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Prove that $\sum_{n=2}^{\infty} \frac{z^{n-1}}{\alpha(n-1)+1}$ is equivalent to $\frac{1}{\alpha} \displaystyle \int_{0}^{1}{ \frac{z t^{\frac{1}{\alpha}}}{1-tz}} dt$?

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@Norylda I wonder if you read this answer of mine to your earlier question. – Sasha Nov 30 '11 at 18:40
You question concerns $\sum_{n=2}^{\infty} \frac{z^{n-1}}{\alpha(n-1)+1}=z \sum_{n=0}^\infty \frac{z^{n}}{\alpha n + \alpha + 1}$, hence, according to my prior answer, it equals $z \frac{1}{\alpha} \int_0^1 \frac{u^{\frac{\alpha+1}{\alpha}-1}}{1-z u} \mathrm{d} u$, which is $z \frac{1}{\alpha} \int_0^1 \frac{u^{\frac{1}{\alpha}} }{1-z u} \mathrm{d} u$. – Sasha Nov 30 '11 at 19:14
@Sasha: I see.. sorry for asking the same question but frankly, I am not familiar with zeta function before this.. so, I need a lot of practice. thank you so much! – DRN Dec 1 '11 at 0:30


Write $\frac1{1-tz}$ as $1+tz + t^2z^2 + \cdots$ and integrate term by term.

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