# Evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}$

How to prove that

$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$

where $$H_n=1+\frac1{2}+\frac1{3}+...+\frac1{n}$$ is the $$n$$th harmonic number.

This sum was proposed by Cornel and I solved it using integration but can we solve it using series manipulation?

The integral representation of the sum is $$\ \displaystyle\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx$$ in case it is needed.

• I'm surprised nobody posted a solution using contour integration. Let $f(z)=\frac{\pi\csc(\pi z) \left(\gamma+\psi_0(-z) \right)}{(2z+1)^3}$. One can show that $\lim_{N\to \infty }\int_{C_N}f(z)dz = 0$ where $C_N$ is a positively oriented square with vertices $\pm \left(N+\frac{1}{2} \right)\pm i \left(N+\frac{1}{2} \right)$. So, the sum of residues of $f(z)$ is equal to 0. I can write rest of the details if you're still interested. Aug 8, 2020 at 11:02
• Sure @Shobhit Bhatnagar different ideas are always welcomed. Aug 9, 2020 at 10:17

What follows in an evaluation of the Euler sum that uses its equivalent integral representation.

We begin by noting that

$$\int_0^1 x^{2n} \ln^2 x \, dx = \frac{d^2}{ds^2} \left [\int_0^1 x^{2n + s} \, dx \right ]_{s = 0} = \frac{2}{(2n + 1)^3}.$$

Thus $$\sum_{n = 1}^\infty \frac{(-1)^{n} H_n}{(2n + 1)^3} = \frac{1}{2} \int_0^1 \ln^2 x \sum_{n = 1}^\infty (-1)^n H_n x^{2n} \, dx\tag1$$ From the Generating function for the Harmonic numbers, namely $$\sum_{n = 1}^\infty H_n x^n = -\frac{\ln (1 - x)}{1 - x},$$ enforcing a substitution of $$x \mapsto -x^2$$ one has $$\sum_{n = 1}^\infty (-1)^n H_n x^{2n} = -\frac{\ln (1 + x^2)}{1 + x^2},$$ allowing us to rewrite (1) as $$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{(2n + 1)^3} = -\frac{1}{2} \int_0^1 \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx.$$

Evaluating the integral we have \begin{align} I &= \int_0^1 \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx\\ &= \int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx - \underbrace{\int_1^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx}_{\displaystyle x \mapsto 1/x}\\ &= \int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx - \int_0^1 \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx + 2 \int_0^1 \frac{\ln^3 x}{1 + x^2} \, dx \tag2\\ &= \int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx - 12 \beta (4) - I\\ 2 I &= \int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx - 12 \beta (4). \end{align} Note the right most integral in (2) was found as follows: $$\int_0^1 \frac{\ln^3 x}{1 + x^2} \, dx = \sum_{n = 0}^\infty (-1)^n \frac{d^3}{ds^3} \left [\int_0^1 x^{2n + s} \, dx \right ]_{s = 0} = -6 \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)^4} = -6 \beta (4).$$ Here $$\beta (x)$$ is the Dirichlet beta function and has a known value in terms of the polygamma function of order 3 of: $$\beta (4) = \frac{1}{768} \left [\psi^{(3)} \left (\frac{1}{4} \right ) - 8 \pi^4 \right ].$$ Thus $$I = \frac{1}{2}\int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx - 6 \beta (4).$$

To find the last outstanding integral, set $$x = \tan \theta$$, then \begin{align} I_1 &= \int_0^\infty \frac{\ln^2 x \ln (1 + x^2)}{1 + x^2} \, dx\\ &= -2 \int_0^{\frac{\pi}{2}} \ln^2 (\tan \theta) \ln (\cos \theta) \, d\theta\\ &= - 2 \int_0^{\frac{\pi}{2}} \Big{(} \ln (\sin \theta) - \ln (\cos \theta) \Big{)}^2 \ln (\cos \theta) \, d\theta\\ &= -2 \int_0^{\frac{\pi}{2}} \ln^2 (\sin \theta) \ln (\cos \theta) \, d\theta + 4 \int_0^{\frac{\pi}{2}} \ln (\sin \theta) \ln^2 (\cos \theta) \, d\theta - 2 \int_0^{\frac{\pi}{2}} \ln^3 (\cos \theta) \, d\theta\\ &= I_\alpha + I_\beta + I_\gamma. \end{align}

Each of the above three integrals can be found by taking third derivatives of the Beta function.

