Solve the integral using beta functions : $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx$

Question

I was solving questions on beta and gamma functions and then I came across this question.

$$ \int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx $$

Generally in questions of beta functions integrals of the type $\int x^n(1−x^m) \, dx$ can be solved by substituting $x^m = t$, but in this case substitution is unconventional because if I put $1+x^4=t$, then I will not get the required form of beta function

I tried substituting $x^2 = \tan t$ and also tried substituting $x^2 = \cos t$ but on simplifying each of them further I could not reduce them to standard form of beta function.

Answer 1

Apply the substitution $x^4 = \frac{1-u}{1+u}$. Then

$$ I := \int_{0}^{1} \frac{(1-x^4)^{3/4}}{(1+x^4)^2} \, dx = \frac{1}{2^{9/4}} \int_{0}^{1} u^{3/4}(1-u)^{-3/4} \, du = \frac{1}{2^{9/4}} B(\tfrac{7}{4},\tfrac{1}{4}) = \frac{3\pi}{2^{15/4}}.$$

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Explanations. First substitute $x^4 = t$ to obtain $ I = \frac{1}{4} \int_{0}^{1} \frac{(1-t)^{3/4}t^{-3/4}}{(1+t)^2} \, dx$. To transform this integral into the form of beta integral, we notice that

• $-1$, $0$ and $1$ are the only branch points/poles of the integrand,
• $0$, $1$, $\infty$ are the only branch points/poles of the integrand of $\int_{0}^{1} u^{\alpha-1}(1-u)^{\beta-1} \, du$.

This leads to consider a Möbius transformation $t = f(u) = \frac{au+b}{cu+d}$ satisfying either

1. $f(\infty) = -1$, $f(0) = 0$ and $f(1) = 1$ and $ad-bc > 0$, or
2. $f(\infty) = -1$, $f(0) = 1$ and $f(1) = 0$ and $ad-bc < 0$.

The first condition yields $f(u) = \frac{u}{2-u}$ and the second condition yields $f(u) = \frac{1-u}{1+u}$. Both can be used, since one condition is simply transformed to the other by $z \mapsto 1-z$.

Answer 2

We can use the Binomial Theorem in the form $$ (1+x)^{-\alpha}=\sum_{k=0}^\infty\frac{(-1)^k\Gamma(k+\alpha)}{\Gamma(\alpha)\Gamma(k+1)}x^k\tag1 $$ to get $$ \begin{align} \int_0^1\frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,\mathrm{d}x &=\frac14\int_0^1\frac{(1-x)^{3/4}}{(1+x)^2}x^{-3/4}\,\mathrm{d}x\tag2\\ &=\frac14\int_0^1(1-x)^{3/4}x^{-3/4}\sum_{k=0}^\infty(k+1)(-x)^k\,\mathrm{d}x\tag3\\ &=\frac14\sum_{k=0}^\infty(-1)^k(k+1)\int_0^1(1-x)^{3/4}x^{k-3/4}\,\mathrm{d}x\tag4\\ &=\frac14\sum_{k=0}^\infty(-1)^k(k+1)\frac{\Gamma\!\left(\frac74\right)\Gamma\left(k+\frac14\right)}{\Gamma(k+2)}\tag5\\ &=\underbrace{\frac14\Gamma\!\left(\frac74\right)\Gamma\!\left(\frac14\right)\vphantom{\sum_{k=0}^\infty}}_{\frac{3\sqrt2}{16}\pi}\underbrace{\sum_{k=0}^\infty\frac{(-1)^k\Gamma\left(k+\frac14\right)}{\Gamma\left(\frac14\right)\Gamma(k+1)}}_{(1+1)^{-1/4}}\tag6\\ &=\frac{3\sqrt[4]2}{16}\pi\tag7 \end{align} $$ Explanation:
$(2)$: substitute $x\mapsto x^{1/4}$
$(3)$: apply $(1)$ with $\alpha=2$ and use $(k+1)\Gamma(k+1)=\Gamma(k+2)$
$(4)$: collect the integral for the Beta Function
$(5)$: evaluate the integral
$(6)$: rearrange factors and use $(k+1)\Gamma(k+1)=\Gamma(k+2)$
$(7)$: $\frac14\Gamma\!\left(\frac74\right)\Gamma\!\left(\frac14\right) =\frac14\cdot\frac34\Gamma\!\left(\frac34\right)\Gamma\!\left(\frac14\right) =\frac3{16}\pi\csc\left(\frac\pi4\right) =\frac{3\sqrt2}{16}\pi$
$\phantom{(7)\text{:}}$ apply $(1)$ with $\alpha=\frac14$ and $x=1$