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How to show that

$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac{\psi^{(3)}\left(\frac14\right)}{384}-\frac{\pi^4}{48}-\frac{35\pi}{128}\zeta(3)$$

without using the generating function:

\begin{align} \sum^\infty_{n=1}\frac{H_n}{n^3}z^n =&2{\rm Li}_4(z)+{\rm Li}_4\left(\tfrac{z}{z-1}\right)-{\rm Li}_4(1-z)-{\rm Li}_3(z)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\tfrac{z}{z-1}\right)\\ &+\frac{1}{2}{\rm Li}_2(z)\ln^2(1-z)+\frac{1}{2}{\rm Li}_2^2(z)+\frac{1}{6}\ln^4(1-z)-\frac{1}{6}\ln{z}\ln^3(1-z)\\ &+\frac{\pi^2}{12}\ln^2(1-z)+\zeta(3)\ln(1-z)+\frac{\pi^4}{90} \end{align}

The common proof is to use the series property $$\sum_{n=0}^\infty (-1)^n f(2n+1)=\Im \left\{\sum_{n=1}^\infty i^n f(n)\right\}$$

then we apply the generating function above by setting $z=i$ but as you can see too much tedious calculations involved which is the reason I am asking for a different approach. By the way, you can find here a similar question that could be helpful. All methods are appreciated.

Thank you.


Edit: This question was solved here but I am looking for an elegant method as question title says.

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We can use contour integration to evaluate this sum. The paper Euler sums and contour integral representations by Philippe Flajolet and Bruno Salvy gives many examples of Euler sums that can be evaluated using this approach.

Proof

First note that

$$ \sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3} = \sum_{n=0}^\infty (-1)^n\frac{H_{2n} + \frac{1}{2n+1}}{(2n+1)^3} = -\sum_{n=1}^\infty (-1)^{n+1}\frac{H_{2n}}{(2n+1)^3} +\beta(4) $$ Now, we'll integrate the function $f(z) = \pi \csc(\pi z) \frac{\gamma+\psi_0(-2z+1)}{(-2z+1)^3}$ around the positively oriented square, $C_N$, with vertices $\pm \left(N+\frac{1}{4} \right)\pm \left(N+\frac{1}{4} \right)i$. It is easy to show that $$ \lim_{N\to \infty}\int_{C_N}f(z)\; dz = 0 $$ See Appendix A for the proof. Hence, the sum of all residues of $f(z)$ at its poles is equal to $0$. The calculation of residues is a tedious but straightforward exercise. The list of local expansions of different kernel functions given on page 20 of the paper mentioned above are quite useful for carrying out these computations.

We have

\begin{align*} \mathop{\text{Res}}\limits_{z=-n} f(z) &= (-1)^n \frac{\psi_0(2n+1)+\gamma}{(2n+1)^3} = (-1)^n \frac{H_{2n}}{(2n+1)^3} , \quad n=0,1,2,\cdots \\ \mathop{\text{Res}}\limits_{z=n} f(z) &= \frac{(-1)^{n+1}H_{2n-1}}{(2n-1)^3}- 3\frac{(-1)^{n+1}}{(2n-1)^4}, \quad n=1,2,3,\cdots \\ \mathop{\text{Res}}\limits_{z=\frac{2n+1}{2}} f(z) &= \frac{(-1)^{n+1} \pi}{16 n^3} , \quad n=1,2,3,\cdots \\ \mathop{\text{Res}}\limits_{z=\frac{1}{2}} f(z) &= \frac{\pi \zeta(3)}{2} \end{align*} Adding up all the residues gives us: \begin{align*} \frac{\pi \zeta(3)}{2}+\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} + \frac{\pi}{16}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3} + \sum_{n=1} ^\infty \frac{(-1)^{n+1}H_{2n-1}}{(2n-1)^3} -3\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-1)^4}&= 0\\ \implies \frac{\pi \zeta(3)}{2}+\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} + \frac{\pi}{16}\left(\frac{3\zeta(3)}{4} \right) + \sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} -2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-1)^4} &= 0 \\ \implies -2\sum_{n=1}^\infty \frac{(-1)^{n+1}H_{2n}}{(2n+1)^3} + \frac{35\pi \zeta(3)}{64} -2\beta(4) = 0 \\ \implies \boxed{\sum_{n=1}^\infty \frac{(-1)^{n+1}H_{2n}}{(2n+1)^3} = -\beta(4) + \frac{35\pi \zeta(3)}{128}} \end{align*}

