How to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}$

Question

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In this post I computed the following integral

$$\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx=\frac{11}{8}\zeta(3)$$

Now I am trying to compute

$$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}dx=\frac{67}{32}\zeta(3)-\frac{\pi}{2}G}$$

What I did is

$$\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}dx=\int_{0}^{1}\frac{\log(1-x)\log[(1-x^2)(1+x^2)]}{x}dx$$

$$=\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx+\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx$$

$$=\frac{11}{8}\zeta(3)+\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx$$

$$=\frac{11}{8}\zeta(3)+I$$

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$$I=\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx=\sum_{k=1}^{\infty}\frac{(-1)^{n-1}}{k}\int_{0}^{1}x^{2k-1}\log(1-x)dx$$

Integrating by parts

$$I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{\frac{\log(1-x)(x^{2n}-1)}{2n}\Big|_{0}^{1}+\frac{1}{2n}\int_{0}^{1}\frac{x^{2n}-1}{1-x}dx \bigg\}$$

$$=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{-\frac{1}{2n}\int_{0}^{1}\frac{1-x^{2n}}{1-x}dx \bigg\}$$

$$=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} }{n}\bigg\{-\frac{1}{2n}H_{2n} \bigg\}$$

$$\boxed{I=\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}}$$

So now, the task is to evaluate this Sum

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Answer 1

To calculate this sum, consider the following generating function

$$\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$$

Letting $x=i$

The left hand side becomes

$$\sum_{n=1}^\infty \frac{i^n}{n^2}H_n=-\frac{H_2}{2^2}+\frac{H_4}{4^2}-\frac{H_6}{6^2}+\frac{H_8}{8^2}- \cdots+i\Big(-\frac{H_1}{1^2}+\frac{H_3}{3^2}-\frac{H_5}{5^2}+\frac{H_7}{7^2}- \cdots \Big)$$

$$\sum_{n=1}^\infty \frac{i^n}{n^2}H_n=\frac{1}{4}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}+i\sum_{n=1}^\infty \frac{(-1)^n}{(2n-1)^{2}}H_{2n-1}$$

We therefore have that

$$Re\Big\{ \sum_{n=1}^\infty \frac{i^n}{n^2}H_n\Big\}=\frac{1}{4}\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}$$

or

$$\boxed{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}=4\Big\{ \sum_{n=1}^\infty \frac{i^n}{n^2}H_n\Big\}}$$ $$Re\Big\{ \sum_{n=1}^\infty \frac{i^n}{n^2}H_n\Big\}=Re\Big\{ \operatorname{Li}_3(i)-\operatorname{Li}_3(1-i)+\operatorname{Li}_2(1-i)\ln(1-i)+\frac{1}{2}\ln (i) \ln^2(1-i)+\zeta(3)\Big\} \quad \quad \tag1$$

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Lets now compute the components of (1) and then, put all together to get the solution.

$$\ln(i)=\ln(e^{i\frac{\pi}{2}})=i\frac{\pi}{2}\quad \quad \tag2$$

$$\ln(1-i)=\ln(\sqrt{2}e^{-i\frac{\pi}{4}})=\frac{1}{2}\ln(2)-i\frac{\pi}{4}\quad \quad \tag3$$

and

$$\ln^2(1-i)=\Big(\frac{1}{2}\ln(2)-i\frac{\pi}{4}\Big)^2=\frac{1}{4}\ln^2(2)-\frac{\pi^2}{16}-i\frac{ \pi}{4}\ln(2)\quad \quad \tag4$$

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$$Li_{2}(i)=\sum_{n=1}^{\infty}\frac{i^n}{n^2}=\frac{i}{1^2}+\frac{(i)^2}{2^2}+\frac{(i)^3}{3^2}+\frac{(i)^4}{4^2}+\cdots$$

$$=\frac{i}{1^2}-\frac{1}{2^2}-\frac{i}{3^2}+\frac{1}{4^2}+\cdots$$

$$=\Big[-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{6^2}+\frac{1}{8^2}-\cdots \Big]+i\Big[\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\cdots \Big]$$

$$Li_{2}(i)=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)^2}+i\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^2}$$

$$Li_{2}(i)=\frac{1}{4}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}+i\beta(2)$$

$$Li_{2}(i)=-\frac{1}{4}\eta(2)+i G$$

$$Li_{2}(i)=-\frac{1}{4} \cdot\frac{\pi^2}{12}+i G$$

$$\boxed{Li_{2}(i)=-\frac{1}{8}\zeta(2)+iG}\quad \quad \tag5$$

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Recall the Dilogarithm reflection formula:

