# Evaluate $\int_{-\pi/2}^{\pi/2} (1+e^{2i\phi})^{\alpha} (1+e^{-2i\phi})^{\beta} \, \mathrm{d}\phi$

I should evaluate:

$$\int_{-\pi/2}^{\pi/2} (1+e^{2i\phi})^{\alpha} (1+e^{-2i\phi})^{\beta} \, \mathrm{d}\phi$$

by using the binomial theorem and the identity:

$${}_2F_1 \left(\begin{array}{c}a , b \\ c \end{array};x\right) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_{0}^{1} t^{b-1}(1-t)^{c-b-1}(1-xt)^{-a} \, \mathrm{d}t$$

So first using binomial theorem I get:

\begin{align*} &\int_{-\pi/2}^{\pi/2} \sum_{k=0}^{\alpha} \binom{\alpha}{k} e^{2i\phi k} \sum_{k=0}^{\beta} \binom{\beta}{k} e^{-2i\phi k} \, \mathrm{d}\phi \\ &= \int_{-\pi/2}^{\pi/2} \sum_{k=0}^{\alpha} \sum_{l=0}^{\beta} \binom{\alpha}{k} e^{2i\phi k} \binom{\beta}{l} e^{-2i\phi l} \, \mathrm{d}\phi \\ &= \int_{-\pi/2}^{\pi/2} \sum_{k=0}^{\alpha} \sum_{l=0}^{\beta} \binom{\alpha}{k} \binom{\beta}{l} e^{2i\phi(k-l)} \, \mathrm{d}\phi \end{align*}

But from here I don't know how to proceed or rather how to use the identity. Any hints?

• What are your $\alpha$ and $\beta$? If they are non-negative integers, then this can be computed fairly in a straightforward way as illustrated in J.G.'s answer. Otherwise, we can still represent the integral using a hypergeometric function. – Sangchul Lee Nov 26 '20 at 19:58
• Good question. It is not specified in the exercise :/ just that it should be solved by the identity. I would assume any integer – craft Nov 26 '20 at 20:01
• @SangchulLee do you know how to evaluate it over the hypergeometric identity? – craft Nov 26 '20 at 22:39
• @craft J.G.'s answer can be generalized, and no hypergeometric formula is needed. Indeed, by Euler formula and parity, $I=\int_{-\pi/2}^{\pi/2} (2e^{ix}\cos(x))^a (2e^{-ix}\cos{x})^bdx=2^{a+b-1}\int_{0}^{\pi/2}\cos^{a+b}x \cos((a-b)x) dx$. Recall that $f(v,a)=\int_0^{\frac{\pi }{2}} \cos ^{v-1}(x) \cos (a x) \, dx=\frac{\pi }{2^v v B\left(\frac{1}{2} (a+v+1),\frac{1}{2} (-a+v+1)\right)}$, let $v\to a+b+1,a\to a-b$ one obtain $I=\frac{\pi \Gamma(a+b+1)}{\Gamma(a+1)\Gamma(b+1)}$. – Iridescent Nov 27 '20 at 3:42
• @craft $1$. G&R denotes Gradshteyn & Ryzhik's Table of integrals, series and products. $2$. See also G&R for expansion of $\cos(ax)$. You may easily find the hypergeometric solution then. – Iridescent Nov 27 '20 at 10:20

If $$\beta$$ is a non-negative integer, with $$z=e^{2i\phi}$$ this becomes$$\oint_{|z|=1}(1+z)^{\alpha+\beta}\frac{dz}{2iz^{\beta+1}}=\pi[z^\beta](1+z)^{\alpha+\beta}=\pi\binom{\alpha+\beta}{\beta}=\frac{\pi\Gamma(\alpha+\beta+1)}{\Gamma(\alpha+1)\Gamma(\beta+1)}.$$Update: @Iridescent has pointed out how we can generalize to complex $$\beta$$. The integral is $$2^{\alpha+\beta-1}\int_0^{\pi/2}\cos^{\alpha+\beta}\phi\cos[(\alpha-\beta)\phi]d\phi$$, since the integrand's imaginary part integrates to $$0$$ on $$[-\tfrac{\pi}{2},\,\tfrac{\pi}{2}]$$. An old question proves this is indeed $$\tfrac{\pi\Gamma(\alpha+\beta+1)}{\Gamma(\alpha+1)\Gamma(\beta+1)}$$.