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Happy New Year 2020, Romania

In this recent post, Evaluate $\int_0^1 \frac{\arctan x\ln^2 x}{1+x^2}\,dx$, the proposed integral reduces to the calculation of $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^3}$ which is known in the literature. Now, what can we say about the more advanced version of it, the one with $n^4$ in the denominator?

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

Can we do it by series manipulations? It looks like a new series in the literature.

A good note: With the previous result in hand which we combine with some of the Cornel's work, we immediately arrive at two delightful series results,

$$i) \ \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_{2 n}^{(2)}}{n^3}=\frac{61 }{192}\pi ^2 \zeta (3)+\frac{1973 }{128}\zeta (5)+\frac{\pi ^5}{16}-\frac{1}{128} \pi \psi ^{(3)}\left(\frac{1}{4}\right);$$ $$ii) \ \sum _{n=1}^{\infty } (-1)^{n-1} \frac{ H_{2 n}^{(3)}}{n^2}=\frac{\pi ^3 G}{8}+\frac{1}{64}\pi ^2 \zeta (3)-\frac{2997 }{256}\zeta (5)-\frac{\pi ^5}{32}+\frac{1}{256} \pi \psi ^{(3)}\left(\frac{1}{4}\right).$$

How would you go proving these last two results?

Another good note: Remaining on the ground with alternating harmonic series of weight $5$ and harmonic numbers of the type $H_{2n}$, we might be curious to know what the value of the series $\displaystyle \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_{2 n}^{(4)}}{n}$ is.

$$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(4)}}{n}$$ $$=4\zeta(5)-\frac{3}{128}\zeta(2)\zeta(3)-\frac{7}{128}\log(2)\zeta(4)+\frac{\pi^5}{192}-\frac{\pi^3}{16}G-\frac{\pi}{1536}\psi^{(3)}\left(\frac{1}{4}\right).$$

This last series is elegantly calculated in A simple strategy of calculating two alternating harmonic series generalizations where one may also find its generalization with respect to the order of the harmonic number.

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    $\begingroup$ Nice and (+1) for the question. $\endgroup$ Commented Jan 2, 2020 at 15:55
  • $\begingroup$ so the integral representation for your sum is $\int_0^1\frac{\arctan x\ln^3x}{1+x^2}dx$ which can be converted to $\int_0^\infty \frac{\arctan x\ln^3x}{1+x^2}dx$ which I think is manageable. $\endgroup$ Commented Jan 2, 2020 at 16:04
  • $\begingroup$ @AliShather Thank you. $\endgroup$ Commented Jan 2, 2020 at 16:15

3 Answers 3

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A solution in large steps by Cornel I. Valean to the main series

The value of the series is $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^4}=\frac{\pi }{192} \psi ^{(3)}\left(\frac{1}{4}\right)-\frac{\pi ^2}{3} \zeta (3)-\frac{437}{64} \zeta (5)-\frac{\pi ^5}{24}, \tag1$$ where $H_n=\sum_{k=1}^n \frac{1}{k}$ is the $n$th harmonic number, $\zeta$ represents the Riemann zeta function, and $\psi^{(n)}$ denotes the Polygamma function.

We will make use of a powerful result with harmonic numbers which is presented in the book, (Almost) Impossible Integrals, Sums, and Series, $$\int_0^{\infty } \tanh (\pi x)\left(\frac{1}{x}-\frac{x}{n^2+x^2}\right) \textrm{d}x=2 H_{2n}-H_n,$$ and that is proved in two ways in the mentioned book on pages $200$ - $203$.

Replacing $n$ by $2n$ in the identity above, multiplying both sides by $1/n^4$, taking the sum from $n=1$ to $\infty$, and rearranging the resulting series, we get

