Integral convergence with parameter Study for which $\beta>0$ the following integral converge:
$$\int_0^\infty \frac{\arctan(\ln^2x)}{\left|\beta-x\right|^e} dx$$
My try:
I managed only to see that:
 $$\int_a^\infty \frac{\arctan(\ln^2x)}{\left|\beta-x\right|^e}<\frac{\pi}{2}\int_a^\infty \frac{1}{\left|\beta-x\right|^e}$$ that converge for each $a>\beta$ because $e>1$. How to go ahead?
 A: The integral does not converges if there's a root of the denominator wich is not a root of the numerator and this can only happen if $\beta\neq 1$. To see this, note that we would have $\arctan(\ln^2(\beta))>0$ so it would be an interval $(a,b)$ with $a>0$ containing $\beta$ and a $\varepsilon>0$ such that $\arctan(\ln^2(x)>\varepsilon$ for every $x\in (a,b)$.
It follows that
$$
\int_a^b\frac{\varepsilon}{|\beta-x|^e}dx\leq\int_a^b\frac{\arctan(\ln^2(x))}{|\beta-x|^e}dx\leq \int_0^\infty\frac{\arctan(\ln^2(x))}{|\beta-x|^e}dx
$$
And since $e>1$, the integral on the left does not converge, so neither your integral.
On the other hand, if $\beta=1$, the integral does converges, but it requires more work to see this:
The first derivative of $\arctan( \ln^2(x))$ vanishes at $x=1$ and the value of the second derivative at that point is 2, so by Taylor's theorem we can write for $x$ in some interval $(c,d)$ around 1
$$
\arctan(ln^2(x))=(x-1)^2+o(x)
$$
Where $o(x)$ is a function that satisfies $\lim\limits_{x\to 1}(x-1)^{-3}o(x)=0$.
So
$$
\int_c^d \frac{\arctan(\ln^2(x))}{|\beta-x|^e}dx =\int_c^d |x-1|^{e-2}+\frac{o(x)}{|x-1|^e} dx<\infty
$$
Since $e-2<1$ and  $\frac{o(x)}{|x-1|^e}$ is bounded in this interval.
Since the integral converges on $(0,c)$ and on $(d,\infty)$ (you have proved this in your try) then the integral converges on $(0,\infty)$.
