How $\int_{-\infty}^{\infty}\frac{dx}{1+x^2}$ exists? How $$\int_{-\infty}^{\infty}\frac{dx}{1+x^2}$$ exists? 
It is difficult question to me. i have tried to evaluate by using fact that $$\int_{-\infty}^{\infty} f(x) \ dx =\int_{-\infty}^{0} f(x)\, dx + \int_{0}^{\infty} f(x)\, dx$$ but i have failed in this one.   any hints on this one?
 A: You have
$$\int_{-X}^{X}\frac1{(1+x^2)}dx=2\int_{0}^{X}\frac1{(1+x^2)}dx= 2 \arctan X$$
then use
$$
\lim_{X\to+\infty}\arctan X=\frac{\pi}{2}
$$
A: You were on the right track. We have 
$$\int_{-\infty}^0 \frac{1}{1 + x^2}\, dx = \lim_{a\to -\infty} \int_a^0 \frac{1}{1 + x^2}\, dx = \lim_{a\to -\infty} \arctan x\bigg|_{x = a}^0 = \lim_{a\to -\infty} (-\arctan a) = \frac{\pi}{2}$$
and
$$\int_0^\infty \frac{1}{1 + x^2}\, dx = \lim_{b \to \infty}\int_0^b \frac{1}{1 + x^2}\, dx = \lim_{b\to \infty} \arctan x\bigg|_{x = 0}^b = \lim_{b\to \infty} \arctan b = \frac{\pi}{2}$$
Therefore $\int_{-\infty}^\infty \frac{1}{1 + x^2}\, dx$ exists and has value $\frac{\pi}{2} + \frac{\pi}{2} = \pi$.
A: Here is an alternative approach:$$\int_{-\infty}^\infty \frac{1}{1+x^2}=\oint_C \frac{1}{1+z^2}=2\pi i\mathrm{Res}(f(z),i)= 2\pi i\times \frac{-i}{2}=\pi$$
where $C$ is a counterclockwise contour.
For your follow up question:
$$\int_{-\infty}^{\infty}\frac1{\sqrt{1+x^3}}$$ is not expressible in terms of elementary functions.
