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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