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.
On
\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.
On
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
On
$$\int_{0}^{\infty }\frac{ln(1+x^2)ln^2x}{1+x^2}dx\\ \\ let\ I(a)=\int_{0}^{\infty }\frac{ln^2(x)ln(1+a^2.x^2)}{1+x^2}\\ \\ \therefore I'(a)=\int_{0}^{\infty }\frac{2ax^2ln^2(x)}{(1+a^2x^2)(1+x^2)}dx=\frac{2a}{1-a^2}\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2}{1+x^2}dx\\ \\ let\ \ G=\int_{0}^{\infty }\frac{ln^2(x))}{1+a^2x^2}dx\ \ \ \ ,\ \ but \ we \ know\ \\ \\ G(a)=\int_{0}^{\infty }\frac{x^p}{(1+x^2)a^{p+1}}dx=\frac{\pi }{2}\frac{sec\frac{\pi p}{2}}{a^{p+1}}\\ \\ \therefore \frac{\partial^2 G(a)}{\partial^2 p}=\frac{\pi }{2}[tan(\frac{\pi p}{2})sec(\frac{\pi p}{2})+ln(a)sec(\frac{\pi p}{2})ln(a)a^{-p-1}]+\frac{1}{a^{p+1}}[\frac{\pi ^{2}}{4}tan^2(\frac{\pi p}{2})sec(\frac{\pi p}{2})+sec^3(\frac{\pi p}{2})-\frac{\pi }{2}tan(\frac{\pi p}{2})sec(\frac{\pi p}{2})ln(a)]\\$$
$$now\ \ take\ \ p=0\ \\ \\ \therefore \frac{\partial ^2 G(a)}{\partial p^2}_{p=0}=\frac{\pi }{2}[\frac{ln^2(a)}{a}+\frac{\pi ^{2}}{4a}]=\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}dx\ ,\ \ \ \ take\ a=1\\ \\ \therefore \int_{0}^{1 }\frac{ln^2(x)dx}{1+x^2}=\frac{\pi ^{3}}{8}, \ \ \ \ \ now\ going\ to \ I\\ \\ \therefore I(a)=\int_{0}^{1}(\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2(x)}{1+x^2})dx=\frac{\pi ^{3}}{8}(\frac{1-a}{a})+\frac{\pi ln^2(a)}{2a}\\ \\ \\ \therefore I'(a)=\frac{2a}{1-a^2}(\frac{\pi ^{3}}{8}(\frac{1-a}{a})+\frac{\pi ln^2(a)}{2a})$$
$$\therefore I(1)=\frac{\pi ^{3}}{4}\int_{0}^{1}\frac{dx}{1+x}+\pi \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx\\ \\ \therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{1}{2}\int_{0}^{1}\frac{ln^2(x)}{1-x}dx+\frac{1}{2}\int_{0}^{1}\frac{ln^2(x)}{1+x}dx\\ \\ \therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{1}{2}[-Ln^2(x)ln(1-x)\tfrac{1}{0}+ln^2(x)ln(1+x)\tfrac{1}{0}+2\int_{0}^{1}\frac{ln(1-x)lnx}{x}dx-2\int_{0}^{1}\frac{ln(1+x)lnx}{x}dx]\\ \\ \\ \therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{7}{4}\zeta (3)\\ \\ \therefore I=\int_{0}^{\infty }\frac{ln^2(x)ln(1+x^2)}{1+x^2}dx=\frac{\pi ^{3}}{4}ln(2)+\frac{7\pi }{4}\zeta (3)$$
On
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
$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^3}=3\beta(4)-\frac{7\pi}{16}\zeta(3)-\frac{\pi^3}{16}\ln(2)$$
$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^5}=5\beta(6)-\frac{31\pi}{64}\zeta(5)-\frac{7\pi^3}{128}\zeta(3)-\frac{5\pi^5}{768}\ln(2)$$
$$\sum_{n=1}^\infty\frac{(-1)^n H_n}{(2n + 1)^7}=7\beta(8)-\frac{127\pi}{256}\zeta(7)-\frac{31\pi^3}{512}\zeta(5)-\frac{35\pi^5}{6144}\zeta(3)-\frac{61\pi^7}{92160}\ln(2)$$
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.