investigating convergence of two integrals Let $a\in\mathbb{R}$ and so i need to find wether any of the following generalized integrals converge.
$$\int_0^{\infty} x^a \sqrt{x+1} \ \ dx$$
$$\int_0^{\infty} \frac{2 \arctan(x)}{(x+1)^a} \ \ dx$$
So i thought i should look at different scenarios of $x$, but i don't know how to start.
$$\int_0^{\infty} \frac{1}{x^{\alpha}} \ \ dx$$
converges for any $\alpha > 1$, so i thought that would help me with both integrals, but i am stuck at applying that into context.
Any help would be appreciated. Thank you in advance.
 A: The integrands are positive and continuous for $x>0$ and you can apply the Asymptotic Comparison Test for improper integrals.
As regards the first integral, you have to check if the integral is convergent "near" infinity and in a right-neighbourhood of zero.
As $x\to +\infty$,
$$x^a \sqrt{x+1}\sim \frac{1}{x^{-(a+1/2)}}\Rightarrow  -(a+1/2)>1$$
(here you use the fact that $\int_1^{\infty} \frac{dx}{x^{\alpha}}$ is convergent iff $\alpha>1$).
As $x\to 0^+$
$$x^a \sqrt{x+1}\sim \frac{1}{x^{-a}}\Rightarrow  -a<1$$
(here you use the fact that $\int_0^{1} \frac{dx}{x^{\alpha}} $ is convergent iff $\alpha<1$).
So the first integral is convergent as soon as both conditions are satisfied: $-(a+1/2)>1$ and $-a<1$ which is impossible.
In a similar way, for the second integral:  as $x\to +\infty$,
$$\frac{2 \arctan(x)}{(x+1)^a}\sim \frac{\pi}{x^{a}}\Rightarrow  a>1.
$$
As $x\to 0^+$,
$$\frac{2 \arctan(x)}{(x+1)^a}\sim 2x \Rightarrow  \forall a\in\mathbb{R}.$$
So the second integral is convergent as soon as $a>1$.