So, we get: $$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3} = -\sum_{n=1}^\infty (-1)^{n+1}\frac{H_{2n}}{(2n+1)^3} +\beta(4) =2\beta(4) - \frac{35\pi \zeta(3)}{128}$$

Appendix A: Proving $\lim_{N\to \infty}\int_{C_N} f(z) dz=0$

(1) First, note that $|\pi \csc(\pi z)| \leq \pi \sqrt{2}$ for all $z\in C_N$ and $N\geq 1$. Indeed, if $\text{Re } z = \pm \left(N+\frac{1}{4}\right)$ and $\text{Im }z = y$, we have \begin{align*} \left|\pi \csc\left(\pm\pi\left(N+\frac{1}{4} \right) +i \pi y\right) \right| &= \left|\frac{\pi }{\sin\left(\pm \frac{\pi}{4} + i\pi y\right)} \right| \\ &= \frac{\pi \sqrt{2}}{|\pm \cosh(\pi y) + i \sinh(\pi y)|} \\ &= \frac{\pi \sqrt{2}}{\sqrt{\cosh(2\pi y)}} \\ &\leq \pi \sqrt{2} \end{align*} Similarly, when $\text{Re }z =x$ and $\text{Im }z = \pm i\left(N+\frac{1}{4} \right)$ we have \begin{align*} \left|\pi \csc \left(\pi x \pm i \pi\left(N+\frac{1}{4} \right)\right)\right| &= \frac{\pi}{\sqrt{\sin^2(\pi x) + \sinh^2(\pi N + \frac{\pi}{4})}}\\&\leq \frac{\pi}{\sinh\left(\pi N + \frac{\pi}{4} \right)} < \pi \sqrt{2} \end{align*}

(2) Using the fact that $|\gamma + \psi_0(-2z+1)|< |-2z+1|$ for all $z\in C_N$ and $N\geq 1$, we can bound the absolute value of our contour integral as follows:

\begin{align*} \left|\int_{C_N} f(z)\; dz \right| &< \pi \sqrt{2} \int_{C_{N}}\frac{1}{|2z-1|^2}|dz| \\ &\leq \frac{4\pi \sqrt{2}}{(4N-1)^2}\int_{C_N} |dz| \\ &= 4\pi \sqrt{2}\frac{8N+2}{(4N-1)^2} \end{align*} Finally, taking the limit $N\to \infty$ on both sides gives: $$\lim_{N\to \infty}\left|\int_{C_N} f(z) dz \right| = 0$$

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  • $\begingroup$ (+1) very nice. Thank you. $\endgroup$ – Ali Shadhar Aug 28 '20 at 5:22
  • $\begingroup$ @Ali Shather I'll try to add some more details like proof of the contour integral going to 0 as $N\to \infty$ and the residue calculations. $\endgroup$ – Shobhit Bhatnagar Aug 28 '20 at 5:46
  • $\begingroup$ Even better. What i like about your solution is that you didn't use any result. Well self-contained. $\endgroup$ – Ali Shadhar Aug 28 '20 at 5:59
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Using the fact that

$$\Re\left\{\frac{\ln(1-ix)}{1-ix}\right\}=\frac{\arctan(x)}{x}-\frac{\arctan(x)}{x(1+x^2)}+\frac{\ln(1+x^2)}{2(1+x^2)}$$

we have

$$\Re\left\{\int_0^1\frac{\ln^2x\ln(1-ix)}{1-ix}dx\right\}$$ $$=\int_0^1\frac{\ln^2x\arctan(x)}{x}dx-\int_0^1\frac{\ln^2x\arctan(x)}{x(1+x^2)}dx+\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}$$

$$=I_1-I_2+\frac12I_3$$


$$I_1=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1 x^{2n}\ln^2xdx=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}=2\beta(4)$$


$I_2$ was evaluated by a friend (Kartick Betal).

