I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.
Could you help me with it?
Real Analysis – Infinite Series Sum of H_n/n^3 2^n
closed-formharmonic-numbersreal-analysissequences-and-serieszeta-functions
Related Solutions
EDITED. Some simplifications were made.
Here is a solution.
1. Basic facts on the dilogarithm. Let $\mathrm{Li}_{2}(z)$ be the dilogarithm function defined by
$$ \operatorname{Li}_{2}(z) = \sum_{n=1}^{\infty} \frac{z^{n}}{n^{2}} = - \int_{0}^{z} \frac{\log(1-x)}{x} \, dx. $$
Here the branch cut of $\log $ is chosen to be $(-\infty, 0]$ so that $\operatorname{Li}_{2}$ defines a holomorphic function on the region $\Bbb{C} \setminus [1, \infty)$. Also, it is easy to check (by differentiating both sides) that the following identities hold
\begin{align*} \operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right) &= -\mathrm{Li}_{2}(z) - \tfrac{1}{2}\log^{2}(1-z); \quad z \notin [1, \infty) \tag{1} \\ \operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right) &= \color{blue}{\boxed{\operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z)}} + \color{red}{\boxed{\log(-z)\log(1-z)}}; \quad z \notin [0, \infty) \tag{2} \end{align*}
Notice that in (2), the blue-colored part is holomorphic on $|z| < 1$ while the red-colored part induces the branch cut $[-1, 0]$.
2. A useful power series. Now let us consider the power series
$$ f(z) = \sum_{n=0}^{\infty} \frac{H_n}{n} z^n. $$
Then $f(z)$ is automatically holomorphic inside the disc $|z| < 1$. Moreover, it is easy to check that
$$ \sum_{n=1}^{\infty} H_{n} z^{n-1} = \frac{1}{z} \left( \sum_{n=1}^{\infty} \frac{z^{n}}{n} \right)\left( \sum_{n=0}^{\infty} z^{n}\right) = -\frac{\log(1-z)}{z(1-z)}. $$
thus integrating both sides, together with the identity $\text{(1)}$, we obtain the following representation of $f(z)$.
$$f(z) = \operatorname{Li}_{2}(z) + \tfrac{1}{2}\log^{2}(1-z) = -\operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right). \tag{3}$$
3. Integral representation and the result. By the Parseval's identity, we have
$$ \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} = \frac{1}{2\pi} \int_{0}^{2\pi} f(e^{it})f(e^{-it}) \, dt = \frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} f\left(\frac{1}{z}\right) \, dz \tag{4} $$
Since $\frac{1}{z}f(z)$ is holomorphic inside $|z| = 1$, the failure of holomorphy of the integrand stems from the branch cut of
\begin{align*} f\left(\tfrac{1}{z}\right) &= -\operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right) \\ &= -\color{blue}{\left( \operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z) \right)} - \color{red}{\log(-z)\log(1-z)}, \end{align*}
which is $[0, 1]$. To resolve this, we utilize the identity $\text{(2)}$. Note that the blue-colored portion does not contributes to the the integral $\text{(4)}$, since it remains holomorphic inside $|z| < 1$. That is, only the red-colored portion gives contribution to the integral. Consequently we have
\begin{align*} \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} &= -\frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} \color{red}{\log(-z)\log(1-z)} \, dz. \tag{5} \end{align*}
Since the integrand is holomorphic on $\Bbb{C} \setminus [0, \infty)$, we can utilize the keyhole contour wrapping around $[0, 1]$ to reduce $\text{(5)}$ to
\begin{align*} \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} &=-\frac{1}{2\pi i} \Bigg\{ \int_{0^{-}i}^{1+0^{-}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \\ &\qquad \qquad + \int_{1+0^{+}i}^{+0^{+}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \Bigg\} \\ &=-\frac{1}{2\pi i} \Bigg\{ \int_{0}^{1} \frac{f(x)(\log x + i\pi)\log(1-x)}{x} \, dx \\ &\qquad \qquad - \int_{0}^{1} \frac{f(x)(\log x - i\pi)\log(1-x)}{x} \, dx \Bigg\} \\ &=-\int_{0}^{1} \frac{f(x)\log(1-x)}{x} \, dx. \tag{5} \end{align*}
Plugging $\text{(3)}$ to the last integral and simplifying a little bit, we have
\begin{align*} \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} &= - \int_{0}^{1} \frac{\operatorname{Li}_2(x)\log(1-x)}{x} \, dx - \frac{1}{2}\int_{0}^{1} \frac{\log^{3}(1-x)}{x} \, dx \\ &= \left[ \frac{1}{2}\operatorname{Li}_2(x)^2 \right]_0^1 - \frac{1}{2} \int_{0}^{1} \frac{\log^3 x}{1-x} \, dx \\ &= \frac{1}{2}\zeta(2)^{2} + \frac{1}{2} \Gamma(4)\zeta(4) \\ &= \frac{17\pi^{4}}{360} \end{align*}
as desired.
