Does $\int_0^\infty\frac{x^\theta}{1+x^2}\ \mathrm dx$ converge? I'm wondering 
$$\int_0^\infty\frac{x^\theta}{1+x^2}dx$$
equals infinity or just a finite number? Surely when $\theta \geq 1$, it's infinity. But I cannot figure out when less than 1, what will happen.
I think I partially get the answer. Substitute the x with $\tan(t)$, then we get
$$\int_0^{\frac{\pi}{2}}\tan^\theta(t)dt$$, it's infinite for $0<\theta<1$
...actually $$\int_0^{\frac{\pi}{2}}\tan^\theta(t)dt$$ is not infinite. See Rran Gordon's comments.
 A: Just for laughs, let's evaluate this one when $\theta \in (-1,1)$.  The integral may be evaluated by using the residue theorem on the contour integral
$$\oint_C dz \frac{z^{\theta}}{1+z^2}$$
wheere $C$ is a keyhole contour about the positive real axis.  In this case, we have
$$\left ( 1-e^{i 2 \pi \theta}\right ) \int_0^{\infty} dx \frac{x^{\theta}}{1+x^2} = i 2 \pi \left (\frac{e^{i \pi \theta/2}}{2 i} - \frac{e^{i 3 \pi \theta/2}}{2 i}  \right )$$
or
$$\int_0^{\infty} dx \frac{x^{\theta}}{1+x^2} = \frac{\pi}{2 \cos{(\pi \theta/2)}}$$
Note that the integral blows up when $|\theta|=1$.
A: If $-1< \theta < 1$ the integral is finite.  Otherwise, it is not.  If $\theta < -1$ the integral fails to integrate at 0. If $\theta \ge 1$ it fails to integrate at $\infty$.
A: I am sorry that I am using all the concept through words,I actually I don't know how to write integral and other symbols
anyway coming to question for 
Theta<(-1)
 use limit comparisons test with g(x)=gamma(theta+1), where gamma is  the function in gamma function,since limit is infinite implies if intrural(g(x)) diverges then given integral diverges,applying condition for gamma function to divergent,from that we get theta<=(-1)
For proving at -1