$$Li_{2}(x)+Li_{2}(1-x)=\zeta(2)-\ln(x)\ln(1-x)$$

Plug $x=i$

$$Li_{2}(1-i)=\zeta(2)-\ln(i)\ln(1-i)-Li_{2}(i)$$

$$Li_{2}(1-i)=\zeta(2)-\left(i\frac{\pi}{2}\right) \cdot\left( \frac{\ln(2)}{2}-i\frac{\pi}{4}\right)+\frac{1}{8}\zeta(2)-iG$$

$$Li_{2}(1-i)=\zeta(2)+\frac{1}{8}\zeta(2)-\frac{\pi^2}{8}-i\left(\frac{\pi ln(2)}{4}+G\right)$$

$$Li_{2}(1-i)=\zeta(2)+\frac{1}{8}\frac{\pi^2}{6}-\frac{\pi^2}{8}-i\left(\frac{\pi ln(2)}{4}+G\right)$$

$$Li_{2}(1-i)=\zeta(2)-\frac{5}{8}\frac{\pi^2}{6}-i\left(\frac{\pi ln(2)}{4}+G\right)$$

$$\boxed{Li_{2}(1-i)=\frac{3}{8}\zeta(2)-i\left(\frac{\pi ln(2)}{4}+G\right)}\quad \quad \tag6$$

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$$Li_{3}(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^3}$$

$$Li_{3}(i)=\sum_{k=1}^{\infty}\frac{i^k}{k^3}=\frac{i}{1^3}+\frac{i^2}{2^3}+\frac{i^3}{3^3}+\frac{i^4}{4^3}+\cdots$$

$$=\frac{i}{1^3}-\frac{1}{2^3}-\frac{i}{3^3}+\frac{1}{4^3}+\cdots$$

$$=-\left[\frac{1}{2^3}-\frac{1}{4^3}+\frac{1}{6^3}-\cdots \right]+i\left[\frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3}-\cdots \right]$$

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

$$=-\frac{1}{8}\eta(3)+i\beta(3)$$

$$\boxed{Li_{3}(i)=-\frac{3}{32}\zeta(3)+i\beta(3)}\quad \quad \tag7$$

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$$Li_{3}(x)+Li_{3}(1-x)+Li_{3}\Big(1-\frac{1}{x}\Big)=\zeta(3)+\frac{\ln^3(x)}{6}+\frac{\pi^2\ln(x)}{6}-\frac{\ln^2(x)\ln(1-x)}{2}$$

$$x=i=\frac{i \pi}{2}$$

$$1-\frac{1}{i}=1+i=(1-i)^{*}$$

therefore

$$Li_{3}(x)+Li_{3}(1-i)+Li_{3}\big((1-i)^{*}\big)=\zeta(3)+\frac{\ln^3(i)}{6}+\frac{\pi^2\ln(i)}{6}-\frac{\ln^2(i)\ln(1-i)}{2}\quad \quad \tag8$$

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but

$$1-i=\sqrt{2}e^{-\frac{i\pi}{4}}$$

and

$$1+i=\sqrt{2}e^{\frac{i\pi}{4}}$$

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$$Li_{3}(1-i)+Li_{3}\big((1-i)^{*}\big)$$

$$Li_{3}(1-i)=\sum_{n=1}^{\infty}\frac{2^{k/2}e^{-\frac{ik \pi}{4}}}{k^3}$$

$$Li_{3}\big((1-i)^{*}\big)=\sum_{n=1}^{\infty}\frac{2^{k/2}e^{\frac{ik \pi}{4}}}{k^3}$$

$$Li_{3}(1-i)+Li_{3}\big((1-i)^{*}\big)=\sum_{n=1}^{\infty}\frac{2^{k/2}e^{\frac{ik \pi}{4}}}{k^3}+\sum_{n=1}^{\infty}\frac{2^{k/2}e^{-\frac{ik \pi}{4}}}{k^3}$$

$$\boxed{Li_{3}(1-i)+Li_{3}\big((1-i)^{*}\big)=2\sum_{n=1}^{\infty}\frac{2^{k/2} \cos\big( \frac{k\pi}{4}\big)}{k^3}}\quad \quad \tag9$$

on the other hand

$$Re\{ Li_{3}(1-i)\}= Re\left \{\sum_{n=1}^{\infty}\frac{2^{k/2} e^{-\frac{i \pi}{4}}}{k^3} \right\}$$

$$\boxed{Re\{ Li_{3}(1-i)\}=\sum_{n=1}^{\infty}\frac{2^{k/2} \cos\big( \frac{k\pi}{4}\big)}{k^3}}\quad \quad \tag{10}$$

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From (9) and (10) we conclude that

$$\boxed{Li_{3}(1-i)+Li_{3}\big((1-i)^{*}\big)=2Re\{ Li_{3}(1-i)\}}\quad \quad \tag{11}$$