So basically, I'll evaluate a bunch of integrals, trying to avoid polylogs as much as possible.
First thing is to notice that $\displaystyle H_n-2H_{2n}+H_{4n}=\int_0^1 \frac{x^{2n}-x^{4n}}{1+x}dx$. I noticed that $H_n-2H_{2n}+H_{4n}=H_{4n}-H_{2n}-(H_{2n}-H_n)=H_{4n^{-}}-H_{{2n}^{-}}$, where $H_{n^{-}}=\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}$ is called a skew harmonic number (at least by Khristo N. Boyadzhiev. link.) Knowing they have a simple intergal representation I found the above. My answer is influenced by Boyadzhiev's work. If I make any unexplainable substitution, it's most likely $t=\frac{1-x}{1+x}$. Also, I'm not very good with Latex, so alignment should be awful. Hopefully there are no typos.
Below, easy enough to prove, is what I take for granted: $ -\ln\sin x=\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n} ,-\ln\cos x=\ln2+\sum_{n=1}^{\infty} \frac{(-1)^n\cos(2nx)}{n} \tag{1}$
$$ \int_0^{\frac{\pi}{2}} \cos x \cos(nx)dx=\begin{cases} \frac{\pi}{4} &n=1\\0 &n \,\,\text{odd}\\ \frac{(-1)^{1+n/2}}{n^2-1} &n \,\,\text{even} \end{cases} \tag{2}$$
$$ \int_0^1 \frac{\ln(1-x)}{a+x}dx=-\operatorname{Li_2}\left(\frac1{a+1}\right)\tag{3}$$ Starting, $$\sum_{n=0}^{\infty}(H_{n}-2H_{2n}+H_{4n})^2=\sum_{n=0}^{\infty}\int_0^1\int_0^1\frac{(x^{2n}-x^{4n})(u^{2n}-u^{4n})}{(1+x)(1+u)}dxdu \\=\small\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^2u^2)}-2\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^2u^4)}+\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^4u^4)} \\=I_{22}-2I_{24}+I_{44}$$
Computing $I_{22}$.
Substitute $u=\frac{y}{x}$ ,change the order of integration, evaluate the inner integral, and substitue $t=\frac{1-x}{1+x}$ to get $$\begin{align} I_{22}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^2u^2)}=\int_0^1\int_0^x\frac{dydx}{(1+x)(x+y)(1-y^2)} \\=\int_0^1 \frac1{1-y^2}\int_1^y \frac{dx}{(1+x)(x+y)} dy=\int_0^1 \frac{\ln\left(\frac{(1+x)^2}{4x}\right)}{(1+x)(1-x^2)}\,dx \\=\frac{-1}{4}\int_0^1 \frac{(1+t)}{t^2}\ln(1-t^2)dt=-\frac14\int_0^1\frac{\ln(1-t^2)}{t^2}dt-\frac14\int_0^1\frac{\ln(1-t^2)}{t}dt \\=\frac14\sum_{n=0}^{\infty} \frac1{(n+1)(2n+1)}+\frac14\sum_{n=0}^{\infty} \frac1{(n+1)(2n+2)}=\frac{\ln2}{2}+\frac{\pi^2}{48}.\end{align}$$
Computing $I_{44}$.
