Peculiar (convergent?) definite integral

Question

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!

Answer 1

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.

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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)

Answer 2

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}