Conjectured value of a harmonic sum ∑∞n=1(Hn−2H2n+H4n)2\sum_{n=1}^\infty\left(H_n-\,2H_{2n}+H_{4n}\right)^2

There is a known asymptotic expansion of harmonic numbers Hn for n:
Hn=γ+lnn+k=1(Bkknk)=γ+lnn+12n112n2+1120n41252n6+,
where Bk are Bernoulli numbers.
We can take a linear combination of harmonic numbers to cancel constant and logarithmic terms, compensate for O(n1) term, and get the following series that is possible to evaluate in a closed form (e.g. using generating function):
k=1(Hn2H2n+H4n18n)=18π16.
Rather than compensating for O(n1) term, we can take a series with alternating signs, that is also possible to evaluate in a closed form:
n=1(1)n(Hn2H2n+H4n)=3π16π42ln28.
Generalizing, we can consider two families of series:
Am=n=1(1)n(Hn2H2n+H4n)m,
Bm=n=1(Hn2H2n+H4n)m,
and try to evaluate them in a closed form.


So far I have the following conjectured result:

n=1(Hn2H2n+H4n)2?=π8π16ln2π296+316ln22G4,

where G is the Catalan constant.

Could you please help me to prove this result and, possibly, find other values of Am,Bm?

Answer

So basically, I’ll evaluate a bunch of integrals, trying to avoid polylogs as much as possible.

First thing is to notice that Hn2H2n+H4n=10x2nx4n1+xdx.
I noticed that Hn2H2n+H4n=H4nH2n(H2nHn)=H4nH2n, where Hn=nk=1(1)k+1k 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=1x1+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:
lnsinx=ln2+n=1cos(2nx)n,lncosx=ln2+n=1(1)ncos(2nx)n

π20cosxcos(nx)dx={π4n=10nodd(1)1+n/2n21neven

10ln(1x)a+xdx=Li2(1a+1)
Starting,
n=0(Hn2H2n+H4n)2=n=01010(x2nx4n)(u2nu4n)(1+x)(1+u)dxdu=1010dxdu(1+x)(1+u)(1x2u2)21010dxdu(1+x)(1+u)(1x2u4)+1010dxdu(1+x)(1+u)(1x4u4)=I222I24+I44


Computing I22.

Substitute u=yx ,change the order of integration, evaluate the inner integral, and substitue t=1x1+x to get
I22=1010dxdu(1+x)(1+u)(1x2u2)=10x0dydx(1+x)(x+y)(1y2)=1011y2y1dx(1+x)(x+y)dy=10ln((1+x)24x)(1+x)(1x2)dx=1410(1+t)t2ln(1t2)dt=1410ln(1t2)t2dt1410ln(1t2)tdt=14n=01(n+1)(2n+1)+14n=01(n+1)(2n+2)=ln22+π248.


Computing I44.

Start the same as with I22 to get I44=10ln((1+x)24x)(1x)(1x4)dx=1810ln(1t2)t2(1+t2)(1+t)3dt.
We can calculate these integrals:
10ln(1x2)1+x2dx=10ln(1+x)1+x2dx+10ln(1x)1+x2dx=10ln(1+x)1+x2dx+10ln(2t1+t)1+t2dt=π4ln2+n=0(1)n10ln(t)t2ndt=π4ln2G.
10ln(1x2)x2(1+x2)dx=10ln(1x2)x2dx10ln(1x2)1+x2dx=n=01n+110x2ndxπ4ln2+G=Gπ4ln22ln2.
10xln(1x2)1+x2dx=1210ln(1x)1+xdx=12Li2(12)=ln224π224.
10ln(1x2)x(1+x2)dx=10ln(1x2)xdx10xln(1x2)1+x2dx=n=01n+110x2n+1dxln224+π224=π224ln224.
Altogether,
I44=1810ln(1x2)x2(1+x2)(1+3x+3x2+x3)dx=π16ln2+ln24+ln2216+π248+G4.


Computing I24.

Substitute u=yx2, change the order of integration, let yy2, evaluate the inner integral,and substitue t=1x1+x:
I24=1010dxdu(1+x)(1+u)(1x4u2)=10x20dydx(1+x)(x2+y)(1y2)=1011y21ydx(1+x)(x2+y)dy=210y1y41ydx(1+x)(x2+y2)dy=210tan1(1x1+x)(1+x2)(1x4)dx210xln((1+x)1+x222x)(1+x2)(1x4)dx=I241I242.


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.

Attribution
Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : Noam Shalev – nospoon

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