Start the same as with $I_{22}$ to get $\displaystyle I_{44}=\int_0^1 \frac{\ln\left(\frac{(1+x)^2}{4x}\right)}{(1-x)(1-x^4)}\,dx=\frac{-1}{8}\int_0^1 \frac{\ln(1-t^2)}{t^2(1+t^2)}(1+t)^3dt$. We can calculate these integrals: $$\begin{align} \int_0^1 \frac{\ln(1-x^2)}{1+x^2}dx=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx+\int_0^1 \frac{\ln(1-x)}{1+x^2}dx \tag{4} \\=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx +\int_0^1 \frac{\ln\left(\frac{2t}{1+t}\right)}{1+t^2}dt \\=\frac{\pi}{4}\ln2+\sum_{n=0}^{\infty} (-1)^n\int_0^1\ln(t) t^{2n}dt=\frac{\pi}{4}\ln2-G. \end{align}$$ $$\begin{align} \int_0^1 \frac{\ln(1-x^2)}{x^2(1+x^2)}dx=\int_0^1 \frac{\ln(1-x^2)}{x^2}dx-\int_0^1 \frac{\ln(1-x^2)}{1+x^2}dx \tag{5} \\=-\sum_{n=0}^{\infty} \frac1{n+1}\int_0^1 x^{2n}dx-\frac{\pi}{4}\ln2+G=G-\frac{\pi}{4}\ln2-2\ln2.\end{align}$$ $$\begin{align} \int_0^1 \frac{x\ln(1-x^2)}{1+x^2}dx=\frac12\int_0^1 \frac{\ln(1-x)}{1+x}dx \tag{6} \\=-\frac12 \operatorname{Li_2}\left(\frac12\right)=\frac{\ln^2 2}{4}-\frac{\pi^2}{24}.\end{align}$$ $$\begin{align} \int_0^1 \frac{\ln(1-x^2)}{x(1+x^2)}dx=\int_0^1 \frac{\ln(1-x^2)}{x}dx-\int_0^1 \frac{x\ln(1-x^2)}{1+x^2}dx \tag{7} \\=-\sum_{n=0}^{\infty}\frac1{n+1}\int_0^1 x^{2n+1}dx-\frac{\ln^2 2}{4}+\frac{\pi^2}{24}=-\frac{\pi^2}{24}-\frac{\ln^2 2}{4}.\end{align}$$ Altogether, $$I_{44}=\frac{-1}{8}\int_0^1 \frac{\ln(1-x^2)}{x^2(1+x^2)}(1+3x+3x^2+x^3)dx \\=-\frac{\pi}{16}\ln2+\frac{\ln2}{4}+\frac{\ln^2 2}{16}+\frac{\pi^2}{48}+\frac{G}{4}.$$
Computing $I_{24}$.
Substitute $u=\frac{y}{x^2}$, change the order of integration, let $y\to y^2$, evaluate the inner integral,and substitue $t=\frac{1-x}{1+x}$: $$\begin{align*} I_{24}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^4u^2)}=\int_0^1\int_0^{x^2} \frac{dydx}{(1+x)(x^2+y)(1-y^2)} \\=\int_0^1 \frac1{1-y^2}\int_{\sqrt{y}}^1\frac{dx}{(1+x)(x^2+y)}dy=2\int_0^1\frac{y}{1-y^4}\int_{y}^1\frac{dx}{(1+x)(x^2+y^2)}dy \\=2\int_0^1\frac{\tan^{-1}\left(\frac{1-x}{1+x}\right)}{(1+x^2)(1-x^4)}dx-2\int_0^1\frac{x\ln\left(\frac{(1+x)\sqrt{1+x^2}}{2\sqrt{2}x}\right)}{(1+x^2)(1-x^4)}dx =I_{241}-I_{242}. \end{align*}$$
Evaulation of $I_{241}$.
Substitute $t=\frac{1-x}{1+x}$ to get $\displaystyle I_{241}=\frac14\int_0^1 \frac{\tan^{-1}(t)}{t(1+t^2)^2}(1+t)^4dt$.
