Prove that $\intop_{-\infty}^{\infty}\frac{1}{\sqrt{1+x^{4}}}dx$ converges

Question

$\intop_{-\infty}^{\infty}\frac{1}{\sqrt{1+x^{4}}}dx$

I've seen many post about $\int\frac{1}{1+x^{4}}dx$ but I don't know how to proceed with this one, with just real methods. Any suggestion? Thanks!

Answer 1

Note that since the integrand is even $$ \int_{-\infty}^{\infty}\frac{1}{\sqrt{1+x^4}}dx = 2 \int_0^{\infty}\frac{1}{\sqrt{1+x^4}}dx $$ and $$ \int_1^{\infty}\frac{1}{\sqrt{1+x^4}}dx < \int_1^{\infty}\frac{1}{\sqrt{x^4}}dx = \int_1^{\infty}x^{-2}dx $$ Can you finish?

Answer 2

For all $x$,

$$\frac1{\sqrt{1+x^4}}\le\min\left(1,\frac1{x^2}\right).$$

You can integrate the $RHS$ explicitly (though it is obvious that this converges).