We can now rewrite (8) as

$$Li_{3}(x)+2Re\{ Li_{3}(1-i)\}=\zeta(3)+\frac{\ln^3(i)}{6}+\frac{\pi^2\ln(i)}{6}-\frac{\ln^2(i)\ln(1-i)}{2}$$

$$2Re\{ Li_{3}(1-i)\}=\zeta(3)+\frac{\ln^3(i)}{6}+\frac{\pi^2\ln(i)}{6}-\frac{\ln^2(i)\ln(1-i)}{2}-Li_{3}(x)$$

$$Re\{ Li_{3}(1-i)\}=\frac{\zeta(3)}{2}+\frac{\ln^3(i)}{12}+\frac{\pi^2\ln(i)}{12}-\frac{\ln^2(i)\ln(1-i)}{4}-\frac{Li_{3}(x)}{2}$$

$$Re\{ Li_{3}(1-i)\}=\frac{\zeta(3)}{2}+\frac{1}{12}\frac{-i\pi^3}{48}+\frac{\pi^2}{12}\frac{i \pi}{2}+\frac{1}{4}\frac{\pi^2}{4}\Big( \frac{\ln(2)}{2}+i\frac{\pi}{4}\Big)-\frac{Li_{3}(x)}{2}$$

$$Re\left\{ Li_{3}(1-i)\right\}=\frac{35}{64}\zeta(3)-\frac{\pi^2}{32}\ln(2)$$

$$\boxed{Re\left\{ Li_{3}(1-i)\right\}=\frac{35}{64}\zeta(3)-\frac{3}{16}\zeta(2)\ln(2)}\quad \quad \tag{12}$$

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From (1) and (3) we have

$$\ln(i)\ln^2(1-i)=\left(i\frac{\pi}{2} \right)\left(\frac{1}{4}\ln^2(2)-\frac{\pi^2}{16}-i\frac{ \pi}{4}\ln(2) \right)$$

$$\boxed{\ln(i)\ln^2(1-i)=\frac{\pi^2}{8}\ln(2)-i\left[\frac{\pi^3}{32}-\frac{\pi}{8}\ln^2(2) \right]}\quad \quad \tag{13}$$

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$$\ln(1-i)Li_{2}(1-i)=\left(\frac{1}{2}\ln(2)-i\frac{\pi}{4} \right)\left(\frac{3}{8}\zeta(2)-i\left(\frac{\pi ln(2)}{4}+G\right) \right)$$

$$\ln(1-i)Li_{2}(1-i)=\frac{3}{16}\zeta(2)\ln(2)-\frac{\pi^2 \ln(2)}{16}-\frac{\pi }{4}G-i\frac{1}{2}\ln(2)\left(\frac{\pi ln(2)}{4}+G\right)-i\frac{3 \pi}{32}\zeta(2)$$

$$\ln(1-i)Li_{2}(1-i)=\frac{\pi^2}{32}\ln(2)-\frac{\pi^2 \ln(2)}{16}-\frac{\pi }{4}G-i\frac{1}{2}\ln(2)\left(\frac{\pi ln(2)}{4}+G\right)-i\frac{3 \pi}{32}\zeta(2)$$

$$\boxed{\ln(1-i)Li_{2}(1-i)=-\frac{\pi^2}{32}\ln(2)-\frac{\pi }{4}G-i\left(\frac{\pi ln^2(2)}{8}+\frac{1}{2}\ln(2)G+\frac{ \pi^3}{64}\right)}\quad \quad \tag{14}$$

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Plugging the real part of (7), (12), (13) and (14) in (1) we get

$$Re\left\{ \sum_{n=1}^\infty \frac{i^n}{n^2}H_n \right \}=-\frac{3}{32}\zeta(3)-\frac{35}{64}\zeta(3)-\frac{\pi^2}{32}\ln(2)-\frac{\pi^2}{32}\ln(2)-\frac{\pi}{4}G+\frac{\pi^2}{16}\ln(2)+\zeta(3)$$

$$\boxed{Re\left\{ \sum_{n=1}^\infty \frac{i^n}{n^2}H_n \right \}=\frac{23}{64}\zeta(3)-\frac{\pi}{4}G}$$

Consequently we get the beautifull result!

$$\boxed{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}=\frac{23}{16}\zeta(3)-\pi G}$$

$$\int_{0}^{1}\frac{\log(1-x)\log(1+x^2)}{x}dx=\frac{23}{32}\zeta(3)-\frac{\pi}{2}G$$

and finally we get our integral

$$J=\frac{11}{8}\zeta(3)+\frac{23}{32}\zeta(3)-\frac{\pi}{2}G$$

$$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}dx=\frac{67}{32}\zeta(3)-\frac{\pi}{2}G}$$

Answer 2

First note that $$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^2}=4\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^2}=4\Re\sum_{n=1}^\infty\frac{i^nH_{n}}{n^2}.$$

