Is there a way, using the definition,ζ(s)=∑i=1∞i−sto find this?

Consider the integral∫0∞xtx−1et+1dt=∫0∞xtx−11+e−te−tdt=x∑k=1∞(−1)k−1∫0∞tx−1e−ktdt=x∑k=1∞(−1)k−1k−x∫0∞tx−1e−tdt=xη(x)Γ(x)(1−21−x)ζ(x)Γ(x+1)

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Bradley Barron

2022-04-13

Is there a way, using the definition,

ζ (s) = \sum_{i = 1}^{\infty} i^{- s}

to find this?

mortaraskxnl

Beginner2022-04-14Added 12 answers

Consider the integral

\int_{0}^{\infty} \frac{x t^{x - 1}}{e^{t} + 1} d t

= \int_{0}^{\infty} \frac{x t^{x - 1}}{1 + e^{- t}} e^{- t} d t

= x \sum_{k = 1}^{\infty} {(- 1)}^{k - 1} \int_{0}^{\infty} t^{x - 1} e^{- k t} d t

= x \sum_{k = 1}^{\infty} {(- 1)}^{k - 1} k^{- x} \int_{0}^{\infty} t^{x - 1} e^{- t} d t

= x η (x) Γ (x)

(1 - 2^{1 - x}) ζ (x) Γ (x + 1)

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