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 Post subject: Product and $\Gamma$ functionPosted: Tue May 03, 2016 6:12 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 457
Location: Ioannina, Greece
If $x$ be positive, prove that $x\,\mathop{\prod}\limits_{n=1}^{+\infty}\bigg(\frac{n+x}{n+1}\bigg)^{\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}}=\exp\bigg(\sqrt{\pi}\,\int_{1}^{x}\frac{\Gamma(t)}{\Gamma\big(t+\frac{1}{2}\big)}\,{\rm{d}}t\bigg)\,.$

(I have no solution.)

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Grigorios Kostakos

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 Post subject: Re: Product and $\Gamma$ functionPosted: Tue May 03, 2016 6:04 pm

Joined: Thu Dec 10, 2015 1:58 pm
Posts: 59
Location: India
Taking logarithm on both sides,

\begin{align*}&\log x + \sum\limits_{n=1}^{\infty}\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)} \log \left(\frac{n+x}{n+1}\right) \\&= \sum\limits_{n=0}^{\infty} \frac{1}{4^n}\binom{2n}{n} \int_{0}^{\infty} \frac{e^{-(n+1)t} - e^{-(n+x)t}}{t}\,dt\\&= \int_0^{\infty} \frac{e^{-t} - e^{-xt}}{t\sqrt{1-e^{-t}}}\,dt\\&= \int_0^{\infty} \frac{1}{\sqrt{1-e^{-t}}}\int_1^{x} e^{-ty}\,dy\,dt\\ &= \int_1^{x} \int_0^{1} \frac{t^{y-1}}{\sqrt{1-t}}\,dt\,dy\\&= \int_1^{x} B\left(\frac{1}{2},y\right)\,dy\\&= \sqrt{\pi} \int_1^{x} \frac{\Gamma\left(y\right)}{\Gamma\left(\frac{1}{2}+y\right)}\,dy \end{align*}

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