Evaluating $\lim_{a \to \infty}\int_{0}^\infty\frac{\arctan x}{1+x^a}\,dx$ This is out of pure curiosity. I'm not sure how to start even remotely evaluating this:
$$\lim_{a\to\infty} \int_{0}^{\infty}\frac{\arctan(x)}{1+x^a}\,dx $$
but I think some of you guys might have a better time. Whenever plugged into a calculator you get approximately $0.438\,824\,573\,117$.
Also, I guess one could also rewrite the integral as
$$\lim_{a\to\infty} \int_{0}^{1}\frac{\arctan(x)}{1+x^a}\,dx $$
since when $x>1$, $f(x)$ tends to zero as $a\to\infty$.
There very well could be no closed form of an answer, but I would be interested to see if there is.
Thanks in advance!

By Dominated Convergence, this is just the same as the definite integral of $\arctan(x)$ from 0 to 1. As you say, when $x>1$, the integrand goes to 0, and the denominator tends to 1 when $x\leq 1$. 
