# 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

## 3 Answers


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!

• 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 – asdf Apr 3 '18 at 15:43
• @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 $\Gamma\left(\frac{1}{6}\right)$.