An attempt to prove the generalization of $\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^{2a}}$

Question

The following classical generalization

$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2a}}=-\left(a+\frac 12\right)\eta(2a+1)+\frac12\zeta(2a+1)+\sum_{j=1}^{a-1}\eta(2j)\zeta(2a+1-2j)$$ where $\eta(a)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^a}=(1-2^{1-a})\zeta(a)$ is the Dirichlet eta function.

was proved by G. Bastien here page 7 Eq. 17 and also by Cornel here.

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I am trying to prove it in a different way but came across an integral that can be calculated by Beta function but I want it in $\zeta$ if possible to get the right result.

Here is my approach which follows from the same idea of my solution here:

By using $$\frac1{n^{2a}}=-\frac1{(2a-1)!}\int_0^1x^{n-1}\ln^{2a-1}(x)\ dx$$

we can write

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^{2a}}=-\frac1{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(x)}{x}\left(\sum_{n=1}^\infty(-x)^nH_n\right)\ dx$$

$$=\frac1{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx=\frac1{(2a-1)!}I_a\tag1$$

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$$I_a=\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx-\underbrace{\int_1^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx}_{x\mapsto 1/x}$$

$$=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx+\color{blue}{\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{1+x}dx}-\int_0^1\frac{\ln^{2a}(x)}{1+x}dx$$

By adding

$$I_a=\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx=\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x}dx-\color{blue}{\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{1+x}dx}$$

to both sides, the blue integral nicely cancels out and we get

$$2I_a=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx+\underbrace{\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{x}dx}_{IBP}-\int_0^1\frac{\ln^{2a}(x)}{1+x}dx$$

$$=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx-\frac{1+2a}{2a}\int_0^1\frac{\ln^{2a}(x)}{1+x}dx$$

where

$$\int_0^1\frac{\ln^{2a}(x)}{1+x}dx=\sum_{n=1}^\infty(-1)^{n-1}\int_0^1 x^{n-1}\ln^{2a}(x)dx=(2a)!\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^{2a+1}}=(2a)!\eta(2a+1)$$

so

$$I_a=\frac12\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx-\left(a+\frac12\right)(2a-1)!\eta(2a+1)\tag2$$

Plug $(2)$ in $(1)$

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^{2a}}=-\left(a+\frac12\right)\eta(2a+1)+\frac1{2(2a-1)!}\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx\tag{3}$$

So any idea how to evaluate the integral in $(3)$ in a way that completes my proof?

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Answer 1

In the question body in Eq $(3)$, we reached

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^{2a}}=-\left(a+\frac12\right)\eta(2n+1)+\frac1{2(2a-1)!}\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx\tag{1}$$

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From following the same approach of this solution, we have

\begin{align} I_a=\int_0^\infty\frac{\ln^{2a-1}(x)\ln(1+x)}{x(1+x)}dx&= - \frac{\partial^{2a-1}}{\partial m^{2a-1}} \frac{\partial}{\partial n} \operatorname{B}(m,n-m) \, \Bigg \rvert_{m=0, \, n=1} \\ &= - \frac{\partial^{2a-1}}{\partial m^{2a-1}} \operatorname{\Gamma}(m) \frac{\partial}{\partial n} \frac{\operatorname{\Gamma}(n-m)}{\operatorname{\Gamma}(n)} \, \Bigg \rvert_{m=0,\, n=1} \\ &= - \frac{\partial^{2a-1}}{\partial m^{2a-1}} \operatorname{\Gamma}(m) \operatorname{\Gamma}(1-m) [\operatorname{\psi}^{(0)} (1-m) + \gamma] ~\Bigg \rvert_{m=0} \\ &= - \frac{\partial^{2a-1}}{\partial m^{2a-1}}\frac{\pi}{\sin(\pi m)} [\operatorname{\psi}^{(0)} (1-m) + \gamma] ~\Bigg \rvert_{m=0} \\ \end{align}

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Edit: Thanks to @Gary for finding the closed form of this limit. I will mention it here with more details:

