Definition
$$\mathbf{H}_{m}^{(n)}(x) = \sum_{k=1}^\infty \frac{H_k^{(n)}}{k^m} x^k\tag{1}$$
We define $$\mathbf{H}_{m}^{(1)}(x) = \mathbf{H}_{m}(x)=\sum_{k=1}^\infty \frac{H_k}{k^m} x^k \tag{2}$$
Note the alternating general formula
$$\mathbf{H}_{m}(-1) = \sum_{k=1}^\infty (-1)^k \frac{H_k}{k^m} \tag{3}$$
Motivation
(1) seems to be impossible to track so we focus on (2) and (3). It has been proven in [5] and [6] that the form $\mathbf{H}_{2m}(-1)$ has a general formula in terms of zeta functions
$$\begin{align*}
\mathbf{H}_{2m}(-1)
&=\frac{2m+1}{2}\left(1-2^{-2m}\right)\zeta(2m+1)-\frac{1}{2}\zeta(2m+1)\\
&\qquad-\sum_{k=1}^{m-1}\left(1-2^{1-2k}\right)\zeta(2k)\zeta(2m+1-2k)
\end{align*}$$
Up to my knowledge the literature lacks any general formula for $\mathbf{H}_{2m+1}(-1)$. The odd formula seems to contain a finite combination of zeta and polylogs and their multiplication.
Examples
In [1] we see different evaluations for
$$\mathbf{H}_{1}(-1) = \frac{1}{2} \log^2 (2)-\frac{1}{2} \zeta(2)$$
In [2] we have
$$\mathbf{H}_{3}(-1)=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3)$$
In [3] we have some impressive calculations leading to
$$\begin{align}
\color{blue}{\mathbf{H}_{3}(x)}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\color{blue}{\mathbf{H}_{2}(x)}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}.
\end{align}$$
Also in [8]
\begin{align}
\color{blue}{\mathbf{H}_{4}(x)}
=&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x\\&+\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5)\
\end{align}
In [4] I showed
$$\int\limits_0^1 \dfrac{\log^2 (1+x)\log^n x}{x}\; dx =2 (-1)^n(n!)
\left[ \mathbf{H}_{n+2}(-1) +
\left(1-2^{-n-2} \right) \zeta(n+3) \right]$$
Questions
- Can we evaluate
$$\mathbf{H}_{5}(x) , \mathbf{H}_{5}(-1)$$
- Can we show the following has no simple general formula ? $$\mathbf{H}_{2n+1}(x),\mathbf{H}_{2n+1}(-1)$$
Conjectures
- Interestingly the evaluations of $\mathbf{H}_{m}^{(n)}(-1)$ are related to $\mathbf{H}_{m}^{(n)}\left(\frac{1}{2}\right)$ with the same complexity.
- The form $\mathbf{H}_{m}^{(n)}(x)$ seem to involve a finite sum of products of logs,polylogs and zeta values.
- There can exist a recursive formula that connects
$$\mathbf{H}_{m}^{(n)}(x) = \sum_{1\leq s,t < m} (a_{s,t})\,\mathbf{H}_{s}^{(t)}(x)$$
References
[2] Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$
[3] Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$
[4] Evaluating $\int_0^1 \frac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$
[5] https://arxiv.org/pdf/1301.7662.pdf
[6] http://algo.inria.fr/flajolet/Publications/FlSa98.pdf
Related
[8] How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$
[9] Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$
Answer
Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have:
\begin{eqnarray}
{\bf H}_n^{(1)}(t):=
\sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\
&=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t)
\end{eqnarray}
Here PolyLog[,,] is the Nielsen generalised poly-logarithm.
In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have:
\begin{eqnarray}
&&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\
&& \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\
&& \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\
&&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\
&&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=
\end{eqnarray}
In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have:
\begin{eqnarray}
{\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\
&=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\
&=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\
&=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\
&=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1)
\right)
\end{eqnarray}
In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following:
\begin{eqnarray}
{\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1)
\end{eqnarray}
for $n\ge 2$. Otherwise by going back to the original integral representation we have:
\begin{equation}
{\bf H}^{(1)}_1(-1) = -\frac{\pi^2}{12} + \frac{1}{2} \log(2)^2
\end{equation}
It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have:
\begin{eqnarray}
&&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\
&&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\
&&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\
&&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta
\end{eqnarray}
Here $n\ge 2$ and $q \ge 1$.
The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations.
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