For what range of $n$ does improper integral $\int_0^{+\infty}\frac{\arctan(x)}{x^n}dx$ converge

Question

Here is the improper integral: $$\int_0^{+\infty}\frac{\arctan(x)}{x^n}dx$$ Determine the range of n to make it converge.

I tried to divide it into two parts: $\int_{0}^{1}\frac{\arctan(x)}{x^n}dx$ + $\int_{1}^{+\infty}\frac{\arctan(x)}{x^n}dx$

For the latter part, $\int_{1}^{+\infty}\frac{\arctan(x)}{x^n}dx$. According to p-test, obviously when $n>1$, it is converge.

For the former part $\int_{0}^{1}\frac{\arctan(x)}{x^n}dx$, I need find the range of n which makes the $\lim_{t \to 0^+}\int_t^1\frac{\arctan(x)}{x^n}dx$ exist.

Then I have no idea how to do that. Could anybody help me? Excuse my English. Thanks!

P.S. The answer is $1<n<2$, but how to...

Answer 1

Note that for $x\to 0^+$

$$\frac{\arctan(x)}{x^n}\sim\frac x{x^n}=\frac 1{x^{n-1}}$$

and $\int_{0}^{1}\frac{1}{x^{n-1}}dx$ converges for $n-1<1$.