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20 votes
4 answers
1k views

Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$

How can I prove that $$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$ I think this post can help me, but I'm not sure.
user91500's user avatar
  • 5,626
23 votes
1 answer
533 views

Closed form for $\sum_{n=0}^\infty\frac{\operatorname{Li}_{1/2}\left(-2^{-2^{-n}}\right)}{\sqrt{2^n}}$

Let $$S=\sum_{n=0}^\infty\frac{\operatorname{Li}_{1/2}\left(-2^{-2^{-n}}\right)}{\sqrt{2^n}},\tag1$$ where $\operatorname{Li}_a(z)$ is the polylogarithm. For $a=1/2$ it can be represented as $$\begin{...
Laila Podlesny's user avatar
10 votes
1 answer
600 views

Indefinite integral $\int \arcsin \left(k\sin x\right) dx$

It would take too long to explain the context reasonably well - but in short, this integral, or rather its equivalent $$\int\frac{x\cos x\,dx}{\sqrt{1-k^2\sin^2x}},\qquad 0<k<1$$ is related to ...
Start wearing purple's user avatar
8 votes
7 answers
907 views

Evaluating $\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$

How would you solve the following $$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$$ I might be able to relate the integral to Euler sums .
Zaid Alyafeai's user avatar
9 votes
6 answers
702 views

Computing the value of $\operatorname{Li}_{3}\left(\frac{1}{2} \right) $

How to prove the following identity $$ \operatorname{Li}_{3}\left(\frac{1}{2} \right) = \sum_{n=1}^{\infty}\frac{1}{2^n n^3}= \frac{1}{24} \left( 21\zeta(3)+4\ln^3 (2)-2\pi^2 \ln2\right)\,?$$ Where ...
Chon's user avatar
  • 6,050

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