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Jul 15, 2020 at 15:25 history edited zeraoulia rafik CC BY-SA 4.0
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Jul 15, 2020 at 15:22 comment added zeraoulia rafik @ChipHurst, I had a wrong linked question, it fixed now , I already asked this question yesterday here
Jul 15, 2020 at 15:19 history edited zeraoulia rafik CC BY-SA 4.0
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Jul 15, 2020 at 13:57 comment added Greg Hurst This doesn't seem to work for $n = 2$. Do you mean to have $\displaystyle \frac1n\sum_{k=0}^n(-1)^k\binom{n}{k}\log^{n-k}(n)\int_0^\infty e^{-x}\log^k(x)\,dx$?
Jul 15, 2020 at 13:37 history answered zeraoulia rafik CC BY-SA 4.0