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 1d awarded Nice Question May23 awarded Yearling May23 revised Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$ deleted 18 characters in body May23 comment Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$ @VladimirReshetnikov You're right! Thanks. I fix that now. May18 accepted Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ May13 comment Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ Thank you very much for your work to my question. May11 comment Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ @JohannesTrost the fact that the inverse symbolic calculator in advanced mode doesn't find anything means almost nothing (I tell you that from my experience). May11 revised Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ added 166 characters in body May9 awarded Popular Question May7 comment Calculating $\int_0^{\infty } \frac{\log (v+1)}{\sqrt{(v+1)^2+1} \sqrt{(v+1)^2+4 \sqrt{(v+1)^2+1} (v+1)+4}} \, dv$ @DavidH Thank you for the feedback. May6 awarded Good Answer May6 awarded Popular Question May6 revised Calculating $\int_0^{\infty } \frac{\log (v+1)}{\sqrt{(v+1)^2+1} \sqrt{(v+1)^2+4 \sqrt{(v+1)^2+1} (v+1)+4}} \, dv$ edited title May6 asked Calculating $\int_0^{\infty } \frac{\log (v+1)}{\sqrt{(v+1)^2+1} \sqrt{(v+1)^2+4 \sqrt{(v+1)^2+1} (v+1)+4}} \, dv$ May6 awarded Nice Question May5 comment Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ @ Raymond Manzoni Thank you! :-) Indeed, I'm deeply involved in personal research that covers such integrals, series and limits. May4 comment Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ Thank you for giving insights on the problem (+1). Very nice the formula by Nected Batir I wasn't aware of. May4 comment Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ (+1) A very good starting point! This is the way to go (at least for the first part of the way). However it might be an illusion that the last integral is easier than the initial one. :-) May3 awarded Notable Question May2 comment Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$ @user17762 long time I haven't seen you around. Nice to meet you again! :-)