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.

2019-05-09 11:25:17
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Answers: 2

Take an appearance at formula 34 on 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.

2019-05-11 23:42:12

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.

2019-05-10 07:02:53