We can calculate these integrals.In the following, let $x=\tan\theta$: $$\begin{align} \int_0^1 \frac{\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\cos^2(\theta)d\theta=\frac{\pi^2}{64}+\frac{\pi}{16}-\frac18.\tag{8}\end{align}$$ $$\begin{align} \int_0^1 \frac{x\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\tan\theta\cos^2{\theta}d\theta=\frac12\int_0^{\frac{\pi}{4}}\theta\sin(2\theta)d\theta=\frac18.\tag{9}\end{align}$$ $$\begin{align} \int_0^1 \frac{x^2\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\sin^2(\theta)d\theta=\frac{\pi^2}{64}-\frac{\pi}{16}+\frac18.\tag{10}\end{align}$$ $$\begin{align} \int_0^1 \frac{x^3\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\sin^3(\theta)\sec{\theta}\,d\theta \tag{11} \\=\int_0^{\frac{\pi}{4}}\theta\tan\theta \,d\theta-\int_0^{\frac{\pi}{4}}\theta\sin\theta\cos\theta \,d\theta=\frac{\pi}{8}\ln2-\frac18+\int_0^{\frac{\pi}{4}}\ln\cos\theta \,d\theta \\=\frac{\pi}{8}\ln2-\frac18-\int_0^{\frac{\pi}{4}}\ln2 \,d\theta-\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\int_0^{\frac{\pi}{4}}\cos(2n\theta)\,d\theta \\=-\frac{\pi}{8}\ln2-\frac18+\frac12\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\sin(\frac{\pi n}{2})=\frac{G}{2}-\frac{\pi}{8}\ln2-\frac18.\end{align}$$ $$\begin{align} \int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\cos^3(\theta)\csc{\theta}\,d\theta \tag{12} \\=\int_0^{\frac{\pi}{4}}\theta\cot\theta \,d\theta-\int_0^{\frac{\pi}{4}}\theta\sin\theta\cos\theta \,d\theta=-\frac18-\frac{\pi}{8}\ln2-\int_0^{\frac{\pi}{4}}\ln\sin\theta \,d\theta \\=-\frac18-\frac{\pi}{8}\ln2+\int_0^{\frac{\pi}{4}}\ln2 \,d\theta+\sum_{n=1}^{\infty}\frac1{n}\int_0^{\frac{\pi}{4}}\cos(2n\theta)\,d\theta \\=-\frac18+\frac{\pi}{8}\ln2+\frac12\sum_{n=1}^{\infty} \frac{\sin(\frac{\pi n}{2})}{n^2}=\frac{G}{2}+\frac{\pi}{8}\ln2-\frac18.\end{align}$$ Altogether, $$I_{241}=2\int_0^1\frac{\tan^{-1}\left(\frac{1-x}{1+x}\right)}{(1+x^2)(1-x^4)}dx= \frac14\int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)^2}(1+4x+6x^2+4x^3+x^4)dx \\=\frac{\pi^2}{32}+\frac18+\frac{G}{4}$$
Evaulation of $I_{242}$.
Substitute $t=\frac{1-x}{1+x}$ to get $$ I_{242}=\frac14\int_0^1 \frac{\ln\left(\frac{\sqrt{1+t^2}}{1-t^2}\right)}{t(1+t^2)^2}(1-t^2)(1+t)^2dt \\=\frac12\int_0^1 \frac{\ln\left(\frac{\sqrt{1+t^2}}{1-t^2}\right)}{(1+t^2)^2}(1-t^2)dt+\frac14\int_0^1 \frac{\ln\left(\frac{\sqrt{1+t^2}}{1-t^2}\right)}{t(1+t^2)}(1-t^2)dt \\=\frac12\int_0^1 \frac{\ln\left(\frac{1+x^2}{1-x^2}\right)}{(1+x^2)^2}(1-x^2)dx-\frac14\int_0^1\frac{\ln(1+x^2)}{(1+x^2)^2}(1-x^2)dx\\+\frac18\int_0^1\frac{\ln(1+x^2)(1-x^2)}{x(1+x^2)}dx-\frac14\int_0^1\frac{\ln(1-x^2)(1-x^2)}{x(1+x^2)}dx $$ Calculating these integrals: $$\begin{align} \int_0^1 \frac{\ln\left(\frac{1+x^2}{1-x^2}\right)}{(1+x^2)^2}(1-x^2)dx=-\frac12\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)}{\sqrt{x}(1+x)}\frac{1-x}{1+x}dx \tag{13} \\=-\frac12\int_0^1\frac{t\ln t}{\sqrt{1-t^2}}dt=-\frac18\int_0^1\frac{\ln t}{\sqrt{1-t}}dt=-\frac18\int_0^1 t^{-1/2}\ln(1-t)dt \\=\frac14\sum_{n=0}^{\infty} \frac1{(n+1)(2n+3)}=\frac12-\frac{\ln2}{2}.\end{align}$$ $$\begin{align} \int_0^1\frac{\ln(1+x^2)(1-x^2)}{x(1+x^2)}dx=\frac12\int_0^1\frac{\ln(1+x)(1-x)}{x(1+x)}dx \tag{14} \\=\frac12\int_0^1\frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx \\=\frac12\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1}\int_0^1 x^n dx-\frac12\ln^2(1+x)\bigg{|}_0^1=\frac{\pi^2}{24}-\frac{\ln^2 2}{2}.