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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

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

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

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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!

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  • $\begingroup$ Can I prove its convergence or non convergence without using the elliptic integral notation? The ellpiptic integrals are non in the course material, that ' s why $\endgroup$ – asdf Apr 3 '18 at 15:43
  • $\begingroup$ @fuzz $\displaystyle \int_1^{\infty} \frac{dx}{\sqrt{1+x^4}} < \int_1^{\infty} \frac{dx}{\sqrt{x^4} }= 1.$ $\endgroup$ – Ryan Apr 3 '18 at 15:57
  • $\begingroup$ No $\int_1^\infty \frac{1}{\sqrt2}dx$ diverges. $\endgroup$ – GEdgar Apr 3 '18 at 17:02
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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 $\Gamma\left(\frac{1}{6}\right)$.

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