2
$\begingroup$

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?

$\endgroup$
13
  • 1
    $\begingroup$ 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. $\endgroup$ – Sangchul Lee Nov 26 '20 at 19:58
  • $\begingroup$ Good question. It is not specified in the exercise :/ just that it should be solved by the identity. I would assume any integer $\endgroup$ – craft Nov 26 '20 at 20:01
  • $\begingroup$ @SangchulLee do you know how to evaluate it over the hypergeometric identity? $\endgroup$ – craft Nov 26 '20 at 22:39
  • $\begingroup$ @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 $I=\frac{\pi \Gamma(a+b+1)}{\Gamma(a+1)\Gamma(b+1)}$. $\endgroup$ – Iridescent Nov 27 '20 at 3:42
  • 1
    $\begingroup$ @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. $\endgroup$ – Iridescent Nov 27 '20 at 10:20
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.