\end{align}$$ $$\begin{align} \int_0^1\frac{\ln(1-x^2)(1-x^2)}{x(1+x^2)}dx=\frac12\int_0^1\frac{\ln(1-x)}{x}dx-\int_0^1\frac{\ln(1-x)}{1+x}dx \tag{15} \\=-\frac{\pi^2}{12}-\left(\frac{\ln^2 2}{2}-\frac{\pi^2}{12}\right)=-\frac{\ln^2 2}{2}.\end{align}$$ $$ \int_0^1\frac{\ln(1+x^2)}{(1+x^2)^2}(1-x^2)dx=-2\int_0^{\frac{\pi}{4}}\cos^2(\theta)(1-\tan^2\theta)\ln\cos\theta\,d\theta \tag{16} \\=-2\int_0^{\frac{\pi}{4}}\cos(2\theta)\ln\cos\theta\,d\theta=2\ln2\int_0^{\frac{\pi}{4}}\cos(2\theta)d\theta+\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\int_0^{\frac{\pi}{2}}\cos\theta \cos(n\theta)d\theta \\=\ln2-\frac{\pi}{4}+\sum_{n=1}^{\infty} \frac{(-1)^{2n}}{2n}\frac{(-1)^{n+1}}{(2n)^2-1}=\frac{\ln2}{2}-\frac{\pi}{4}+\frac12.$$ Altogether, $\displaystyle I_{242}=\frac{\pi^2}{192}+\frac{\pi}{16}+\frac{\ln^2 2}{16}-\frac{3\ln2}{8}+\frac18$,
leading to $\displaystyle I_{24}=I_{241}-I_{242}=\frac{5\pi^2}{192}-\frac{\pi}{16}-\frac{\ln^2 2}{16}+\frac{3\ln2}{8}+\frac{G}{4}$, and finally, confirming the conjecture, $$\sum_{n=0}^{\infty}(H_{n}-2H_{2n}+H_{4n})^2=I_{22}-2I_{24}+I_{44}=\frac{\pi}{8}-\frac{\pi}{16}\ln2-\frac{\pi^2}{96}+\frac{3\ln^2 2}{16}-\frac{G}{4}.$$
I don't know about higher powers. I guess the case $\mathcal A_2$ can also be done. If we start the same as with $\mathcal B_2$, writing $\mathcal A_2=J_{22}-2J_{24}+J_{44}$ we can find that $\displaystyle J_{22}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1+x^2u^2)}=2I_{44}-I_{22}=-\frac{\pi}{8}\ln2+\frac{G}{2}+\frac{\pi^2}{48}+\frac{\ln^2 2}{8}$
$J_{44}$ can be reduced to $\displaystyle =-\frac12\int_0^1 \frac{\ln(1-x^2)}{x(x^4+6x^2+1)}(1+x)^3 dx$. already this i can't evaulate fully, as polylogs are inescapable. Factorizing $x^2+6x+1=(x+3+2\sqrt{2})(x+3-2\sqrt{2}).$
I can get $\displaystyle \int_0^1 \frac{\ln(1-x^2)}{x(x^4+6x^2+1)}dx=-\frac{\pi^2}{12}+\frac{4-3\sqrt{2}}{16}\operatorname{Li_2}\left(\frac{2-\sqrt{2}}{4}\right)+\frac{4+3\sqrt{2}}{16}\operatorname{Li_2}\left(\frac{2+\sqrt{2}}{4}\right)$
but nothing more.
Edit 1.
After some more work and a fair amount of cancellation, we obtain $$\int_0^1 \frac{\ln(1-x^2)}{x(x^4+6x^2+1)}(1+x)^3 dx=\frac{1+2\sqrt{2}}{4}\pi\ln2-\frac{\pi^2}{24}-\frac14\ln\left(\frac{2+\sqrt{2}}{4}\right)\ln\left(\frac{2-\sqrt{2}}{4}\right) -\frac{\sqrt{2}+1}{2}\Im\operatorname{Li_2}\left(\frac{2+\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\right)-\frac{\sqrt{2}-1}{2}\Im\operatorname{Li_2}\left(\frac{2-\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\right)$$
I obtained it by calculating $\displaystyle \int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2\ln\left(\frac{a+1}{a-1}\right)+\operatorname{Li_2}\left(\frac2{1-a}\right)-\operatorname{Li_2}\left(\frac1{1-a}\right)$, which together with $(3)$ can be used to give a closed form for $\displaystyle \int_0^1\frac{\ln(1-x^2)}{x+a}dx$, which in turn, through partial fractions, can be used to give a closed form for $\displaystyle \int_0^1\frac{\ln(1-x^2)}{x^2+a^2}dx$. Fortunately, things didn't get too ugly as both $3+2\sqrt{2}$ and $3-2\sqrt{2}$ have nice square roots. I will fill in details as soon as I can.
