Does $\int_0^{2\pi} \cos\left(\frac{x}{2}\right)^{2a}\mathrm dx = 2^{1-2a}\frac{(2a)!}{(a!)^2}\pi$ always hold for $a\in\mathbb{N}$?

Question

It does seem to hold for $0\leq a\leq 18$, but I was unable to find a way to prove if it is true for higher values of $a$, and I have absolutely no idea how I should go about and prove that. For anyone curious, I came up with this formula by using the Vietoris coefficients as seen in https://arxiv.org/pdf/1608.02579.pdf equation 1.

Answer 1

Yes, the equality should hold. Here is how I prove the identity. Let $a \in \mathbb{N}$. We will begin by integrating by substituting $x \mapsto 2x$: \begin{align*} \Omega:=\int_0^{2\pi} \cos^{2a}\left(\frac{x}{2}\right)\,dx \end{align*} \begin{align*} \Omega &\stackrel{x \mapsto 2x}{=} 2\int_0^{\pi} \cos^{2a}(x)\,dx\\ &= 2\int_0^\frac{\pi}{2}\cos^{2a}(x)\,dx+ \underbrace{2\int_\frac{\pi}{2}^\pi\cos^{2a}(x)\,dx}_{x \mapsto \pi - x}\\ &= 2\int_0^\frac{\pi}{2}\cos^{2a}(x)\,dx + 2\int_0^\frac{\pi}{2} (-\cos(x))^{2a}\,dx\\ &= 4\int_0^\frac{\pi}{2}\cos^{2a}(x)\,dx. \end{align*}

From here, we will use the beta function, which is represented by this integral: $$\mathrm{B}(x,y) = 2\int_0^\frac{\pi}{2} \sin^{2x - 1}(\theta) \cos^{2y - 1}(\theta)\,d\theta$$ such that $\Re(x), \Re(y) > 0$. We also know that $\mathrm{B}(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x + y)}$. Using these properties of the beta function, we get \begin{align*} \Omega &= 2 \cdot 2\int_0^\frac{\pi}{2}\sin^{2 \cdot \frac{1}{2} - 1}\cos^{2\left(a + \frac{1}{2}\right) - 1}(x)\,dx\\ &= 2 \cdot \mathrm{B}\left(\frac{1}{2}, a + \frac{1}{2}\right)\\ &= 2 \cdot \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(a + \frac{1}{2}\right)}{\Gamma\left(a + 1\right)}\\ &= \frac{2\sqrt{\pi}}{a!} \cdot \Gamma\left(a + \frac{1}{2}\right). \end{align*}

By Legendre duplication, we finally get \begin{align*} \Omega &=\frac{2\sqrt{\pi}}{a!} \cdot \frac{\sqrt{\pi}2^{1-2 \left(a + \frac{1}{2} \right)} \Gamma\left(2 \left(a + \frac{1}{2} \right) \right)}{\Gamma\left(a + \frac{1}{2} + \frac{1}{2}\right)}\\ &= \frac{2\sqrt{\pi}}{a!} \cdot \frac{\sqrt{\pi} 2^{-2a} \Gamma\left(2a + 1 \right)}{\Gamma\left(a + 1\right)}\\ &= \frac{\pi 2^{1 - 2a}(2a)!}{(a!)^2}. \end{align*}

With this proof, we can see that $a$ can be an integer greater than $18$ and the equality you derived would still hold!

Answer 2

Letting $\frac{x}{2}\mapsto x $ yields $$ \begin{aligned} I & =2 \int_0^\pi \cos ^{2 a}x d x \\ & =4 \int_0^{\frac{\pi}{2}} \cos ^{2 a} x d x \\ & =2 B\left(a+\frac{1}{2}, \frac{1}{2}\right) \\ & =2 \frac{\Gamma\left(a+\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma(a+1)} \\ &= \frac{2 \sqrt{\pi}}{a !} \Gamma\left(a+\frac{1}{2}\right) \\ &= \frac{2 \sqrt{\pi}}{a !}\left(a-\frac{1}{2}\right) \Gamma\left(a-\frac{1}{2}\right) \\ &= \frac{2 \sqrt{\pi}}{a !}\left(a-\frac{1}{2}\right)\left(a-\frac{3}{2}\right) \cdots\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)\\& =\frac{2 \pi}{a ! 2^a}(2 a-1)(2 a-3) \cdots 1\\& =\frac{2 \pi}{a ! 2^a} \frac{(2 a) !}{(2 a)(2 a-2) \ldots(2)}\\&= \frac{2^{1-2a}(2 a) !}{(a !)^2} \end{aligned} $$