# Question Relating Gamma Function to Riemann Zeta function reviewed at integers

I was simply reviewing a paper of Ramanujan qualified " On inquiry 330 of Professor Sanjana" when I obtained perplexed pertaining to a suggestion which I am incapable to address. The suggestion is:

$\displaystyle f(p) = \frac{\pi}{2^{p+1}} \frac{\Gamma(p+1)}{ \Bigl[ \Gamma(\frac{P+2}{2}) \Bigr]^{2}}$, after that $$\log{f(p)}= \log(\frac{\pi}{2}) - p\log{2} + \frac{p^2}{2} \Bigl(1-\frac{1}{2}\Bigr)S_{2} - \frac{p^3}{3} \Bigl(1-\frac{1}{2^2}\Bigr)S_{3} + \cdots$$

where $\displaystyle S_{n} = \frac{1}{1^n} + \frac{1}{2^n} + \cdots \text{ad inf.}$

Could any person clarify me regarding just how can i reason this.

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2019-05-09 11:25:17
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Take an appearance at formula 34 on http://mathworld.wolfram.com/GammaFunction.html After taking the log of both sides, the zeta function shows up in basically the very same kind as in your formula. They reference a 1968 paper by Wrench for additional information. I have not resolved the information myself, yet the resemblances stand out, and also it resembles an encouraging strategy.

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2019-05-11 23:42:12
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I assume this is a Maclaurin collection. I do not recognize just how it comes around, yet my reaction would certainly be to expand out the boundless item for $f(p)$ (which comes to be a boundless amount for $\log f(p)$) in powers of $p$.

Included (13/ 8/2010) As a tip : recall that $$\Gamma(p+1)=p\Gamma(p) =e^{-\gamma p}\prod_{k=1}^\infty\left(\frac{n}{n+p}\right)e^{p/n}$$ by the boundless item for the gamma function. Using this to $\Gamma(p+1)/\Gamma(p/2+1)^2$ one obtains some wonderful termination. After that take logs and also use the collection for $\log(1+x)$. You get a double amount ; scuff of summation and also wish that the internal amount involves something like a zeta collection.

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2019-05-10 07:02:53
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