$$120 \sum _{n=1}^{\infty } \frac{H_n}{n^4}-120\sum _{n=1}^{\infty } (-1)^{n-1} \frac{ H_n}{n^4}-16\sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_{2 n}}{n^4}$$ $$=\frac{2}{3}\int_0^{\infty } \frac{\tanh (\pi x)}{x^5} \left(\pi ^2 x^2-6 \pi x \coth \left(\frac{\pi x}{2}\right)+12\right) \textrm{d}x$$ $$=\frac{1}{36}\int _0^{\infty }\left(\int _0^{\infty }\tanh (\pi x) \left(\pi ^2 x^2-6 \pi x \coth \left(\frac{\pi x}{2}\right)+12\right)y^4 e^{-xy} \textrm{d}y\right)\textrm{d}x$$ $$=\frac{1}{36}\int _0^{\infty }\left(\int _0^{\infty }\tanh (\pi x) \left(\pi ^2 x^2-6 \pi x \coth \left(\frac{\pi x}{2}\right)+12\right)y^4 e^{-xy} \textrm{d} x\right)\textrm{d}y$$ $$\small=\int_0^{\infty } \biggr(\frac{y^4 }{6 \pi }\psi \left(\frac{1}{4} \left(\frac{y}{\pi }+2\right)\right)-\frac{y^4 }{6 \pi }\psi\left(\frac{y}{4 \pi }\right)+\frac{y^4}{48 \pi } \psi ^{(1)}\left(\frac{1}{4} \left(\frac{y}{\pi }+3\right)\right)-\frac{y^4}{48 \pi } \psi ^{(1)}\left(\frac{y+\pi }{4 \pi }\right) $$ $$ +\frac{y^4 }{1152 \pi }\psi ^{(2)}\left(\frac{1}{4} \left(\frac{y}{\pi }+2\right)\right)-\frac{y^4}{1152 \pi } \psi ^{(2)}\left(\frac{y}{4 \pi }\right)-\frac{y^3}{3}-\frac{\pi y^2}{6}-\frac{\pi ^2 y}{18}\biggr) \textrm{d}y$$ $$=248 \zeta (5)-\frac{14}{3} \pi ^2 \zeta (3)+\frac{2 }{3}\pi ^5-\frac{1}{12} \pi \psi ^{(3)}\left(\frac{1}{4}\right),$$

where for avoiding some tedious calculations I used Mathematica (but these can also be done perfectly manually). For example, the last integral has a nonelementary antiderivative in terms of Polygamma and Negapolygamma functions (that is, for the latter the order becomes negative).

Since we know the classical results,

$$\small 2\sum_{k=1}^\infty \frac{H_k}{k^m}=(m+2)\zeta(m+1)-\sum_{k=1}^{m-2} \zeta(m-k) \zeta(k+1), m\ge2;$$ $$\small \sum_{k=1}^{\infty} (-1)^{k-1}\frac{H_k}{k^{2m}}=\left(m+\frac{1}{2}\right)\eta(2m+1)-\frac{1}{2}\zeta(2m+1)-\sum_{i=1}^{m-1}\eta(2i)\zeta(2m-2i+1), m\ge1$$ where $H_n=\sum_{k=1}^n \frac{1}{k}$ is the $n$th harmonic number, $\zeta$ represents the Riemann zeta function, and $\eta$ represents the Dirichlet eta function,

then the stated result follows.

First note: Be careful Mathematica goes crazy during the calculations and you might think something is wrong, but it's not. All the calculations should be done very carefully.

Second note: Essentially, this strategy opens the gates to evaluate very advanced harmonic series like

$$\sum_{n=1}^{\infty}\frac{(H_{kn}^{(m)})^r}{n^p}, \ \sum_{n=1}^{\infty} (-1)^{(n-1)}\frac{(H_{kn}^{(m)})^r}{n^p}.$$

Third note: To make everything easier in the work with Negapolygamma it's important to groups everything strategically. I'll give an example.

Let's prove that $$\psi ^{(-2)}\left(\frac{3}{4}\right)-\psi ^{(-2)}\left(\frac{1}{4}\right)=\frac{1}{4}\left(\log(2\pi)-\frac{2}{\pi}G\right).$$

Proof. Let's denote $I=\psi ^{(-2)}\left(\frac{3}{4}\right)-\psi ^{(-2)}\left(\frac{1}{4}\right)= \int_{1/4}^{3/4} \log (\Gamma (x)) \textrm{d}x$.

