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

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.

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

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

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 $$z \mapsto 1-z$$.

• (+1) Nice substitution! I tried a different approach using the binomial theorem a couple of times. – robjohn Sep 24 '18 at 6:32

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 $$x=1$$