We have $$ \psi(1-z)+\gamma=\sum\limits_{n= 1}^\infty {\frac{1}{{n!}}\left[ {\frac{{d^n\psi(1-z)}}{{dz^n }}}\right]_{z=0}z^n}=\sum\limits_{n = 1}^\infty {(-1)^n \frac{{\psi^{(n)}(1)}}{{n!}}z^n} $$ for $|z|<1$. Thus, $$ \frac{{\psi(1-z)+\gamma}}{z}=\sum\limits_{n=0}^\infty {(-1)^{n+1} \frac{{\psi^{(n+1)}(1)}}{{(n+1)!}}z^n} $$ for $|z|<1$ (the left-hand side is defined as a limit when $z=0$). We also have $$ \pi z\csc(\pi z)=\sum\limits_{n=0}^\infty {\frac{{(-1)^{n-1}2\pi^{2n} (2^{2n -1}-1)B_{2n}}}{{(2n)!}}z^{2n}}=2\sum\limits_{n=0}^\infty {\eta (2n)z^{2n} } $$ for $|z|<1$ (the left-hand side is defined as a limit when $z=0$).

so we have $$F(z)=\pi\csc(\pi z)[\psi(1-z)+\gamma]=\left(2\sum\limits_{n = 0}^\infty {\eta (2n)z^{2n}}\right)\left(\sum\limits_{n=0}^\infty {(-1)^{n+1} \frac{{\psi^{(n+1)}(1)}}{{(n+1)!}}z^n}\right).$$ Apply Cauchy product:

$$\left(\sum_{n=0}^\infty a_{2n} z^{2n}\right)\left(\sum_{n=0}^\infty b_{n+1} z^n\right)=\sum_{n=0}^\infty\left(\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}a_{2k} b_{n-2k+1}\right)z^n$$

we get

$$F(z)=2\sum_{n=0}^\infty\left(\underbrace{\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor} \eta(2k) (-1)^{n-2k+1}\frac{\psi^{(n-2k+1)}}{(2-2k+1)!}}_{f_n}\right)z^n=2\sum_{n=0}^\infty f_n z^n.$$

Note that

\begin{align} \lim_{z\to0}\frac{\partial^{2a-1}}{\partial z^{2a-1}}F(z)&=2\lim_{z\to0}\frac{\partial^{2a-1}}{\partial z^{2a-1}} \sum_{n=0}^\infty f_n z^n\\ &=2\lim_{z\to0}\frac{\partial^{2a-1}}{\partial z^{2a-1}}\left(f_0 z^0+f_1 z^1+f_2 z^2+...\right)\\ &=2(2a-1)! f_{2a-1}\\ &=2(2a-1)!\sum_{k=0}^{\lfloor \frac{2a-1}{2}\rfloor} \eta(2k)(-1)^{2a-2k}\frac{\psi^{(2a-2k)}}{(2a-2k)!}\\ &=2(2a-1)!\sum_{k=0}^{a-1}\frac{\eta(2k)}{(2a-2k)!}\psi^{(2a-2k)}(1). \end{align} Substitute $\psi^{(a)}(1)=(-1)^{a-1}a!\zeta(a+1)$,

$$\lim_{z\to 0}\frac{\partial^{2a-1}}{\partial z^{2a-1}}F(z)=-2(2a-1)!\sum_{k=0}^{a-1}\eta(2k)\zeta(2a-2k+1)=-I_a.\tag{2}$$

Plug $(2)$ in $(1)$ we get

$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2a}}=-\left(a+\frac 12\right)\eta(2a+1)+\sum_{k=0}^{a-1}\eta(2k)\zeta(2a-2k+1).$$

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acknowledgement:

Big thanks to @ComplexYetTrivial and @Gary for their solutions here and here. With their help, the proof is now completed rigorously which I think is a new proof in the literature.