If we let the variable change $x\mapsto 1-x$ in the last integral, we get $\displaystyle I=\int_{1/4}^{3/4} \log\left(\Gamma (1-x)\right) \textrm{d}x$, that if we add to the initial integral and then combine with the Euler's reflection formula for Gamma function and the Fourier series of $\log(\sin(x))$, we get \begin{equation*} I=\frac{1}{2}\int_{1/4}^{3/4} \log\left(\Gamma(x)\Gamma (1-x)\right) \textrm{d}x=\frac{1}{2}\int_{1/4}^{3/4} \log\left(\frac{\pi}{\sin(\pi x)}\right) \textrm{d}x \end{equation*} \begin{equation*} =\frac{1}{4}\log(\pi)-\frac{1}{2}\int_{1/4}^{3/4} \log\left(\sin(\pi x)\right) \textrm{d}x=\frac{1}{4}\left(\log(2\pi)-\frac{2}{\pi}G\right). \end{equation*}

When taken separately we'll also have to deal with the derivative of the Riemann zeta function, and it is not hard to see the presence of the Glaisher–Kinkelin constant in every of the two Negapolygamma particular values like here https://www.wolframalpha.com/input/?i=PolyGamma%5B-2%2C+1%2F4%5D. So, you want to better group everything strategically.


A solution in large steps by Cornel to the supplementary series

$$i) \ \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_{2 n}^{(2)}}{n^3}=\frac{61 }{192}\pi ^2 \zeta (3)+\frac{1973 }{128}\zeta (5)+\frac{\pi ^5}{16}-\frac{1}{128} \pi \psi ^{(3)}\left(\frac{1}{4}\right);$$ $$ii) \ \sum _{n=1}^{\infty } (-1)^{n-1} \frac{ H_{2 n}^{(3)}}{n^2}=\frac{\pi ^3 G}{8}+\frac{1}{64}\pi ^2 \zeta (3)-\frac{2997 }{256}\zeta (5)-\frac{\pi ^5}{32}+\frac{1}{256} \pi \psi ^{(3)}\left(\frac{1}{4}\right).$$

With the right results in hands, the extraction will happen very fast. So, here is what we need:

$a)$ The series result in $(1)$

$b)$ The Cauchy product of two series, $$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=6\sum_{n=1}^{\infty} x^n\frac{H_n}{n^4}+3\sum_{n=1}^{\infty} x^n\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^{\infty} x^n\frac{H_n^{(3)}}{n^2}-10\operatorname{Li}_5(x).$$

$c)$ The special integral

$$\int_0^1 \frac{\log(1-x) \log ^2(x) \log(1+x^2)}{x} \textrm{d}x$$ $$=\frac{7}{48} \pi ^2 \zeta (3)-\frac{1}{4} \sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_{2 n}}{n^4}-\frac{1}{2} \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_{2 n}^{(2)}}{n^3}-\sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_{2 n}^{(3)}}{n^2},$$

which is calculated in the preprint The derivation of eighteen special challenging logarithmic integrals by Cornel Ioan Valean.

Done.

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    $\begingroup$ beautiful (+1). $\endgroup$ Commented Jan 2, 2020 at 17:26
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Lets start with the integral representation of the sum, $$I=\int_0^1\frac{\arctan x\ln^3x}{1+x^2}dx=\int_0^\infty\frac{\arctan x\ln^3x}{1+x^2}dx-\underbrace{\int_1^\infty\frac{\arctan x\ln^3x}{1+x^2}dx}_{\large1/x\ \mapsto x}$$

$$=\int_0^\infty\frac{\arctan x\ln^3x}{1+x^2}dx+\int_0^1\frac{(\frac{\pi}{2}-\arctan x)\ln^3x}{1+x^2}dx$$

$$\Longrightarrow 2I=\int_0^\infty\frac{\arctan x\ln^3x}{1+x^2}dx+\frac{\pi}2\underbrace{\int_0^1\frac{\ln^3x}{1+x^2}dx}_{-6\beta(4)}$$

$$I=\frac12\int_0^\infty\frac{\arctan x\ln^3x}{1+x^2}dx-\frac{3\pi}{2}\beta(4)$$

where $\beta(4)=\frac{\psi_3(1/4)}{768}-\frac{\pi^4}{96}$ is the Dirichlet beta function.


