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For n >1 Show what Laplace transform of t^(n-1) /(1-e^(-t))
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I want to show that Laplace of \[\frac{ t^{(n-1) }}{ 1-e^(-t) }\] = \[\Gamma (n) \sum_{k=0}^{\infty} \frac{ 1 }{ (s+k)^n} \]
\[ \int_0^\infty \frac{t^{(n-1)}}{1 - e^{-t}} e^{-st}dt = \int_0^\infty t^{(n-1)} \sum_{k = 0}^\infty e^{-kt} e^{-st}dt \\ = \sum_{k=0}^\infty \int_0^\infty t^{n-1}e^{-(k+s)t}ds = \Gamma(n-1)\sum_{k=0}^\infty \frac{1}{(k+s)^{n}} \]
Does the \[\Gamma (n-1) \] different from \[\Gamma(n)\] ??
\[ \Gamma (n) = (n-1) \Gamma(n-1) \]
woops!! that's supposed to be \[ \Gamma (n) \sum_{k=0}^{\infty} \frac{ 1 }{ (s+k)^n}\]
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