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 Post subject: Root integralPosted: Wed Mar 01, 2017 3:19 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
Do not think dirty about the "root" here.

Let $k \in \mathbb{N} \cup \{0\}$. Evaluate the parametric integral:

$$\mathcal{J}(k)= \int_1^2 \frac{x^{2k+1}}{\sqrt{(x^2-1)(4-x^2)}} \, {\rm d}x$$

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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 Post subject: Re: Root integralPosted: Mon May 29, 2017 7:28 am

Joined: Thu Dec 10, 2015 1:58 pm
Posts: 59
Location: India
Making the change of variable $x \mapsto x^2$, the integral changes to: $\displaystyle \mathcal{J}(k) = \frac{1}{2}\int_1^4 \frac{x^k}{\sqrt{(x-1)(4-x)}}\,dx$.

Let, $\displaystyle f(z) = \frac{z^k}{\sqrt{(z-1)(4-z)}}$ and consider the integral of $f(z)$ about a positively oriented dumbell contour ($\gamma_{\epsilon}$) around the points $1$ and $4$ and the branch-cut along the line $[1,4]$. Let, $\gamma_{\epsilon}$ shrink 'nicely' to the line $[1,4]$ as $\epsilon \to 0^{+}$ we have:

$$\displaystyle \lim\limits_{\epsilon \to 0^{+}} \int_{\gamma_{\epsilon}} f(z)\,dz = 2\mathcal{J}(k)(-1 + e^{\pi i}) = -2\pi i \text{Res}_{z = \infty} \left(f(z)\right)$$
Hence, \begin{align*} \mathcal{J}(k) = \frac{1}{2}\pi i \text{Res}_{z = \infty} \left(e^{-i\pi/2}z^{k-1}\left(1 - \frac{1}{z}\right)^{-1/2}\left(1 - \frac{4}{z}\right)^{-1/2}\right) &= \frac{\pi}{2} \text{Res}_{z = 0} \left(z^{-(k+1)}\left(1 - z\right)^{-1/2}\left(1 - 4z\right)^{-1/2}\right) \\&= \frac{\pi}{2} \sum_{j=0}^{k} \binom{2j}{j}\binom{2(k-j)}{k-j}\frac{1}{4^{j}} \end{align*}

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