To complete my solution, I am going to borrow @user97357329's result in the comment below,

$$\int_0^{\infty } \frac{ \arctan x\ln ^3x}{1+x^2} \ dx=\frac{1}{8} \int_0^1 \frac{\left(2 \ln ^2x+\pi ^2\right)\ln ^2x }{1-x^2} \ dx.$$

$$=\frac14\underbrace{\int_0^1\frac{\ln^4x}{1-x^2}\ dx}_{\frac{93}4\zeta(5)}+\frac{\pi^2}{8}\underbrace{\int_0^1\frac{\ln^2x}{1-x^2}\ dx}_{\frac74\zeta(3)}$$

$$=\frac{93}{16}\zeta(5)+\frac{7\pi^2}{32}\zeta(3)$$

$$\Longrightarrow I=\frac{93}{32}\zeta(5)+\frac{7\pi^2}{64}\zeta(3)+\frac{\pi^5}{64}-\frac{\pi}{512}\psi_3(1/4)\tag1$$


Now, from here we have

$$\frac{\arctan x}{1+x^2}=\frac12\sum_{n=1}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}$$

which gives

$$\int_0^1\frac{\arctan x\ln^3x}{1+x^2}\ dx=\frac12\sum_{n=1}^\infty(-1)^n\left(H_n-2H_{2n}\right)\int_0^1 x^{2n-1}\ln^3x\ dx$$

$$=-3\sum_{n=1}^\infty(-1)^n\frac{H_n-2H_{2n}}{(2n)^4}=-\frac3{16}\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}+\frac3{8}\sum_{n=1}^\infty(-1)^n\frac{H_{2n}}{n^4}\tag2$$

By combining $(1)$ and $(2)$ and substituting $\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$ we reach the closed form of $\sum_{n=1}^\infty(-1)^n\frac{H_{2n}}{n^4}$.


Explanation for $\displaystyle\int_0^{\infty } \frac{ \arctan x\ln ^3x}{1+x^2} \ dx=\frac{1}{8} \int_0^1 \frac{\left(2 \ln ^2x+\pi ^2\right)\ln ^2x }{1-x^2} \ dx$,

Write $\tan^{-1}x=\int_0^1\frac{x}{1+x^2y^2}dy$

$$\Longrightarrow\int_0^{\infty } \frac{ \arctan x\ln ^3x}{1+x^2}dx=\int_0^\infty\frac{\ln^3x}{1+x^2}\left(\int_0^1\frac{x}{1+x^2y^2}dy\right)dx$$

$$=\int_0^1\underbrace{\left(\int_0^\infty\frac{x\ln^3x}{(1+x^2)(1+x^2y^2)}dx\right)}_{x^2\mapsto x}dy=\frac1{16}\int_0^1\left(\int_0^\infty\frac{\ln^3x}{(1+x)(1+y^2x)}dx\right)dy$$

The inner integral was nicely calculated by @Zacky in the comments below and as follows,

$$\int_0^\infty\frac{\ln^3x}{(1+x)(1+y^2x)}dx=\int_0^1\frac{\ln^3x}{(1+x)(1+y^2x)}dx+\underbrace{\int_1^\infty\frac{\ln^3x}{(1+x)(1+y^2x)}dx}_{\large x\mapsto 1/x}$$

$$=\int_0^1\frac{\ln^3x}{(1+x)(1+y^2x)}dx-\int_0^1\frac{\ln^3x}{(1+x)(x+y^2)}dx$$

$$=\frac{1}{16(1-y^2)}\int_0^1 \ln^3 x\left(\frac{2}{1+x}-\frac{1}{y^2+x}-\frac{1}{\frac{1}{y^2}+x}\right)dx$$

$$=-\frac{3}{8(1-y^2)}\left(\color{red}{\frac74\zeta(4)+\operatorname{Li}_4\left(-\frac{1}{y^2}\right)+\operatorname{Li}_4\left(-y^2\right)}\right)$$

Now use the identity

$$\frac74\zeta(4)+\operatorname{Li}_4(1/x)+\operatorname{Li}_4(x)=-\frac{\pi^2}{12}\ln^2(-x)-\frac1{24}\ln^4(-x)$$

replace $x$ with $-y^2$ to get

$$\color{red} {\frac74\zeta(4)+\operatorname{Li}_4\left(-\frac{1}{y^2}\right)+\operatorname{Li}_4\left(-y^2\right)}=-\frac{\pi^2}{3}\ln^2y-\frac23\ln^4y$$

giving us

$$\int_0^{\infty } \frac{ \arctan x\ln ^3x}{1+x^2} \ dx=\frac{1}{8} \int_0^1 \frac{\left(2 \ln ^2x+\pi ^2\right)\ln ^2x }{1-x^2} \ dx$$