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More rigorous proof of $\displaystyle\pi z\csc(\pi z)=\sum_{n=0}^\infty \eta(2n) x^{2n}:$

Set $x=0$ in the Fourier series of $\cos(z x)$: $$\cos(zx)=\frac{2z\sin(\pi z)}{\pi}\left[\frac1{2z^2}-\sum_{n=1}^\infty\frac{(-1)^n\cos(nx)}{n^2-z^2}\right],$$ we get \begin{align} \pi z \csc(\pi z)&=1-2\sum_{n=1}^\infty\frac{(-1)^n z^2}{n^2-z^2}\\ &=1-2\sum_{n=1}^\infty (-1)^n \frac{z^2/n^2}{1-z^2/n^2}\\ &=1-2\sum_{n=1}^\infty (-1)^n \left(\sum_{k=1}^\infty (z^2/n^2)^k\right)\\ &=1-2\sum_{k=1}^\infty z^{2k} \left(\sum_{n=1}^\infty \frac{(-1)^n}{n^{2k}}\right)\\ &=1-2\sum_{k=1}^\infty z^{2k} \left(-\eta(2k)\right)\\ &=2\left(\frac12+\sum_{k=1}^\infty \eta(2k) z^{2k}\right)\\ &=2\sum_{k=0}^\infty \eta(2k) z^{2k}. \end{align}

Answer 2

A different proof with a big bonus:

By the definition of the skew harmonic number we have \begin{gather} \sum_{n=1}^\infty\frac{(-1)^n \overline{H}_n}{n^{2q}}=\sum_{n=1}^\infty \frac{(-1)^n}{n^{2q}}\left(\ln(2)-\int_0^1 \frac{(-x)^n}{1+x}\mathrm{d}x\right)\\ =\ln(2)\sum_{n=1}^\infty \frac{(-1)^n}{n^{2q}}-\sum_{n=1}^\infty\frac{1}{n^{2q}}\int_0^1\frac{x^n}{1+x}\mathrm{d}x\\ =-\ln(2)\eta(2q)-\sum_{n=1}^\infty\frac{1}{n^{2q}}(\ln(2)+H_{\frac n2}-H_n)\\ =-\ln(2)\eta(2q)-\ln(2)\zeta(2q)-\sum_{n=1}^\infty\frac{H_{\frac n2}}{n^{2q}}+\sum_{n=1}^\infty\frac{H_n}{n^{2q}}.\label{fbi} \end{gather} To get the first sum, set $p=2$ and replace $q$ by $2q$ in (1), \begin{gather} \sum_{n=1}^\infty\frac{H_{\frac n2}}{n^{2q}}=2 \sum_{n=1}^\infty\frac{H_{2n}}{(2n)^{2q}}-\sum_{j=1}^{2q-2}(-2)^{-j}\zeta(2q-j)\zeta(j+1)\\ =\sum_{n=1}^\infty\frac{H_{n}}{n^{2q}}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^{2q}}-\sum_{j=1}^{2q-2}(-2)^{-j}\zeta(2q-j)\zeta(j+1). \end{gather} Thus we have \begin{gather} \sum_{n=1}^\infty\frac{(-1)^n \overline{H}_n}{n^{2q}}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^{2q}}=\sum_{j=1}^{2q-2}(-2)^{-j}\zeta(2q-j)\zeta(j+1)\\ -\ln(2)(\eta(2q)+\zeta(2q)).\tag{a} \end{gather} To establish another relation, let $a_n=\frac{\overline{H}_{n}}{n^{2q}}$ in $$\sum_{n=1}^\infty a_{2n}=\frac12\sum_{n=1}^\infty a_n+\frac12\sum_{n=1}^\infty (-1)^n a_n,$$ we obtain \begin{gather*} \sum_{n=1}^\infty \frac{(-1)^n\overline{H}_{n}}{n^{2q}}+\sum_{n=1}^\infty \frac{\overline{H}_{n}}{n^{2q}}=2\sum_{n=1}^\infty \frac{\overline{H}_{2n}}{(2n)^{2q}}\\ \{\text{substitute $\overline{H}_{2n}=H_{2n}-H_n$}\}\\ =2\sum_{n=1}^\infty \frac{H_{2n}}{(2n)^{2q}}-2\sum_{n=1}^\infty \frac{H_{n}}{(2n)^{2q}}\\ =\sum_{n=1}^\infty \frac{(-1)^nH_{n}}{n^{2q}}+\sum_{n=1}^\infty \frac{H_{n}}{n^{2q}}-2\sum_{n=1}^\infty \frac{H_{n}}{(2n)^{2q}}\\ =\sum_{n=1}^\infty \frac{(-1)^nH_{n}}{n^{2q}}+(1-2^{1-2q})\sum_{n=1}^\infty \frac{H_{n}}{n^{2q}}. \end{gather*} Rearrange the terms, \begin{equation} \sum_{n=1}^\infty \frac{(-1)^n\overline{H}_{n}}{n^{2q}}-\sum_{n=1}^\infty \frac{(-1)^nH_{n}}{n^{2q}}=(1-2^{1-2q})\sum_{n=1}^\infty \frac{H_{n}}{n^{2q}}-\sum_{n=1}^\infty \frac{\overline{H}_{n}}{n^{2q}}.\tag{b} \end{equation} Taking the difference of $(a)$ and $(b)$ then dividing by $2$ gives \begin{gather*} \sum_{n=1}^\infty \frac{(-1)^nH_{n}}{n^{2q}}=-\frac12\ln(2)(\zeta(2q)+\eta(2q))+\frac12\sum_{j=1}^{2q-2}(-2)^{-j}\zeta(2q-j)\zeta(j+1)\\ +\frac12\sum_{n=1}^\infty \frac{\overline{H}_{n}}{n^{2q}}-\frac{1-2^{1-2q}}{2}\sum_{n=1}^\infty \frac{H_{n}}{n^{2q}} \end{gather*}

