Is $\int_{0}^{\infty}\frac1{\sqrt{x^{2n}+1}}dx$ finite?

Question

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. :)

$\frac1{\sqrt{x^{2n}+1}} \sim \frac{1}{x^n}$ for $x \to \infty$, so ... — Martin R, Jan 09 '21 at 20:46
Note that $\frac1{\sqrt{x^{2n}+1}}\leq \frac1{\sqrt{x^{2n}}}= \frac1{x^{n}}$ for $x>0$ — Robert Z, Jan 09 '21 at 20:47
But $\int_{0}^{\infty}(1/x^{n})dx$ is for big n divergent, isn't it? — Mathpiano, Jan 09 '21 at 20:53
At zero your integral is fine at any n, so all restrictions for n come from the region of big x. — Svyatoslav, Jan 09 '21 at 21:00

score 1 · Answer 1 · answered Jan 09 '21 at 20:58

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.

score 1 · Accepted Answer · answered Jan 09 '21 at 21:00

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.

Thanks! That definitely makes sense :) – Mathpiano Jan 09 '21 at 21:05 — Mathpiano, Jan 09 '21 at 21:05

Is $\int_{0}^{\infty}\frac1{\sqrt{x^{2n}+1}}dx$ finite?

2 Answers2