Generalized hypergeometric functions VIII

Nice integral related to the Gauss hypergeometric function $F(a,b;c;z)$

Today we show the proof of this nice integral without contour integration as proposed by @AlbahariRicardo. \[ \int_{0}^{\frac{\pi}{2}} \cos(a\theta) \cos^b(\theta) d\theta = \frac{\pi\Gamma(b+1)}{2^{b+1}\Gamma\left(\frac{2-a+b}{2}\right)\Gamma\left(\frac{2+a+b}{2}\right)} \quad a,b\in \mathbb{C}, \quad \Re(b)>-1 \] The proof relies on the Gauss theorem for the function that have his name $F(a,b;c;z)$, the proof of this theorem neither relies on contour integration.

Update: We updated this post to cover the case $a,b\in \mathbb{C}$.

Proof:

Let $a,b \in \mathbb{C}$ \begin{align*} I= \int_{0}^{\frac{\pi}{2}} \cos(a\theta) \cos^b(\theta) d\theta = &\frac{1}{2} \int_{0}^{\frac{\pi}{2}} (e^{ia\theta}+e^{-ia\theta})\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta\\ =& \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{ia\theta}\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{-ia\theta}\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta\\ =& \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{ia\theta}\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta - \frac{1}{2} \underbrace{\int_{0}^{-\frac{\pi}{2}} e^{ia\vartheta}\left( \frac{e^{i\vartheta }+e^{-i\vartheta}}{2}\right)^b d\vartheta}_{\vartheta \mapsto -\theta}\\ =& \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{ia\theta}\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta + \frac{1}{2} \int_{-\frac{\pi}{2}}^{0} e^{ia\vartheta}\left( \frac{e^{i\vartheta }+e^{-i\vartheta}}{2}\right)^b d\vartheta\\ =& \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{ia\theta}\left( \frac{e^{i\theta }+e^{-i\theta}}{2}\right)^b d\theta \\ =& \frac{1}{2^{b+1}} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{ia\theta}\sum_{j=0}^{\infty}\binom{b}{j} e^{i\theta(b-2j)} d\theta \quad \textrm{(Generalized binomial theorem)} \\ =& \frac{1}{2^{b+1}}\sum_{j=0}^{\infty}\binom{b}{j} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{i\theta(a+b-2j)} d\theta \\ =& \frac{1}{2^{b+1}}\sum_{j=0}^{\infty}\binom{b}{j} \frac{1}{i(a+b-2j)} \int_{-\frac{i\pi(a+b-2j)}{2}}^{\frac{i\pi(a+b-2j)}{2}} e^{w} dw \quad (w \mapsto i\theta(a+b-2j), \quad a+b\neq 2j \quad j\in \mathbb{N})\\ =& \frac{1}{2^{b+1}}\sum_{j=0}^{\infty}\binom{b}{j} \frac{e^{\frac{i\pi(a+b-2j)}{2}}-e^{\frac{-i\pi(a+b-2j)}{2}}}{i(a+b-2j)}\\ =& \frac{1}{2^{b+1}}\sum_{j=0}^{\infty}\binom{b}{j} \frac{e^{\frac{i\pi(a+b)}{2}}e^{-i\pi j} -e^{\frac{-i\pi(a+b)}{2}}e^{i\pi j}}{i(a+b-2j)}\\ =& \frac{1}{2^{b}}\sum_{j=0}^{\infty}\binom{b}{j} \frac{(-1)^j}{(a+b-2j)} \frac{e^{\frac{i\pi(a+b)}{2}} -e^{\frac{-i\pi(a+b)}{2}}}{2i}\\ =& \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}}\sum_{j=0}^{\infty}\binom{b}{j} \frac{(-1)^j}{(a+b-2j)} \\ =& \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}}\sum_{j=0}^{\infty}\frac{(b-j+1)_{j}}{j!} \frac{(-1)^j}{(a+b-2j)} \\ =& \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}}\sum_{j=0}^{\infty} \frac{(-b)_{j}}{(a+b-2j)j!} \\ \end{align*} Note that \[ 2j-a-b = \frac{-(a+b)\left(\frac{2-a-b}{2}\right)_{j}}{\left(\frac{-a-b}{2}\right)_{j}}\] Hence \begin{align*} I = \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}}\sum_{j=0}^{\infty} \frac{(-b)_{j}}{(a+b-2j)j!} = & \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}(a+b)}\sum_{j=0}^{\infty} \frac{(-b)_{j}\left(\frac{-a-b}{2}\right)_{j}}{\left(\frac{2-a-b}{2}\right)_{j}}\frac{1}{j!} \\ =& \frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^{b}(a+b)}F\left(-b,\frac{-a-b}{2};\frac{2-a-b}{2};1\right) \\ \end{align*} where $F(a,b;c;z)$ is the Gauss hypergeometric function which satisfies th Gauss theorem: \[F(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} \quad \Re(c-a+b)>0\] Applying the theorem to $\displaystyle F\left(-b,\frac{-a-b}{2};\frac{2-a-b}{2};1\right)$ and supposing $\Re(b)>-1$ \[ F\left(-b,\frac{-a-b}{2};\frac{2-a-b}{2};1\right) = \frac{\Gamma\left(\frac{2-a-b}{2}\right)\Gamma(b+1)}{\Gamma\left(\frac{2-a+b}{2}\right)\Gamma(1)}\] Hence \begin{align*} I =&\frac{\sin\left(\frac{\pi(a+b)}{2}\right)}{2^b(a+b)}\frac{\Gamma\left(\frac{2-a-b}{2}\right)\Gamma(b+1)}{\Gamma\left(\frac{2-a+b}{2}\right)}\\ =& \frac{1}{2^b(a+b)}\left[\frac{\pi}{\Gamma\left(\frac{2-a+b}{2}\right) \Gamma\left(\frac{a+b}{2}\right)}\right]\frac{\Gamma\left(\frac{2-a-b}{2}\right)\Gamma(b+1)}{\Gamma\left(\frac{2-a+b}{2}\right)} \quad \textrm{(Euler reflection formula)} \\ =& \frac{\pi\Gamma(b+1)}{2^b(a+b)\Gamma\left(\frac{2-a+b}{2}\right)\Gamma\left(\frac{a+b}{2}\right)}\\ =& \frac{\pi\Gamma(b+1)}{2^{b+1}\Gamma\left(\frac{2-a+b}{2}\right)\Gamma\left(\frac{2+a+b}{2}\right)} \end{align*} The cases $b=a=2n$, $2m=a>b=2n$ and $2m=a\lt b=2n$ can easily be obtained as limiting cases of \[ \int_{0}^{\frac{\pi}{2}} \cos(a\theta) \cos^b(\theta) d\theta = \frac{\pi\Gamma(b+1)}{2^{b+1}\Gamma\left(\frac{2-a+b}{2}\right)\Gamma\left(\frac{2+a+b}{2}\right) } \quad \Re(b)>-1, a+b\neq 2j \quad j\in \mathbb{N}\] Hence, we can conclude \[ \boxed{\int_{0}^{\frac{\pi}{2}} \cos(a\theta) \cos^b(\theta) d\theta = \frac{\pi\Gamma(b+1)}{2^{b+1}\Gamma\left(\frac{2-a+b}{2}\right)\Gamma\left(\frac{2+a+b}{2}\right)} \quad a,b\in \mathbb{C}, \Re(b)>-1} \]