Integral of $\frac{1}{\sqrt{1+x^4}}$ How do you compute $$\int_1^\infty \frac{1}{\sqrt{1+x^4}}\mathrm dx?$$
I thought of series expansion but the sum is too complicated.
Any help appreciated
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$$
\bbx{\mbox{Note that}\quad
\int_{1}^{\infty}{\dd x \over \root{1 + x^{4}}}  =
{1 \over 2}\int_{0}^{\infty}{\dd x \over \root{1 + x^{4}}}}
$$
Then,
\begin{align}
\int_{1}^{\infty}{\dd x \over \root{1 + x^{4}}} & =
{1 \over 8}\int_{0}^{\infty}{x^{-3/4} \over \root{1 + x}}\,\dd x =
{1 \over 8}\int_{1}^{\infty}{\pars{x - 1}^{-3/4} \over \root{x}}\,\dd x
\\[5mm] & =
{1 \over 8}\int_{1}^{0}{\pars{1/x - 1}^{-3/4} \over x^{-1/2}}\,\pars{-\,{\dd x \over x^{2}}} =
{1 \over 8}\int_{0}^{1}x^{-3/4}\pars{1 - x}^{-3/4}\,\dd x
\\[5mm] & =
{1 \over 8}\,{\Gamma\pars{1/4}\Gamma\pars{1/4} \over
\Gamma\pars{1/2}} = \bbx{\Gamma^{2}\pars{1/4} \over 8\root{\pi}}
\approx 0.9270
\end{align}
A: As noted, an elliptic integral.
$$
\int_1^\infty \frac{1}{\sqrt{1+x^4}}dx = \frac{1}{2}\;\mathrm K\left(\frac{1}{\sqrt2}\right)
$$
For elliptic integral K see https://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_first_kind
Note, in Wolfram notation, write $K(1/2)$ instead of $K(1/\sqrt{2})$.
added
OK, prove or disprove: this integral converges.  That is a much easier question than evaluate the integral.   You can think of a good comparison to make for that .... try it!
A: Joining GEdgar and Felix Marin's answers, we have that $\int_{1}^{+\infty}\frac{dx}{\sqrt{1+x^4}}$ is simultaneously related to the complete elliptic integral of the first kind and to the lemniscate constant via the Beta and $\Gamma$ functions. Since complete elliptic integrals of the first kind can be efficiently evaluated via the AGM mean we have:
$$ \int_{1}^{+\infty}\frac{dx}{\sqrt{1+x^4}} = \frac{\pi}{4\,\text{AGM}\left(1,\frac{1}{\sqrt{2}}\right)}$$
i.e. an explicit and efficient computation algorithm, leading to the accurate double bound
$$ \frac{\pi}{4}2^{1/4}\geq \int_{1}^{+\infty}\frac{dx}{\sqrt{1+x^4}}\geq \frac{\pi}{2+\sqrt{2}}.$$
See also this answer where the same is done for $\Gamma\left(\frac{1}{6}\right)$.
