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I'd like to have an expression of the following integral: $$\int_0^{+\infty} \left(\sqrt{1+x^4} - x^2\right) dx$$ in terms of some special functions (but not in the form given by Wolfram Alpha).

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up vote 2 down vote accepted

The simplified answer seems to be : $$\frac 23K\left(\frac 1{\sqrt{2}}\right)$$ This means $\frac 23K\left(\frac 12\right)$ using Alpha's convention with a rather interesting alternate form :

\begin{align} \frac{\Gamma\bigl(\frac 14\bigr)^2}{6\;\sqrt{\pi}}&=\frac{\Gamma\bigl(\frac 14\bigr)^2}{6\;\Gamma\bigl(\frac 12\bigr)}\\ &=\frac 16 B\left(\frac 14,\frac 14\right)\\ &=\frac 13 \int_0^\infty \frac {t^{1/4-1}}{\sqrt{1+t}}\;dt\\ \end{align} with $B(x,y)$ the Euler integral of the first kind ('beta function') and using the DLMF classical relation $(5.12.3)$. Setting $x:=t^{1/4}$ should bring you nearer to the initial question...

As noted by O.L. the rewriting of the elliptic integral $K(z)$ in terms of $\Gamma$ function is due to $z=\frac 1{\sqrt{2}}$ being one of the singular values of the elliptic integral. It is in fact the first one (see equation $(11)$ here). Your specific case appears in page $3$ of the original article of Chowla and Selberg.

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rhooo, I wish I could upvote once more... btw shouldn't it be $1/\sqrt{2}$ instead of $1/4$ in the 1st answer? Also maybe it could be helpful to mention singular values of elliptic integrals. – Start wearing purple May 24 '13 at 12:24
@O.L. Yes you are right of course (the $1/4$ was from squaring two times the initial answer... ;-)). – Raymond Manzoni May 24 '13 at 12:35
@RaymondManzoni thank you. – qoqosz May 24 '13 at 12:50
You are welcome @qoqosz ! – Raymond Manzoni May 24 '13 at 12:52
Thanks for the edit @O.L. (quite a loss of precision while answering here...) – Raymond Manzoni May 24 '13 at 13:50

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