# Can anyone solve this integral? $\int_{-\infty}^{\infty}\sqrt{x^4+1}-x^2dx$

I thought of this peculiar integral: $$\int_{-\infty}^{\infty}\sqrt{x^4+1}-x^2dx$$ and I'm wondering whether or not it has a closed-form solution. I can show that this integral converges using the comparison theorem, since $$\sqrt{x^4+1}-x^2 = \frac{1}{\sqrt{x^4+1}+x^2} < \frac{1}{1+x^2}$$ but with the $\sqrt{x^4+1}$ , we also know that the function itself has no elementary integral. Any ideas?

• Can someone help me? I'm confused. If you let $f(z) = \frac{1}{(z^4+1)^{1/2}+z^2}$ (with, say, the principal branch of $\sqrt{.}$), then the integral over an upper semicircle with radius $R$ goes to $0$ as $R \to \infty$; but also, $f$ has no poles because if $(z^4+1)^{1/2} = -z^2$, then $z^4+1 = z^4$, impossible. So, we should end up with $\int_{-\infty}^{\infty} f(x)dx = 0$, which is obviously false. Where did I go wrong? – mathworker21 Jul 8 '18 at 0:05
• @mathworker21, $\sqrt{z^4+1}$ has the usual square-root branch problems wherever $z^4=-1$, which occurs twice on the upper half plane (and twice more on the lower half plane), so it is not holomorphic on the upper half plane, and thus Cauchy's residue theorem doesn't apply. (I.e., what saultpatz said in fewer words!) – Barry Cipra Jul 8 '18 at 0:14
• @mathworker It has branch points at the fourth roots of -1. – saulspatz Jul 8 '18 at 0:14
• One would expect to be able to deform the contour to one around a branch cut between the two roots of $\sqrt{z^4+1}$ in the upper half-plane. Whether this is actually a helpful approach is another matter. – Chappers Jul 8 '18 at 0:16
• Using integration by parts I found a simpler equivalent integral $$\frac{2}{3} \int_{-\infty }^{\infty } \frac{1}{\sqrt{x^4+1}} \, dx$$, assuming $\left[ \frac{x^3}{3}\left(\sqrt{1+\frac{1}{x^4}}-1 \right)\right]_{-\infty}^{\infty}=0$ – James Arathoon Jul 8 '18 at 2:38

HINT: Start with @JamesArathoon's integral, obtained by IBP: $$\frac{4}{3}\int_0^\infty \frac{dx}{\sqrt{x^4+1}}$$ Then, after an appropriate substitution, apply this well known formula using the Gamma function: $$\int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}}dx=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=\mathrm{B}(m,n)$$