and combining $(a)$ and $(b)$ then dividing by $2$ gives \begin{gather*} \sum_{n=1}^\infty \frac{(-1)^n\overline{H}_{n}}{n^{2q}}=-\frac12\ln(2)(\zeta(2q)+\eta(2q))+\frac12\sum_{j=1}^{2q-2}(-2)^{-j}\zeta(2q-j)\zeta(j+1)\\ -\frac12\sum_{n=1}^\infty \frac{\overline{H}_{n}}{n^{2q}}+\frac{1-2^{1-2q}}{2}\sum_{n=1}^\infty \frac{H_{n}}{n^{2q}}. \end{gather*}

Substitute

$$\sum_{m=1}^\infty\frac{H_m}{m^q} =\frac{q+2}{2}\zeta(q+1)-\frac12\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)$$

and

$$\sum_{m=1}^\infty\frac{\overline{H}_m}{m^q}=\left(1-2^{-q}-\frac{q}{2}\right)\zeta(q+1)+(2-2^{1-q})\ln(2)\zeta(q)$$ $$+\frac12\sum_{j=1}^{q-2}(1-2^{1-q+j})(1-2^{-j})\zeta(q-j)\zeta(j+1)$$

we get $$ \sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2q}}=-\,\frac{2^{2q+1}q-2q-1}{2^{2q+1}}\zeta(2q+1)$$ $$+\frac14\sum_{j=1}^{2q-2}\left[2-2^{-j}-(-2)^{1-j}-2^{j-2q+1}\right]\zeta(2q-j)\zeta(j+1)\tag{c}$$

and

$$\sum_{n=1}^\infty\frac{(-1)^n \overline{H}_n}{n^{2q}}=(2^{1-2q}-2)\ln(2)\zeta(2q)+\frac{2^{2q+1}q-2q-1}{2^{2q+1}}\zeta(2q+1)$$ $$+\frac14\sum_{j=1}^{2q-2}\left[2^{j-2q+1}-(-2)^{1-j}+2^{-j}-2\right]\zeta(2q-j)\zeta(j+1)\tag{d}.$$

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Note: we can see that the the form of $(c)$ is different than the form in the question body but for sure they are equivalent as they both give the right results for any $q\in N.$