $\ln(b/a)$ diverges if either $a\rightarrow 0$ or $b\rightarrow\infty.$
So:
If $a > 0$ and $b<\infty$ then it always converges.
If $a=0$ but $b < \infty$ then it converges for $n<1$ and diverges for $n\ge 1.$
If $b=\infty$ but $a > 0$ it converges for $n>1$ and diverges for $n\le1.$
If both $a=0$ and $b=\infty$ then it never converges.