# 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

$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$ $$\bbx{\mbox{Note that}\quad \int_{1}^{\infty}{\dd x \over \root{1 + x^{4}}} = {1 \over 2}\int_{0}^{\infty}{\dd x \over \root{1 + x^{4}}}}$$ 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}

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 $K(1/\sqrt{2})$.

• @fuzz $\displaystyle \int_1^{\infty} \frac{dx}{\sqrt{1+x^4}} < \int_1^{\infty} \frac{dx}{\sqrt{x^4} }= 1.$ – Ryan Apr 3 '18 at 15:57
• No $\int_1^\infty \frac{1}{\sqrt2}dx$ diverges. – GEdgar Apr 3 '18 at 17:02
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 $\Gamma\left(\frac{1}{6}\right)$.