Integral $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}$, may involve beta function

Alternative approaches? The integral is a hypergeometric functions, and I would immediately try to get it because then I could always use some identities to simplify if needed.

Let's use the substitution $x^4=v$, which the OP rejected.

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

We know the general integral for the hypergeometric function:

$$\mathrm {B} (b,c-b)\,_{2}F_{1}(a,b;c;z)=\int _{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\,dx\qquad \Re (c)>\Re (b)>0$$

In this case $a=2$, $b=1/4$, $c=2$, so we get:

$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{1}{4} \mathrm {B} \left(\frac14 , \frac74 \right) {_2 F_1} \left(2,\frac14; 2;-1 \right)$$

Using some simplifications:

$$\frac{1}{4}\mathrm {B} \left(\frac14 , \frac74 \right)=\frac{1}{4}\Gamma \left( \frac14 \right) \Gamma \left(\frac74 \right)=\Gamma \left(\frac54 \right)\Gamma \left(\frac74 \right)$$

$${_2 F_1} \left(2,\frac14; 2;-1 \right)={_1 F_0} \left(\frac14; ;-1 \right)=(1-(-1))^{-1/4}=\frac{1}{2^{1/4}}$$

(It's known that ${_1 F_0} \left(a; ;x \right)=(1-x)^{-a}$).

So we get:

$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{1}{2^{1/4}}\Gamma \left(\frac54 \right)\Gamma \left(\frac74 \right)$$

Which is the same as the OP's result, but obtained in a more simple way, using the known properties of hypergeometric functions.

This can be further simplified by using the reflection formula for the Gamma function (thanks to @Szeto for reminding me):

$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{3}{16 \cdot 2^{1/4}}\Gamma \left(\frac14 \right)\Gamma \left(\frac34 \right)$$

$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{3\pi}{16 \cdot 2^{1/4} \sin \frac{\pi}{4}}$$

$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{3\pi \cdot 2^{1/4}}{16}$$

Tags:

Integration

Definite Integrals

Beta Function

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