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1d
comment a problem about averages of fractional parts
@user115608: the answer to $(4)$ is a good part of de la Vallée Poussin's (very long) linked paper proving the first part of the Prime Number Theorem so no hope for that! As indicated Pillichshammer proves a more general formula (than the one you ask) another way. You may try this one! Of you could consider too the claim as a theorem from de la Vallée Poussin (which it really is).
2d
comment Differentiating with respect to the limit of integration
In case of doubts you may go back to the bases. Your first answer appears right, the second doesn't make sense to me (integration variables are 'dummy' and the answer should be $0$), the third should be right (you may too put $x$ out of the integral).
2d
comment sth interesting about $\gamma$ the euler's Constant
There is even an (excellent) dedicated book by Julian Havil "Gamma" (some opinions) (a large part of the book is about the $\,\zeta\,$ function).
Oct
19
comment a problem about averages of fractional parts
with $\,\theta(n):=\sum_{p\le n}\log(p)\;$ the first Chebyshev function... Perhaps that the proof from the reference at the end will be simpler after all!
Oct
19
comment a problem about averages of fractional parts
make many manipulations to obtain (at the limit $s\to 1$) $$\sum_{p\le y} \frac{\log(p)}{p-1}-\frac 1y\sum_{p\le y}\log(p)=\log(y)-1-\lim_{s\to 1}\left[\frac{\zeta'(s)}{\zeta(s)}+\frac 1{s-1}\right]+\epsilon$$ The limit of the [ ] term will be $\gamma$ because $\zeta(1+e)=1/e+\gamma+O(e)\,$ from the Stieltjes expansion while $ \frac 1y\sum_{p\le y}\log(p)$ goes to $1$ (your question 2.) getting thus $(4)$.$$-$$ 2. de la Vallée Poussin proved that the PNT was equivalent to $\;\lim_{n\to\infty}\dfrac {\theta(n)}n=1\,$
Oct
19
comment a problem about averages of fractional parts
1. De la Vallée Poussin's original (very long) proof may be consulted here (from $24.$ page 99 to $28.$ p.34, $52.$ p.61, $55.$ p.64 and $57.$ P.69). The idea is to start with $\;\log \zeta(s)=-\sum_p \log\left(1-\frac 1{p^s}\right)$ for $s>1$, differentiate it to get $\displaystyle \,\dfrac{\zeta'(s)}{\zeta(s)}= -\sum_p \frac{\log(p)}{p^s-1}=-\sum_p \frac{\log(p)}{p^s}-\sum_p \frac{\log(p)}{p^{2s}}-\cdots$, use the logarithmic derivative of the functional equation to rewrite this in another way (in $8.$ p.9)...
Oct
19
revised a problem about averages of fractional parts
deleted 3 characters in body
Oct
19
comment a problem about averages of fractional parts
@user115608: I edited my answer and hope it will be clearer. Fine continuation,
Oct
19
revised a problem about averages of fractional parts
Adding details to the proof.
Oct
19
comment a problem about averages of fractional parts
Glad it helped @user115608. For $(3)$ use simply $\dfrac n{p-1}=\dfrac np+\dfrac n{p^2}+\cdots$.
Oct
19
answered a problem about averages of fractional parts
Oct
18
comment Closed form of the sequence ${_2F_1}\left(\begin{array}c\tfrac12,-n\\\tfrac32\end{array}\middle|\,\frac{1}{2}\right)$
@user153012: As explained by Lucian in the comments we may rewrite it as an incomplete beta function (with the substitution $t:=\sqrt{2}\sin(x)$) getting : $$a_n=\sqrt{2}\int_0^{\pi/4}\cos(x)^{2n+1}\;dx=\frac 1{\sqrt{2}}B\left(\frac 12,\frac 12, n+1\right)$$ as you may try in Wolfram (which doesn't seem to help...)
Oct
18
revised Closed form of the sequence ${_2F_1}\left(\begin{array}c\tfrac12,-n\\\tfrac32\end{array}\middle|\,\frac{1}{2}\right)$
simpler form
Oct
18
revised Closed form of the sequence ${_2F_1}\left(\begin{array}c\tfrac12,-n\\\tfrac32\end{array}\middle|\,\frac{1}{2}\right)$
sign error
Oct
18
revised Closed form of the sequence ${_2F_1}\left(\begin{array}c\tfrac12,-n\\\tfrac32\end{array}\middle|\,\frac{1}{2}\right)$
hypergeometric tag
Oct
18
answered Closed form of the sequence ${_2F_1}\left(\begin{array}c\tfrac12,-n\\\tfrac32\end{array}\middle|\,\frac{1}{2}\right)$
Oct
15
awarded  Nice Answer
Oct
13
comment Summation of convergent series
A general method using $\psi$ and the polygamma function is provided in $(6.8)$ of A&S.
Oct
4
comment Shape made by Beltrami
Well representing the complete hyperbolic plane may be 'difficult' from Hilbert's theorem. See this MO thread but 'crocheting representations' were given in this neat paper from Henderson and Taimina.
Oct
4
comment Shape made by Beltrami
!Here is for example a picture for $\,(x,y)\mapsto (x^2+y^2)\sin(12\,\arctan(y/x))$.