For $$I_\alpha$$ \begin{align} I_\alpha &= -2 \int_0^{\frac{\pi}{2}} \ln^2 (\sin \theta) \ln (\cos \theta) \, d\theta\\ &= -\frac{1}{8} \lim_{x,y \to 1/2} \frac{\partial^3}{\partial x^2 \partial y} \operatorname{B} (x,y)\\ &= -\frac{1}{8} \left (2 \pi \zeta (3) - 8\pi \ln^3 2 \right )\\ &= -\frac{\pi}{4} \zeta (3) + \pi \ln^3 2 \end{align}

For $$I_\beta$$ \begin{align} I_\beta &= 4 \int_0^{\frac{\pi}{2}} \ln (\sin \theta) \ln^2 (\cos \theta) \, d\theta\\ &= \frac{1}{4} \lim_{x,y \to 1/2} \frac{\partial^3}{\partial x \partial y^2} \operatorname{B} (x,y)\\ &= \frac{1}{4} \left (2 \pi \zeta (3) - 8\pi \ln^3 2 \right )\\ &= \frac{\pi}{2} \zeta (3) - 2\pi \ln^3 2 \end{align}

For $$I_\gamma$$ \begin{align} I_\gamma &= -2 \int_0^{\frac{\pi}{2}} \ln^3 (\cos \theta) \, d\theta\\ &= -\frac{1}{8} \lim_{y \to 1/2} \frac{\partial^3}{\partial y^3} \operatorname{B} \left (\frac{1}{2},y \right )\\ &= -\frac{1}{8} \left (-12 \pi \zeta (3) - 8\pi \ln^3 2 -2 \pi^3 \ln 2\right )\\ &= -\frac{3\pi}{2} \zeta (3) + \pi \ln^3 2 + \frac{\pi^3}{4} \ln 2. \end{align} Thus $$I_1 = \frac{7}{4} \pi \zeta (3) + \frac{\pi^3}{4} \ln 2,$$ giving $$I = \frac{7}{8} \pi \zeta (3) + \frac{\pi^3}{8} \ln 2 - 6 \beta (4),$$ which finally leads to the following value for the Euler sum of $$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{(2n + 1)^3} = 3 \beta (4) - \frac{7}{16} \pi \zeta (3) - \frac{\pi^3}{16} \ln 2,$$ or $$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{(2n + 1)^3} = \frac{1}{256} \psi^{(3)} \left (\frac{1}{4} \right ) - \frac{\pi^4}{32} - \frac{7 \pi}{16} \zeta (3) - \frac{\pi^3}{16} \ln 2,$$ as claimed.

• nice work. it is funny how your solution is identical to mine facebook.com/cornelioan.valean/posts/1757941204302670 Jun 21, 2019 at 8:07
• @Ali Shather - Really. Believe me, I don't have facebook and have been struggling with the integral $I$ for almost half the day! Jun 21, 2019 at 8:11
• @ omegadot I believe you but i just wanted to point that out. such stuff happens in mathematics. Jun 21, 2019 at 8:15
• @Ali Shather - Have just taken a look at the link and I see Cornel mentions in the question there is no need for an advance generating function! So your question regarding whether the sum can be solved using series manipulation remains open. Jun 21, 2019 at 8:21
• Hi omegadot, I provided a nice generalization above. I hope you like it. Also I am going to delete my comments above. I wonder how come i have not deleted them yet. Mar 5 at 2:40

\begin{align} \sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}&=\sum_{n=1}^\infty(-1)^{n-1} H_n\int_0^1\frac12x^{2n}\ln^2 x\ dx\\ &=-\frac12\int_0^1\ln^2x\sum_{n=1}^\infty(-x^2)H_n\\ &=\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx\tag{1} \end{align}

\begin{align} I&=\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx\\ &=\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+2\int_0^1\frac{\ln^3x}{1+x^2}\ dx\\ 2I=&\frac18\int_0^\infty\frac{\ln^2x\ln(1+x)}{\sqrt{x}(1+x)}\ dx+2(-6\beta(4))\\ I&=\frac1{16}\lim_{a\ \mapsto1/2\\b\ \mapsto1/2}\frac{-\partial^3}{\partial a^2\partial b}\text{B}(a,b)-6\beta(4)\\ &=\frac1{16}(14\pi\zeta(3)+2\pi^3\ln2)-6*\frac1{768}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4\right)\\ &=\frac{7\pi}{8}\zeta(3)+\frac{\pi^3}{8}\ln2-\frac1{128}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4\right)\tag{2} \end{align}

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

$$\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$

Notes:

$$\displaystyle\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\$$ is the Dirichlet beta function and the value of $$\beta(4)$$ can be found here.

$$\displaystyle\text{B}(a,b)=\int_0^\infty\frac{x^{a-1}}{(1+x)^{a+b}}\ dx$$ is beta function.