Now we just need to evaluate $J_{24}$. Starting similarly as with $I_{24}$, we have: $$J_{24}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1+x^4u^2)} \\=2\int_0^1\frac{\tan^{-1}\left(\frac{1-x}{1+x}\right)}{(1+x^2)(1+x^4)}dx-2\int_0^1\frac{x\ln\left(\frac{(1+x)\sqrt{1+x^2}}{2\sqrt{2}x}\right)}{(1+x^2)(1+x^4)}dx \\=J_{241}-J_{242}$$
Through $t=\frac{1-x}{1+x}$, $J_{241}$ turns to $\displaystyle \int_0^1 \frac{\tan^{-1}(x)}{(1+x^2)(x^4+6x^2+1)}(1+x)^4\,dx$. I don't have any idea about that yet. \Edit 1.
Best Answer
In the same spirit as Robert Israel's answer and continuing Raymond Manzoni's answer (both of them deserve the credit because of inspiring my answer) we have $$ \sum_{n=1}^\infty \frac{H_nx^n}{n^2}=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x). $$ Dividing equation above by $x$ and then integrating yields \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\zeta(3)\ln x+\frac12\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}+\color{blue}{\int\frac{\ln(1-x)\operatorname{Li}_2(1-x)}x\ dx}\\&+\operatorname{Li}_4(x)-\color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}.\tag1 \end{align} Using IBP to evaluate the green integral by setting $u=\operatorname{Li}_3(1-x)$ and $dv=\frac1x\ dx$, we obtain \begin{align} \color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}&=\operatorname{Li}_3(1-x)\ln x+\int\frac{\ln x\operatorname{Li}_2(1-x)}{1-x}\ dx\qquad x\mapsto1-x\\ &=\operatorname{Li}_3(1-x)\ln x-\color{blue}{\int\frac{\ln (1-x)\operatorname{Li}_2(x)}{x}\ dx}.\tag2 \end{align} Using Euler's reflection formula for dilogarithm $$ \operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}6-\ln x\ln(1-x), $$ then combining the blue integral in $(1)$ and $(2)$ yields $$ \frac{\pi^2}6\int\frac{\ln (1-x)}{x}\ dx-\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}=-\frac{\pi^2}6\operatorname{Li}_2(x)-\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}. $$ Setting $x\mapsto1-x$ and using the identity $H_{n+1}-H_n=\frac1{n+1}$, the red integral becomes \begin{align} \color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}&=-\int\frac{\ln (1-x)\ln^2 x}{1-x}\ dx\\ &=\int\sum_{n=1}^\infty H_n x^n\ln^2x\ dx\\ &=\sum_{n=1}^\infty H_n \int x^n\ln^2x\ dx\\ &=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\int x^n\ dx\right]\\ &=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\frac {x^{n+1}}{n+1}\right]\\ &=\sum_{n=1}^\infty H_n \left[\frac{x^{n+1}\ln^2x}{n+1}-2\frac{x^{n+1}\ln x}{(n+1)^2}+2\frac{x^{n+1}}{(n+1)^3}\right]\\ &=\ln^2x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{n+1}-2\ln x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^2}+2\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^3}\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^2}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^3}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^4}\right]\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\sum_{n=1}^\infty\frac{x^{n}}{n^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\sum_{n=1}^\infty\frac{x^{n}}{n^4}\right]\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\operatorname{Li}_4(x)\right]. \end{align} Putting all together, we have \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+C.\tag3 \end{align} Setting $x=1$ to obtain the constant of integration, \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^3}&=\operatorname{Li}_4(1)-\frac{\pi^2}{12}\operatorname{Li}_2(1)+C\\ \frac{\pi^4}{72}&=\frac{\pi^4}{90}-\frac{\pi^4}{72}+C\\ C&=\frac{\pi^4}{60}. \end{align} Thus \begin{align} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}.\tag4 \end{align} Finally, setting $x=\frac12$, we obtain \begin{align} \sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\color{purple}{\frac{\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2}8\zeta(3)+\operatorname{Li}_4\left(\frac12\right)}, \end{align} which matches Cleo's answer.
References :
$[1]\ $ Harmonic number
$[2]\ $ Polylogarithm