2

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?

Infiniticism
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craft
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    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 $I=\frac{\pi \Gamma(a+b+1)}{\Gamma(a+1)\Gamma(b+1)}$. – Infiniticism Nov 27 '20 at 03:42
  • Nevertheless, the formula featuring $f(v,a)$ can be established via hypergeometric means rather than contour integration as linked, that is, to expand the $\cos(ax)$ in terms of $\sum_k {a_k} \cos(x)^k$ (see for instance G&R), and use Beta identity & Gauss $_2F_1$ formula. – Infiniticism Nov 27 '20 at 03:45
  • Thanks @Iridescent, I've updated to acknowledge this. – J.G. Nov 27 '20 at 07:35
  • @Iridescent, That is interesting. This shows that $$\sum_{k=0}^{\infty}\binom{\alpha}{k}\binom{\beta}{k}=\binom{\alpha+\beta}{\alpha,\beta}$$ holds even for non-integer $\alpha$ and $\beta$. – Sangchul Lee Nov 27 '20 at 07:41
  • @Iridescent thanks, can you maybe elaborate on the hypergeometric solution or give me a reference? (what is G&R?) – craft Nov 27 '20 at 10:07
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    @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. – Infiniticism Nov 27 '20 at 10:20
  • Thanks! Will do! – craft Nov 27 '20 at 10:22
  • @SangchulLee Indeed, $\sum _{k=0}^{\infty } z^k \binom{a}{k} \binom{b}{k}=, _2F_1(-a,-b;1;z)$ holds. Thus by Gauss formula one have $\sum _{k=0}^{\infty } \binom{a}{k} \binom{b}{k}=\frac{\Gamma (a+b+1)}{\Gamma (a+1) \Gamma (b+1)}$. – Infiniticism Nov 27 '20 at 10:24
  • One may easily generalize this to higher weights, say a $_3F_2$ example $$\sum _{k=0}^{\infty } \binom{\frac{1}{2}}{k}^2 \binom{-\frac{1}{2}}{k}=\frac{\pi ^2+\sqrt{2} \left(3 \Gamma \left(\frac{5}{8}\right)^2 \Gamma \left(\frac{7}{8}\right)^2+\Gamma \left(\frac{3}{8}\right)^2 \Gamma \left(\frac{9}{8}\right)^2\right)}{\pi ^3}$$ One may refer to arXiv $2010.03727$ for more Gamma evaluation of hypergeometric series. – Infiniticism Nov 27 '20 at 10:27
  • Cant find a representation for cos ax in G&R – craft Nov 27 '20 at 10:46

1 Answers1

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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 $\tfrac{\pi\Gamma(\alpha+\beta+1)}{\Gamma(\alpha+1)\Gamma(\beta+1)}$.

J.G.
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