Linked Questions

116
votes
11answers
8k views

Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$

Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$ where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number. The $A(p,q)$'s are known as alternating Euler sums. ...
114
votes
8answers
24k views

How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$

Please help me to find a closed form for this integral: $$I=\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx\tag1$$ I suspect it might exist because there are similar integrals having closed forms: $$\begin{...
66
votes
7answers
3k views

Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$

I found the following formula $$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$ and it is cited that Euler proved the ...
50
votes
7answers
8k views

Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$

I'm trying to find a closed form for the following sum $$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$ where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number. Could you help me with it?
20
votes
5answers
2k views

Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$

How to analytically prove $$\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3) $$ As O.L answer ...
10
votes
4answers
316 views

Closed form for the harmonic approximation sum $\sum _{k=1}^{\infty } \left(H_k^{(2)}-\zeta (2)\right){}^2$

Question Is there a closed form of this harmonic approximation sum $$s=\sum _{k=1}^{\infty } \left(H_k^{(2)}-\zeta (2)\right){}^2\tag{1}$$ The notation is standard. Motivation This question ...
10
votes
3answers
433 views

Evaluation of a dilogarithmic integral

Problem. Prove that the following dilogarithmic integral has the indicated value: $$\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}\stackrel{?}{=}-11\zeta{(5)}+6\zeta{(3)}\...
8
votes
3answers
405 views

Strategies For Summing Harmonic Numbers

Lately, I have found several interesting problems involving Harmonic numbers such as \begin{equation*}\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}=\frac{7\pi^4}{360}\end{equation*} I am not familiar with ...
8
votes
3answers
617 views

Evaluating $\int_0^1 \frac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$

Evaluate $\displaystyle \int\limits_0^1 \dfrac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$ I was wondering if the above had some kind of a closed form, here some of the special cases have ...
7
votes
1answer
5k views

A list of Multiple Zeta values of depth three

The multiple zeta function of depth three has a following integral representation: \begin{eqnarray} \zeta(t^{-1},p,q,r) &:=& \sum\limits_{m_1 > m_2 > m_3 > 0} \frac{t^{m_1}}{m_1^p} \...
7
votes
4answers
445 views

Double harmonic sum $\sum_{n\geq 1}\frac{H^{(p)}_nH_n}{n^q}$

Are there any general formula for the following series $$\tag{1}\sum_{n\geq 1}\frac{H^{(p)}_nH_n}{n^q}$$ Where we define $$H^{(p)}_n= \sum_{k=1}^n \frac{1}{k^p}\,\,\,\,\,H^{(1)}_n\equiv H_n =\...
5
votes
3answers
244 views

Two challenging sums $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}$

where $H_n$ is the harmonic number and can be defined as: $H_n=1+\frac12+\frac13+...+\frac1n$ $H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$ again, my goal of posting these two challenging ...
6
votes
2answers
229 views

On the Euler sum $\sum \limits_{n=1}^{\infty} \frac{H_n^{(4)} H_n^2}{n^6}$

Here is an Euler sum I ran into. $$\mathcal{S} = \sum_{n=1}^{\infty} \frac{\mathcal{H}_n^{(4)} \mathcal{H}_n^2}{n^6}$$ where $\mathcal{H}_n^{(s)}$ is the generalised harmonic number of order $s$. I ...
2
votes
2answers
216 views

Alternating Euler sums

$ \displaystyle \sum_{n=1}^{\infty} (-1)^{(n-1)} \frac{{\rm H}_n}{n^p} $ Does this have a nice closed form? I am trying to evaluate the case of p=4 also
5
votes
3answers
261 views

Closed-forms for the integral $\int_0^1\frac{\rm{Li}_n(x)}{1+x}dx$?

(This is related to this question.) Define the integral, $$I_n = \int_0^1\frac{\rm{Li}_n(x)}{1+x}dx$$ with polylogarithm $\rm{Li}_n(x)$. Given the Nielsen generalized polylogarithm $S_{n,p}(z)$, $$...

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