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1d
awarded  Nice Question
May
23
awarded  Yearling
May
23
revised Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$
deleted 18 characters in body
May
23
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.
May
18
accepted Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
May
13
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.
May
11
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).
May
11
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
May
9
awarded  Popular Question
May
7
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.
May
6
awarded  Good Answer
May
6
awarded  Popular Question
May
6
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
May
6
asked Calculating $\int_0^{\infty } \frac{\log (v+1)}{\sqrt{(v+1)^2+1} \sqrt{(v+1)^2+4 \sqrt{(v+1)^2+1} (v+1)+4}} \, dv$
May
6
awarded  Nice Question
May
5
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.
May
4
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.
May
4
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. :-)
May
3
awarded  Notable Question
May
2
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! :-)