By Cauchy product we have $$-\ln(1-x)\operatorname{Li}_2(x)=2\sum_{n=1}^\infty \frac{H_n}{n^2}x^n+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}x^n-3\operatorname{Li}_3(x).$$

Set $x=i$ and consider the real parts,

$$-\Re\{\ln(1-i)\operatorname{Li}_2(i)\}=2\Re\sum_{n=1}^\infty \frac{i^nH_n}{n^2}+\Re\sum_{n=1}^\infty \frac{i^nH_n^{(2)}}{n}-3\Re\operatorname{Li}_3(i).$$

So we just need to calculate $\Re\sum_{n=1}^\infty \frac{i^nH_n^{(2)}}{n}$which is equivalent to $\frac12\sum_{n=1}^\infty \frac{(-1)^nH_{2n}^{(2)}}{n}$ whose integral representation $\int_0^1\frac{\ln(x)\ln(1+x^2)}{1-x}dx$ is calculated here. To show this conversion, expand $\ln(1+x^2)$ in Taylor series then integrate using $\int_0^1\frac{x^{2n}\ln(x)}{1-x}dx=\zeta(2)-H_{2n}^{(2)}.$

Answer 3

In theorem 6 of this preprint, we have the following generalization:

$$\begin{equation*} \sum_{n=1}^\infty\frac{{(-1)}^{n}H_{2n}}{n^{2a}}=-\frac14(2a+1)\eta(2a+1)-2^{2a-2}\pi\beta(2a)+\sum_{k=0}^{a-1}2^{2a-2k}\eta(2k)\zeta(2a-2k+1), \end{equation*}$$

where $\zeta$ is the Riemann zeta function, $\eta$ is the Dirichlet eta funtion and $\beta$ is the Dirichlet beta funtion.

Answer 4

One idea is to start with the generating function

$$\sum _{n=1}^{\infty } (-1)^n H_{2 n} \,x^{2 n-1}=-\frac{\log \left(x^2+1\right)+2 x \tan ^{-1}(x)}{2x \left(x^2+1\right)}\tag{1}$$

and integrate twice (not forgetting to multiplying the result by $4$). Thus $$2I=\sum _{n=1}^{\infty } (-1)^n \frac{H_{2n}}{n^2} =-4 \int_0^1 \frac{1}{x} \int \frac{\log \left(x^2+1\right)+2 x \tan ^{-1}(x)}{x \left(2 x^2+2\right)} \, dx\, dx$$

which becomes

$$2I=-2 \int_0^1 \frac{1}{x}\left( \frac{1}{2}\text{Li}_2\left(-x^2\right)+\frac{1}{4} \log ^2\left(x^2+1\right)-\tan ^{-1}(x)^2\right)\, dx\tag{2}$$

alternatively we could of started with

$$\sum _{n=1}^{\infty } (-1)^n H_{2 n} x^{n-1}=-\frac{\log (x+1)+2 \sqrt{x} \tan ^{-1}\left(\sqrt{x}\right)}{x (2 x+2)}\tag{3}$$

to give

$$2I=-\int_0^1 \frac{1}{x} \left( \frac{\text{Li}_2(-x)}{2}+\frac{1}{4} \log ^2(x+1)-\tan ^{-1}\left(\sqrt{x}\right)^2 \right)\, dx\tag{4}$$

So you pick either set of three functions to integrate, or mix and match, since only a factor of 2 is between them.

One of the integrals can be easily integrated:

$$-\frac{1}{2} \int \frac{\text{Li}_2(-x)}{x} \, dx=-\frac{\text{Li}_3(-x)}{2}$$

with $$-\frac{1}{2} \int_0^1 \frac{\text{Li}_2(-x)}{x} \, dx=\frac{3 \zeta (3)}{8}\tag{5}$$

The next integral also evaluates to $\zeta(3)$ $$-\frac{1}{4} \int_0^1 \frac{\log ^2(x+1)}{x} \, dx=-\frac{\zeta (3)}{16}\tag{6}$$

So interestingly the problem really boils down to integrating: $$2 \int_0^1 \frac{\left(\tan ^{-1}(x)\right)^2}{x} \, dx \;\;\;\;\; \text{or} \;\;\;\;\; \int_0^1 \frac{\left(\tan ^{-1}\left(\sqrt{x}\right)\right)^2}{x} \, dx$$

With an example equivalent integral being

$$4 \int_0^{\frac{\pi }{4}} y^2 \csc (2 y) \, dy=\pi \, G-\frac{7 \,\zeta (3)}{4}\tag{7}$$

$2I$ is the sum of results $(5)$, $(6)$ and $(7)$

$$2I=\frac{3 \,\zeta (3)}{8}-\frac{\zeta (3)}{16}+ \pi \, G -\frac{7\, \zeta (3)}{4}=\pi \, G-\frac{23\, \zeta (3)}{16}$$