Is $\int_{0}^{\infty}\frac1{\sqrt{x^{2n}+1}}dx$ finite? I want to evaluate the integral $\int_{0}^{\infty}\frac1{\sqrt{x^{2n}+1}}dx$, where $n\in \mathbb{N},n\ge2$ in order to show the integrability of another function. However, this seems rather difficult, but I need only to know whether it has a finite value, not necessarily how to solve it. WolframAlpha didn't help, so I'm gonna ask here. :)
 A: If you break your integral into two pieces, you can see that
$$\int_0^1 \frac{1}{\sqrt{x^{2n}+1}} \; dx $$
Is a bounded function over a finite interval and hence finite.
The other piece is
$$\int_1^\infty \frac{1}{\sqrt{x^{2n}+1}} \; dx \leq \int_1^\infty \frac{1}{x^n} \; dx $$
which is convergent for $n\geq 2$ on that interval.
The sum of two finites is finite.
A: When $0\leq x \leq 1$, the function is bounded, and thus has a finite integral.  When $1\leq x$, we have that $0\leq 1\leq x^{2n}$, and so
$$\int_{1}^{\infty}\frac{1}{\sqrt{x^{2n}+x^{2n}}}dx<\int_{1}^{\infty}\frac{1}{\sqrt{x^{2n}+1}}dx<\int_{1}^{\infty}\frac{1}{\sqrt{x^{2n}}}dx.$$
If $n>1$, the outside integrals both converge by the $p$-test (or you can simply compute their anti-derivative, evaluate, and take the limit).  Similarly, if $n\leq 1$ both outside integrals diverge.  So by the comparison test, the integral converges if and only if $n>1$.
This can be done slightly more simply by using the limit comparison test, since $\lim_{x\to \infty} \frac{1}{\sqrt{1+x^{2n}}} / \frac{1}{x^n} =1$, although care must be taken to break the interval up into two parts, since the function we are comparing with has a singularity at $x=0$, and the limit comparison test only helps us as we go to infinity.
