For which values of $m,n \in \mathbb{Z}$ the integral $\int_2^{\infty} \frac{(x-2)^m \arctan(x)}{(x-1)^n} dx $ converges?

Question

For which values of $m,n \in \mathbb{Z}$ the integral $$\int_2^{\infty} \frac{(x-2)^m \arctan(x)}{(x-1)^n} \,dx $$ converges?

It has been a while since I solved this kind of questions, but here I see there is only problem at $\infty$.

So it seems to me that I only need the power of $x$ in the denominator to be bigger then the power of $x$ in the nominator. So it has to be for $m \le n$, am I right here?

Help would be appreciated.

Answer 1

Hint. Note that as $x\to 2^+$, $$ \frac{(x-2)^m \arctan(x)}{(x-1)^n}\sim\frac{\arctan(2)}{(x-2)^{-m}},$$ hence the integral is convergent in $(2,3]$ if and only if $-m<1$, i.e. $m>-1$.

Moreover, as $x\to +\infty$, $$ \frac{(x-2)^m \arctan(x)}{(x-1)^n}\sim\frac{\pi/2}{x^{n-m}}.$$ What is the convergence condition for the integral over $[3,+\infty)$?

What is the final condition of convergence for the integral over $(2,+\infty)$?

Answer 2

Hint: this integral converges iff $\int_2^{\infty} \frac {(x-2)^{m}} {(x-1)^{n}} dx$ converges iff $\int_2^{3} (x-2)^{m}dx$ and $\int_2^{\infty} x^{m-n}dx$ both converge. .