Improper integral of $\sqrt{x^4 + 1} - x^2$

Question

I'm having a little trouble with this integral:

$\int^\infty_0 (\sqrt{x^4+1} - x^2)\,dx$.

Using the likes of Maple, I can easily find that it takes the form

$-\frac{2}{3}\sqrt{2}(1+i)K(i) - \frac{1}{3}\sqrt{2}(1+i)K(\sqrt{2})$,

where

$K(m) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1 - m\sin^2\theta}}$

is the complete elliptic integral of the first kind. However, I am looking at a textbook which tells me the solution is simply

$\frac{\Gamma^2(1/4)}{6\sqrt{\pi}}$,

with $\Gamma$ denoting the usual gamma function

$\Gamma(t) = \int^\infty_0 e^{-u}u^{t - 1}du$.

I've thrown a fair amount at this: trying to reduce it via various substitutions to, for instance, a combination of beta functions which can be rewritten as gamma functions via

$B(m,n)= \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$,

or just a combination of elliptic $K$ functions. Even after a fair amount of time consulting Gradshteyn and Ryzhik, I can't seem to get anywhere... I'd like to eventually get to the gamma function solution, and I'd be most grateful to anyone who can help me get there, even if it's just via a nudge in the right direction.

The general solution can be found here. – Lucian Jul 20 '15 at 20:27 — Lucian, Jul 20 '15 at 20:27
Thanks all for pointing out the duplicate. – Jakob Lindemann Jul 20 '15 at 22:38 — Jakob Lindemann, Jul 20 '15 at 22:38

score 0 · Accepted Answer · answered Jul 20 '15 at 18:09

$$I=\int_{0}^{+\infty}\left(\sqrt{x^4+1}-x^2\right)\,dx = \int_{0}^{+\infty}\frac{dx}{x^2+\sqrt{x^4+1}}$$ then by setting $u=x^2+\sqrt{x^4+1}$ we have: $$ I = \frac{1}{2\sqrt{2}}\int_{1}^{+\infty}\frac{u^2+1}{u^{5/2}\sqrt{u^2-1}}\,du=\frac{1}{2\sqrt{2}}\int_{0}^{1}\frac{(1+v^2)\,dv}{\sqrt{v}\sqrt{1-v^2}} $$ and by replacing $v$ with $\sqrt{t}$ we get: $$ I = \frac{1}{4\sqrt{2}}\int_{0}^{1}\frac{(1+t)\,dt}{t^{3/4}\sqrt{1-t}}$$ hence: $$ I = \frac{1}{4\sqrt{2}}\left(\frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)}+\frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{7}{4}\right)}\right)=\frac{1}{3\sqrt{2}}\cdot \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)}$$ and: $$ I = \frac{\Gamma\left(\frac{1}{4}\right)^2}{6\sqrt{\pi}} $$ follows from Legendre's duplication formula.

Thanks for answering in spite of the duplicate question! – Jakob Lindemann Jul 20 '15 at 22:37 — Jakob Lindemann, Jul 20 '15 at 22:37

Improper integral of $\sqrt{x^4 + 1} - x^2$

1 Answers1