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7 questions with no upvoted or accepted answers
2 votes
0 answers
140 views

The ultimate polylogarithm ladder

As you can see, here I performed a derivation of a quite simple formula, not much differing from the standard integral representation of the Polylogarithm. Seeking to make it fancier, I arrived at ...
Artur Wiadrowski's user avatar
2 votes
1 answer
225 views

Hyperbolic series similar to Ramanujan’s identities

I want to prove this ,but nothing’s came up in my mind Could Anyone give me a hint or a solution please .i saw another sum looks like this and was solved by hypergeometric function and Residue .i ...
Unik Sillavich's user avatar
1 vote
0 answers
69 views

Polylogarithm further generalized

Here I proposed a generalized formula for the polylogarithm. However, because of a slight mistake towards the end, visible prior to the edit, I was unaware that it yields just a result of an integral ...
Artur Wiadrowski's user avatar
1 vote
0 answers
119 views

Generalized form of this Harmonic Number series $\sum_{n=1}^{\infty} \frac{{H_n}x^{n+1}}{(n+1)^3}$

i've tried to Evaluate $$\int_{0}^{\frac{\pi}{6}}x\ln^2(2\sin(x))dx$$ without using Contour integral first i changed $2\sin(x)$ into polar form ,and i got $$\int_{0}^{\frac{\pi}{6}}x\ln^2(2\sin(x))dx ...
Unik Sillavich's user avatar
0 votes
0 answers
96 views

$\operatorname{Li}_{2} \left(\frac12 \right)$ vs $\operatorname{Li}_{2} \left(-\frac12 \right)$ : some long summation expressions

Throughout this post, $\operatorname{Li}_{2}(x)$ refers to Dilogarithm. While playing with some Fourier Transforms, I came up with the following expressions: $$2 \operatorname{Li}_{2}\left(\frac12 \...
Srini's user avatar
  • 862
0 votes
0 answers
59 views

Stirling number series resummation

\begin{equation}\sum_{m=1}^{\infty}\frac{a_1^3 S_m^{(3)} (u-1)^{m-1} \left(\frac{x}{x-1}\right)^m}{m!}\end{equation} Does somebody know the result of this resummation? Note: $S_m^{(3)} $ belongs to ...
YU MU's user avatar
  • 99
0 votes
0 answers
82 views

General expression of a triangle sequence

\begin{gather*} \frac{1}{4} \\ \frac{1}{4} \quad \frac{1}{4} \\ \frac{11}{48} \quad \frac{1}{4} \quad \frac{11}{48} \\ \frac{5}{24} \quad \frac{11}{48} \quad \frac{11}{48} \quad \frac{5}{24} \\ \frac{...
YU MU's user avatar
  • 99