Prove that $\sum_{n=2}^{\infty} \frac{z^{n-1}}{\alpha(n-1)+1}$ is equivalent to $\frac{1}{\alpha} \displaystyle \int_{0}^{1}{ \frac{z t^{\frac{1}{\alpha}}}{1-tz}} dt$?
How to prove this zeta function?
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$\begingroup$
real-analysis
sequences-and-series
analysis
integration
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0@Sasha: I see.. sorry for aski$n$g the same questio$n$ but frankly, I am not familiar with zeta function before this.. so, I need a lot of practice. thank you so much! – 2011-12-01
1 Answers
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HINT
Write $\frac1{1-tz}$ as $1+tz + t^2z^2 + \cdots$ and integrate term by term.