$$\int_{0}^{\infty }\frac{ln(1+x^2)ln^2x}{1+x^2}dx\\ \\ let\ I(a)=\int_{0}^{\infty }\frac{ln^2(x)ln(1+a^2.x^2)}{1+x^2}\\ \\ \therefore I'(a)=\int_{0}^{\infty }\frac{2ax^2ln^2(x)}{(1+a^2x^2)(1+x^2)}dx=\frac{2a}{1-a^2}\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2}{1+x^2}dx\\ \\ let\ \ G=\int_{0}^{\infty }\frac{ln^2(x))}{1+a^2x^2}dx\ \ \ \ ,\ \ but \ we \ know\ \\ \\ G(a)=\int_{0}^{\infty }\frac{x^p}{(1+x^2)a^{p+1}}dx=\frac{\pi }{2}\frac{sec\frac{\pi p}{2}}{a^{p+1}}\\ \\ \therefore \frac{\partial^2 G(a)}{\partial^2 p}=\frac{\pi }{2}[tan(\frac{\pi p}{2})sec(\frac{\pi p}{2})+ln(a)sec(\frac{\pi p}{2})ln(a)a^{-p-1}]+\frac{1}{a^{p+1}}[\frac{\pi ^{2}}{4}tan^2(\frac{\pi p}{2})sec(\frac{\pi p}{2})+sec^3(\frac{\pi p}{2})-\frac{\pi }{2}tan(\frac{\pi p}{2})sec(\frac{\pi p}{2})ln(a)]\\$$

$$now\ \ take\ \ p=0\ \\ \\ \therefore \frac{\partial ^2 G(a)}{\partial p^2}_{p=0}=\frac{\pi }{2}[\frac{ln^2(a)}{a}+\frac{\pi ^{2}}{4a}]=\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}dx\ ,\ \ \ \ take\ a=1\\ \\ \therefore \int_{0}^{1 }\frac{ln^2(x)dx}{1+x^2}=\frac{\pi ^{3}}{8}, \ \ \ \ \ now\ going\ to \ I\\ \\ \therefore I(a)=\int_{0}^{1}(\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2(x)}{1+x^2})dx=\frac{\pi ^{3}}{8}(\frac{1-a}{a})+\frac{\pi ln^2(a)}{2a}\\ \\ \\ \therefore I'(a)=\frac{2a}{1-a^2}(\frac{\pi ^{3}}{8}(\frac{1-a}{a})+\frac{\pi ln^2(a)}{2a})$$

$$\therefore I(1)=\frac{\pi ^{3}}{4}\int_{0}^{1}\frac{dx}{1+x}+\pi \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx\\ \\ \therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{1}{2}\int_{0}^{1}\frac{ln^2(x)}{1-x}dx+\frac{1}{2}\int_{0}^{1}\frac{ln^2(x)}{1+x}dx\\ \\ \therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{1}{2}[-Ln^2(x)ln(1-x)\tfrac{1}{0}+ln^2(x)ln(1+x)\tfrac{1}{0}+2\int_{0}^{1}\frac{ln(1-x)lnx}{x}dx-2\int_{0}^{1}\frac{ln(1+x)lnx}{x}dx]\\ \\ \\ \therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{7}{4}\zeta (3)\\ \\ \therefore I=\int_{0}^{\infty }\frac{ln^2(x)ln(1+x^2)}{1+x^2}dx=\frac{\pi ^{3}}{4}ln(2)+\frac{7\pi }{4}\zeta (3)$$

A nice generalization proved here:

$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{(2n+1)^{2a+1}}=(2a+1)\beta(2a+2)-\frac{\ln(2)|E_{2a}|}{(2a)!}\left(\frac{\pi}{2}\right)^{2a+1}$$

$$-\frac12\left(\frac{\pi}{2}\right)^{2a+1}\sum_{k=1}^{a} \frac{|E_{2a-2k}|}{(2a-2k)!}{\pi^{-2k}}(2^{2k+1}-1)\zeta(2k+1),$$

where $$E_r$$ is the Euler numbers.

Set $$a=1,2,3$$ we get

$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^3}=3\beta(4)-\frac{7\pi}{16}\zeta(3)-\frac{\pi^3}{16}\ln(2)$$

$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^5}=5\beta(6)-\frac{31\pi}{64}\zeta(5)-\frac{7\pi^3}{128}\zeta(3)-\frac{5\pi^5}{768}\ln(2)$$

$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^7}=7\beta(8)-\frac{127\pi}{256}\zeta(7)-\frac{31\pi^3}{512}\zeta(5)-\frac{35\pi^5}{6144}\zeta(3)-\frac{61\pi^7}{92160}\ln(2)$$

very nice solution Ali , this integral $$I=\int_{0}^{\infty }\frac{ln(1+x^2)ln^2x}{1+x^2}dx$$ i have another approach to evaluate i will post it

• Thank you Ahmed but this should be a comment not a solution. You can delete it after you read my comment. Looking forward to seeing your solution. It's better to put \ in front of the ln to look nicer. Aug 17, 2019 at 14:27