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!

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}