$$I_2=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\underbrace{\int_1^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx}_{\displaystyle x\mapsto 1/x}$$ $$=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\int_0^1 \frac{x\ln^2x\left(\frac{\pi}{2}-\arctan x\right)}{1+x^2}\ dx$$ $$=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{\pi}{2}\int_0^1 \frac{x\ln^2x}{1+x^2}\ dx+\int_0^1 \frac{x\ln^2x\arctan x}{1+x^2}\ dx$$ $$=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{\pi}{2}\cdot\frac3{16}\zeta(3)+\int_0^1 \left(\frac1x-\frac1{x(1+x^2)}\right)\ln^2x\arctan xdx$$ $$=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{3\pi}{32}\zeta(3)+\int_0^1 \frac{\ln^2x\arctan x}{x}\ dx-I$$ $$\Longrightarrow 2I_2=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{3\pi}{32}\zeta(3)+2\beta(4)$$

For the remaining integral, write $\arctan x=\int_0^1\frac{x}{1+x^2y^2}\ dy\ $, we get

$$\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\int_0^\infty \frac{\ln^2x}{x(1+x^2)}\left(\int_0^1\frac{x}{1+x^2y^2}\ dy\right)\ dx$$ $$=\int_0^1\frac{1}{1-y^2}\left(\int_0^\infty\frac{\ln^2x}{1+x^2}\ dx-\int_0^\infty\frac{y^2\ln^2x}{1+x^2y^2}\ dx\right)\ dy$$ $$=\int_0^1\frac{1}{1-y^2}\left(\frac{\pi^3}{8}-\frac{y\pi^3}{8}-\frac{y\pi\ln^2y}{2}\right)\ dy$$ $$=\frac{\pi^3}{8}\int_0^1\frac{1-y}{1-y^2}\ dy-\frac{\pi}2\int_0^1\frac{y\ln^2y}{1-y^2}\ dy$$ $$=\frac{\pi^3}{8}\int_0^1\frac{1}{1+y}\ dy-\frac{\pi}{16}\int_0^1\frac{\ln^2y}{1-y}\ dy$$

$$=\frac{\pi^3}{8}\ln2-\frac{\pi}{8}\zeta(3)$$

Plug in this result, we get $$I_2=\frac{\pi^3}{16}\ln(2)-\frac{7\pi}{32}\zeta(3)+\beta(4)$$


$$I_3=\int_0^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx-\underbrace{\int_1^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx}_{\large x\mapsto1/x}$$ $$=\underbrace{\int_0^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx}_{\large x^2\mapsto x}-I_3+2\int_0^1\frac{\ln^3x}{1+x^2}\ dx$$ $$\Longrightarrow 2I_3=\frac18\int_0^\infty\frac{\ln^2x\ln(1+x)}{\sqrt{x}(1+x)}\ dx+2(-6\beta(4))$$ $$I_3=\frac1{16}\lim_{a\ \mapsto1/2\\b\ \mapsto1/2}\frac{-\partial^3}{\partial a^2\partial b}\text{B}(a,b)-6\beta(4)$$ $$=\frac{7\pi}{8}\zeta(3)+\frac{\pi^3}{8}\ln(2)-6\beta(4)$$


For the LHS integral, write $\frac{\ln(1-ix)}{1-ix}=-\sum_{n=1}^\infty (ix)^{n-1}H_{n-1}$

$$\Re\left\{\int_0^1\frac{\ln^2x\ln(1-ix)}{1-ix}dx\right\}=\Re\left\{-\sum_{n=1}^\infty i^{n-1}H_{n-1}\int_0^1 x^{n-1}\ln^2xdx\right\}$$

$$=\Im\left\{2\sum_{n=1}^\infty \frac{i^{n}H_{n-1}}{n^3}\right\}=\Im\left\{2\sum_{n=1}^\infty \frac{i^{n}H_{n}}{n^3}-2\sum_{n=1}^\infty \frac{i^{n}}{n^4}\right\}$$

use $\Im \left\{\sum_{n=1}^\infty i^n f(n)\right\}=\sum_{n=0}^\infty (-1)^n f(2n+1)$

$$=2\sum_{n=0}^\infty \frac{(-1)^nH_{2n+1}}{(2n+1)^3}-2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^4}$$

$$=2\sum_{n=0}^\infty \frac{(-1)^nH_{2n+1}}{(2n+1)^3}-2\beta(4)$$


Collect all results and use $\beta(4)=\frac1{768}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4\right)$ we find

$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac{\psi^{(3)}\left(\frac14\right)}{384}-\frac{\pi^4}{48}-\frac{35\pi}{128}\zeta(3)$$

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