A_math_ninja
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 Feb 3 awarded Yearling Jan 14 comment Confirmation of a nice closed form Good job there! Jan 14 accepted Confirmation of a nice closed form Jan 13 comment Confirmation of a nice closed form @KellenMyers Right. I just corrected that $\frac{24}{5} \sin \left(\frac{5 \pi }{24}\right) \Gamma \left(\frac{7}{12}\right)$. Jan 13 comment Confirmation of a nice closed form @george2079 yeah, that should be $\frac{2}{15} \sin \left(\frac{5 \pi }{24}\right) \Gamma \left(\frac{7}{12}\right)$. My calculations suggest the same thing. You can post that as an answer. Jan 13 comment Confirmation of a nice closed form @BobHanlon thank you for the helping hand Jan 13 revised Confirmation of a nice closed form deleted 189 characters in body Jan 13 asked Confirmation of a nice closed form Jan 13 revised An integral with a fractional part in 3 dimensions edited title Sep 7 awarded Popular Question Aug 25 awarded Custodian Aug 25 reviewed Approve Computing a binomial series Jun 29 comment Calculating $\lim_{x\to 1} \, \int -\frac{i \text{Li}_2\left(x-x^2\right)}{\sqrt{3} \left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)} \, dx$ Many thanks (+1) Jun 29 comment Calculating $\lim_{x\to 1} \, \int -\frac{i \text{Li}_2\left(x-x^2\right)}{\sqrt{3} \left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)} \, dx$ Many thanks (+1) Jun 29 comment Calculating $\lim_{x\to 1} \, \int -\frac{i \text{Li}_2\left(x-x^2\right)}{\sqrt{3} \left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)} \, dx$ Many thanks (+1) Jun 28 asked Calculating $\lim_{x\to 1} \, \int -\frac{i \text{Li}_2\left(x-x^2\right)}{\sqrt{3} \left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)} \, dx$ May 28 awarded Nice Question May 4 comment About an infinite product @MichaelE2 thank you for information. Glad for this improvement. :-) Feb 3 awarded Yearling Jan 20 accepted A regularized hypergeometric function related question