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Question Number 144928 by ArielVyny last updated on 30/Jun/21
∑n⩾0(−1)n(n+z)n!
Answered by Dwaipayan Shikari last updated on 01/Jul/21
∫01∑n⩾0(−1)nn!xn+z−1dx=∫01e−xxz−1dx=[−e−xxz−1]01+(z−1)∫01e−xxz−2dxI(z−1)=−1e+(z−1)I(z−2)...Or∫01e−xxz−1dz=Γ(1,z)(IncompleteGammafunction)
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