First of all be careful. It is not necessary true that:
$$\int_{-\infty}^{\infty} f(x)dx=\lim_{t \to \infty} \int_{-t}^{t} f(x) dx$$
We need to integrate from left to right:
$$\int_{-\infty}^{\infty} f(x) dx=\lim_{s \to -\infty}\int_{s}^{a} f(x)+\lim_{t \to \infty} \int_{a}^{t} f(x) dx$$
To be convinced that your approach is incorrect as it gives an incorrect answer to some integral so, consider $\int_{-\infty}^{\infty} x^3 dx$.
Try filling that are by taking a interval $x \in [-t,t]$ for your integral bounds and letting $t \to \infty$. Now try taking $x \in [-t,2t]$.
Next to address your question:
$$\lim_{x \to \infty} \arctan(x)$$
The way I view $\arctan(x)$ is a function that given some slope of a ray that points in a direction of the first quadrant and that crosses the origin by, it will return the angle created by the positive $x$ axis and the line. To see this just draw a line, make a right angle, and do some trig. You just just view it as a function that given some slope of a line returns the angle with the positive $x$ axis and the line, but a ray whose endpoint and is at the origin will simplify things. Anyways as $x \to \infty$ that means the slope is approaching $\infty$, i.e. It's line is becoming vertical (in the positive quadrant) and thus $\arctan(x)$ approaching $\frac{\pi}{2}$.

Ignore anything in this picture not in the first quadrant. Picture from geoan.com: http://www.geoan.com/recta/pendiente.html.