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!
 A: 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 $x\leq 1$. 
A: One has, for $a>1$,
$$
0<\int_{1}^\infty\frac{\arctan x}{1+x^a}dx\le \frac \pi2 \int_{1}^\infty\frac{1}{x^a}dx\le \frac \pi2 \frac{1}{a-1}
$$ giving 
$$
\lim_{a \to \infty}\int_{1}^\infty\frac{\arctan x}{1+x^a}dx=0.
$$ There exists  $c \in (0,1)$ such that
$$
\lim_{a \to \infty}\int_{0}^1\frac{\arctan(x)}{1+x^a}dx=\lim_{a \to \infty}\left(\frac{1}{1+c^a}\int_{0}^1\arctan(x)\,dx\right)=\int_{0}^1\arctan(x)\,dx.
$$
Thus
$$
\lim_{a \to \infty}\int_{0}^\infty\frac{\arctan x}{1+x^a}dx=\int_{0}^1\arctan x\,dx.
$$ 
Integrating by parts, one gets
$$
\begin{align}
\int_{0}^1\arctan x \,dx& =\left[x\frac{}{}\arctan x\right]_0^1
-\frac12\int^1_0 \frac{2x}{1+x^2}\,dx
=\frac \pi4-\frac{\ln 2}2.
\end{align}
$$ Finally

$$
\lim_{a \to \infty}\int_{0}^\infty\frac{\arctan x}{1+x^a}dx=\frac \pi4-\frac{\ln 2}2.
$$

