# The shortest way to prove that $\int_1^\infty \frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }}dx$ converges.

I'm trying to show that the integral $$\int_1^\infty \frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }}dx \quad \text{is convergent}.$$

We know that $$\frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }} < \frac{{\sqrt x }}{{\sqrt {{x^4} - 1} }}{\text{ for}}\,\,{\text{all}}\,\,\,x \in \left\langle {1, + \infty } \right\rangle$$ Now this is reduced to prove that $$\int_1^\infty {\frac{{\sqrt x }}{{\sqrt {{x^4} - 1} }}}dx$$ is convergent, but the latter integral is difficult to calculate or prove to be convergent.

I know that this integral is convergent because when I calculate with Maple is $$\int_1^\infty {\frac{{\sqrt x }}{{\sqrt {{x^4} - 1} }}}dx = \frac{{{\pi ^{3/2}}\csc \left( {\frac{\pi }{8}} \right)}}{{4\Gamma \left( {\frac{7}{8}} \right)\Gamma \left( {\frac{5}{8}} \right)}} \approx {\text{2}}{\text{.327185143}}$$

So my question is:

Is there a simpler way of proving that $\int_1^\infty \frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }}dx$ converges?

• $\arctan x$ is bounded, so we can forget about it. May 16, 2015 at 14:30
• @AndréNicolas Yes, but it implies that $\mathop \smallint \limits_1^\infty \frac{{\frac{\pi }{2}}}{{\sqrt {{x^4} - 1} }}\;{\text{d}}x$ converges, which it is still difficult to prove May 16, 2015 at 14:34