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    $\begingroup$ Write $$\int_0^\infty\frac{\arctan x\ln^3x}{1+x^2}dx,$$ like a double integral. You may use that $$\arctan(x)=\int_0^1 \frac{x}{1+x^2 y^2} \textrm{d}y.$$ $\endgroup$ Commented Jan 2, 2020 at 16:37
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    $\begingroup$ Thus, $$\int_0^{\infty } \frac{ \arctan(x)\log ^3(x)}{1+x^2} \textrm{d}x=\frac{1}{8} \int_0^1 \frac{\left(2 \log ^2(x)+\pi ^2\right)\log ^2(x) }{1-x^2} \textrm{d}x.$$ $\endgroup$ Commented Jan 2, 2020 at 16:44
  • $\begingroup$ (+1) for a good starting point. $\endgroup$ Commented Jan 2, 2020 at 16:51
  • $\begingroup$ @user97357329 Thank you $\endgroup$ Commented Jan 2, 2020 at 16:55
  • $\begingroup$ @AliShather do you know why there is $$\int_0^{\infty } \frac{ \arctan x\ln ^3x}{1+x^2}dx=\frac{1}{8} \int_0^1 \frac{\left(2 \ln ^2x+\pi ^2\right)\ln ^2x }{1-x^2}dx?$$ $\endgroup$
    – Zacky
    Commented Jan 2, 2020 at 17:45
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Consider $$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \right) \right){{\left( z-n \right)}^{k}}}, n\ge 0$$ Then $$\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{2z\to 2k+1}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}\pi f\left( k+\tfrac{1}{2} \right)}{2\left( z-\tfrac{2k+1}{2} \right)}+O\left( 1 \right)$$ On the other hand we also have $$\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{2z\to 2k}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}f\left( k \right)}{2{{\left( z-k \right)}^{2}}}+\frac{{{\left( -1 \right)}^{k}}\left\{ {{H}_{2k}}f\left( k \right)+\tfrac{1}{2}f'\left( k \right) \right\}}{z-k}+O\left( 1 \right), k\ge 0$$ Similarly $$\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{z\to -k}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}\left( \psi \left( 2k \right)+\gamma \right)f\left( -k \right)}{z+k}, k>0$$ The only other residues are those due to $f$ which we assume has one pole at the origin of order at least $2$. The sum of residues over the entire plane is zero. Hence $$\begin{align} & \sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}{{H}_{2k}}f\left( k \right)} \\ & =-\frac{1}{2}\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}f'\left( k \right)}-\frac{\pi }{2}\sum\limits_{k=0}^{\infty }{{{\left( -1 \right)}^{k}}f\left( k+\tfrac{1}{2} \right)}-\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}\left( \psi \left( 2k \right)+\gamma \right)f\left( -k \right)}\\&-\underset{z=0}{\mathop{res}}\,\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)} \\ \end{align}$$

This goes a long way to generalise such results as you can now pick any power by simply choosing f appropriately. For example letting $f\left( z \right)=\frac{1}{{{z}^{4}}}$ we find $$\begin{align} & \sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}} \\ & =-\frac{\pi }{192}\left\{ {{\psi }^{\left( 3 \right)}}\left( \tfrac{1}{4} \right)-{{\psi }^{\left( 3 \right)}}\left( \tfrac{3}{4} \right) \right\}+\frac{2}{3}{{\pi }^{2}}\zeta \left( 3 \right)+\frac{113}{8}\zeta \left( 5 \right)-\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}\frac{{{H}_{2k-1}}}{{{k}^{4}}}} \\ \end{align}$$ Now $$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}}=A-\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k-1}}}{{{k}^{4}}}}\Rightarrow \sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\left( {{H}_{2k}}+{{H}_{2k-1}} \right)}{{{k}^{4}}}}=A$$ But we can write this as $$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\left( \frac{1}{2k}+2{{H}_{2k-1}} \right)}{{{k}^{4}}}}=A$$ hence $$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k-1}}}{{{k}^{4}}}}=\frac{1}{2}A+\frac{15}{64}\zeta \left( 5 \right)$$ We have then $$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}}=\frac{\pi }{384}\left\{ {{\psi }^{\left( 3 \right)}}\left( \tfrac{3}{4} \right)-{{\psi }^{\left( 3 \right)}}\left( \tfrac{1}{4} \right) \right\}+\frac{{{\pi }^{2}}}{3}\zeta \left( 3 \right)+\frac{437}{64}\zeta \left( 5 \right)$$

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