As the title suggests, I need help finding $a>0$ for which the following improper integral converges:
$$\int_a^\infty \frac{\mathrm{d}x}{(x^2-a)^{4a}}$$
So, at first I thought I would just do this: $$ f(x)=\frac{1}{(x^2-a)^{4a}}$$ Then I wanted to find a function $$g(x)$$ such that $f(x)\leq g(x)$ for any $x\in [a, \infty)$.
I thought I could use something like $$g(x)=\frac{1}{(a^2-a)^{4a}}$$
but I am not sure if that works?
So if I could prove that $g(x)$ converges for some $a$, I could state that $f(x)$ also converges.
Correct me if I am wrong. Any help would be appreciated.
Your integral converges at "infinity" if $8a>1$, i.e., $a>\frac{1}{8}.$ To see what happens at finite points, first write $$\frac{1}{x^2-a}=\frac{1}{x-\sqrt a} \cdot \frac{1}{x+\sqrt a}.$$ The second fraction does not have any impact on convergence/divergence since it defines a bounded continuous function on $[0,\infty)$. If $\sqrt a<a$, then the improper integral does not have any singularities at finite points from where it follows that the improper integral definitively converges whenever $a>1.$ Suppose now $0<a\leq 1$. Then $\sqrt a\geq a$ and in this case, the condition that the improper integral converges at $x=\sqrt a$ is $4a<1.$ From this we conlcude that the improper integral converges for $a\in (\frac{1}{8},\frac{1}{4})\cup (1,\infty)$.