About improper integrals I have the following function:
$$f(x) = \frac{1}{x^\alpha\sqrt{x^2+1}}$$ and I have to determine the convergence of $\int{}{}_0^{+\infty}f(x)dx$ in relation to $\alpha$
I know that $\int_0^{+\infty}f(x) = \int_0^{1}f(x) + \int_1^{+\infty}f(x)$
Since I didn't understand the problem I looked in the answers and there it read the following:

The first integral converges if $\alpha<1$ since it behaves like
  $\int_0^{1}\frac{1}{x^\alpha}$.
The second integral converges if  $\alpha>0$ since it behaves like
  $\int_1^{+\infty}\frac{1}{x^{\alpha+1}}$.

Can anyone explain this behaviour please?
 A: This is based on the comparison test, one version of which is the following.  If there are positive constants $c, C$ such that $0 < c g(x) < f(x) < C g(x)$ for $a < x < b$, then an improper integral
$ \int_a^b f(x)\; dx$ converges if and only if $\int_a^b g(x)\; dx$ converges.
If $f(x) = 1/(x^\alpha \sqrt{x^2 + 1})$ and $g(x) = 1/x^\alpha$, we have for $0  < x < 1$, $g(x)/2 < f(x) < g(x)$, so $\int_0^1 f(x)\; dx$ converges if and only if $\int_0^1 (1/x^\alpha)\; dx$ converges, and that is true if and only if $\alpha < 1$.
On the other hand, for $x > 1$ we use
$$   f(x) =  \dfrac{1}{x^{\alpha + 1} \sqrt{1 + 1/x^2}}$$
so $$ \dfrac{1}{2 x^{\alpha+1}} < f(x) < \dfrac{1}{x^{\alpha+1}}$$ 
A: Well, whenever $x>1$, you should notice that
$$\sqrt{x^2+1}>x$$
To prove the above, square both sides.  It is then noted that
$$\frac1{x^\alpha\sqrt{x^2+1}}<\frac1{x^\alpha\cdot x}=\frac1{x^{\alpha+1}}$$
Thus, by comparison, $\int_1^\infty\frac1{x^\alpha\sqrt{x^2+1}}dx$ converges whenever $\alpha>0$ due to the p-series.

Whenever $0<x<1$, you should notice that
$$\sqrt{x^2+1}>1$$
Again, just square both sides to see this.  It is then noted that
$$\frac1{x^\alpha\sqrt{x^2+1}}<\frac1{x^\alpha}$$
And I assume you can conclude the rest?

From there, you should test the boundaries.  That is, the cases when $\alpha=0,1$.
