Why $\int_0^1(1-x^4)^{2016}dx=\prod_{j=1}^{2016}\left(1-\frac1{4j}\right)$?

Question

In this answer it states that:

$\int_0^1(1-x^4)^{2016}dx=\prod_{j=1}^{2016}\left(1-\frac1{4j}\right)$

How to prove that?

Answer 1

The given answer is incorrect. To see this, we might as well solve the more general problem of computing

$$I_n = \int_0^1 (1 - x^p)^n \, dx.$$

As I mentioned in the comments, the form of the answer naturally suggests that we should try to find a recursion, so let's try it. We can write

$$I_n = \int_0^1 (1 - x^p) (1 - x^p)^{n-1} \, dx = I_{n-1} - \int_0^1 x^p (1 - x^p)^{n-1} \, dx.$$

Next, we'll use integration by parts with $u = x, dv = x^{p-1} (1 - x^p)^{n-1} \, dx$, giving $v = - \frac{1}{np} (1 - x^p)^n$, to get

$$\int_0^1 x^p (1 - x^p)^{n-1} \, dx = \int_0^1 \frac{1}{np} (1 - x^p)^n \, dx = \frac{1}{np} I_n.$$

(the $uv$ term disappears because it is equal to $0$ at both $x = 0$ and $x = 1$). This gives

$$I_n = I_{n-1} - \frac{1}{np} I_n$$

and a bit of rearranging gives

$$I_n = \left( 1 - \frac{1}{np+1} \right) I_{n-1}.$$

The initial value is $I_0 = 1$ and so we conclude by induction that

$$\boxed{ I_n = \prod_{j=1}^n \left( 1 - \frac{1}{pj+1} \right) }.$$

Note that when $p = 4, n = 2016$ this does not agree with the given answer; instead, we get

$$\boxed{ \int_0^1 (1 - x^4)^{2016} \, dx = \prod_{i=1}^{2016} \left( 1 - \frac{1}{4j+1} \right) }.$$

If we generalize the given answer to arbitrary values of $n$ and $p$ then it would suggest that

$$I_1 = \int_0^1 (1 - x^p) \, dx \stackrel{?}{=} 1 - \frac{1}{p}$$

which is clearly wrong.

Answer 2

By setting $x=z^{1/4}$ and exploiting Euler's Beta function we get:

$$ I = \int_{0}^{1}(1-x^4)^{2016}\,dx = \frac{1}{4}\int_{0}^{1} z^{-3/4}(1-z)^{2016}\,dz = \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(2017\right)}{4\,\Gamma\left(2017+\frac{1}{4}\right)} \tag{1}$$ and the RHS of $(1)$ turns into (something similar to) the given product by recalling that $\Gamma(z+1)=z\,\Gamma(z)$.

Answer 3

In the same spirit as Qiaochu Yuan and Jack D'Aurizio's answers, assuming $m$ and $n$ positive, since $$\int(1-x^m)^n\,dx=x \, _2F_1\left(\frac{1}{m},-n;1+\frac{1}{m};x^m\right)$$ $$\int_0^1(1-x^m)^n\,dx=\frac{\Gamma \left(1+\frac{1}{m}\right) \Gamma (n+1)}{\Gamma \left(n+\frac{1}{m}+1\right)}$$ while (using Pochhammer symbols) $$\prod_{j=1}^n \left(1-\frac 1 {mj} \right)=\frac{\left(1-\frac{1}{m}\right)_n}{n!}=\frac{\Gamma \left(n-\frac{1}{m}+1\right)}{\Gamma \left(1-\frac{1}{m}\right) \Gamma (n+1)}$$ but $$\prod_{j=1}^n \left(1-\frac 1 {mj+1} \right)=\frac{n!}{\left(1+\frac{1}{m}\right)_n}=\frac{\Gamma \left(1+\frac{1}{m}\right) \Gamma (n+1)}{\Gamma \left(n+\frac{1}{m}+1\right)}$$