Peculiar (convergent?) definite integral I have been trying to calculate the integral:
$$\int_1^{\infty} \left(\frac{x^2}{\sqrt{x^4-1}}-1\right)dx$$
A hint is to multiply the whole integral by $x^{\lambda}$, calculate the two terms independently as a function of $\lambda$ and then set $\lambda=0$. But this did not work.
By substituting $x=\frac{1}{u}$ the integral transforms into $$\int_0^1 \left(\frac{1}{\sqrt{1-u^4}}-1\right)\frac{1}{u^2}du$$
which leads to a Beta function with one negative argument. 
The result should be $$1-\frac{\pi}{\Gamma(\frac{1}{4})^2}$$
Thank you very much in advance!
 A: The integral in $u\in[0,1]$ is for me simpler, so let us introduce for a handy notation
$$y=y(u) = \sqrt{1-u^4}\ .
$$
Then for the integral to be calculated we observe first
$$
\frac\partial{\partial u}
\left(\frac{1-y}u\right) = 
\frac{u^2}y-\left(\frac 1y-1\right)\frac 1{u^2}\ .
$$
So we need to calculate
$$
\begin{aligned}
J
&=
\int_0^1 \left(\frac{1}{\sqrt{1-u^4}}-1\right)\frac{1}{u^2}\;du
%\\&
=
\int_0^1 \left(\frac1y-1\right)\frac{1}{u^2}\;du
\\
&=\left[\frac {1-y}u\right]_0^1
-
\int_0^1 \frac{u^2}y\;du
%\\&
=
1
-
\int_0^1 \frac{u^2}{\sqrt{1-u^4}}\;du
\\
&= 1-\frac 14B\left(\frac 12,\frac 34\right)\qquad
\text{ with $B$ being the $\beta$-function}
\\
&=1-\frac 14\cdot\frac  
{\Gamma\left(\frac 12\right)\cdot\Gamma\left(\frac 34\right)}
{\Gamma\left(\frac 54\right)}
%\\&
=1-\frac 14\cdot\frac  
{\sqrt\pi\cdot\Gamma\left(\frac 34\right)}
{\frac 14\Gamma\left(\frac 14\right)}
\\
&=1-\sqrt\pi
\cdot
\frac  
{\Gamma\left(\frac 14\right)\cdot\Gamma\left(\frac 34\right)}
{\Gamma\left(\frac 14\right)^2}
%\\&
=1-\sqrt\pi
\cdot
\frac  
{\sqrt{2\pi}\Gamma\left(2\cdot\frac 14\right)}
{\Gamma\left(\frac 14\right)^2}
\\
&
=
1-
\frac 12
(2\pi)^{3/2}\cdot 
\frac  
1{\Gamma\left(\frac 14\right)^2}
\ .
\end{aligned}
$$
I tried to represent the result closer to the shape of the prediction in the OP.

We have the assisted checks / numerical validations:
sage: value = 1 - (2*pi)^(3/2) / 2 / gamma(1/4)^2
sage: value.n()
0.400929882632204
sage: var('u');
sage: integral( (1/sqrt(1-u^4) - 1)/u^2, u, 0, 1).n()
0.40092987307719175
sage: integral( u^2/sqrt(1-u^4), u, 0, 1)
1/4*beta(1/2, 3/4)

A: Note that
$$ \int_0^1x^{p-1}(1-x)^{q-1}dx=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}, \Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}$$
Under $u=t^{1/4},t=\sin ^{2}v$, one has
\begin{eqnarray}
&&\int_0^1 \left(\frac{1}{\sqrt{1-u^4}}-1\right)\frac{1}{u^2}du\\
&=&\frac14\int_0^{1}\bigg(-t^{-5/4}+t^{-5/4}(1-t)^{-1/2}\bigg)dt.
\end{eqnarray}
Noting that
$$ \int_0^1\left(-t^{-a}+t^{-a}(1-t)^{-1/2}\right)dt=\int_0^1\frac{1}{t^{a-1}(1-t)^{1/2}\left(1+(1-t)^{1/2}\right)}dt$$
converges for $a<2$, so one has
\begin{eqnarray}
&&\int_0^1 \left(\frac{1}{\sqrt{1-u^4}}-1\right)\frac{1}{u^2}du\\
&=&\lim_{a\to\frac54}\frac14\int_0^{1}\bigg(-t^{-a}+t^{-a}(1-t)^{-1/2}\bigg)dt\\
&=&\lim_{a\to\frac54}\frac14\left(\frac{1}{a-1}+\frac{\Gamma(1-a)\Gamma(\frac12)}{\Gamma(\frac32-a)}\right)\\
&=&1+\frac{\sqrt\pi\Gamma(-\frac14)}{\Gamma(\frac14)}\\
&=&1-\frac{\sqrt2\pi^{3/2}}{\Gamma^2(\frac14)}\\
\end{eqnarray}
