Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$

Question

Does the following series or integral have a closed-form

\begin{equation} \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx \end{equation}

where $\Psi_3(x)$ is the polygamma function of order $3$.

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Here is my attempt. Using equation (11) from Mathworld Wolfram: \begin{equation} \Psi_n(z)=(-1)^{n+1} n!\left(\zeta(n+1)-H_{z-1}^{(n+1)}\right) \end{equation} I got \begin{equation} \Psi_3(n+1)=6\left(\zeta(4)-H_{n}^{(4)}\right) \end{equation} then \begin{align} \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=6\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\left(\zeta(4)-H_{n}^{(4)}\right)\\ &=6\zeta(4)\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\ &=\frac{\pi^4}{15}\ln2-6\sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}\\ \end{align} From the answers of this OP, the integral representation of the latter Euler sum is \begin{align} \sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=\int_0^1\int_0^1\int_0^1\int_0^1\int_0^1\frac{dx_1\,dx_2\,dx_3\,dx_4\,dx_5}{(1-x_1)(1+x_1x_2x_3x_4x_5)} \end{align} or another simpler form \begin{align} \sum_{n=1}^\infty\frac{(-1)^{n+1}H_{n}^{(4)}}{n}&=-\int_0^1\frac{\text{Li}_4(-x)}{x(1+x)}dx\\ &=-\int_0^1\frac{\text{Li}_4(-x)}{x}dx+\int_0^1\frac{\text{Li}_4(-x)}{1+x}dx\\ &=\text{Li}_5(-1)-\int_0^{-1}\frac{\text{Li}_4(x)}{1-x}dx\\ \end{align} I don't know how to continue it, I am stuck. Could anyone here please help me to find the closed-form of the series preferably with elementary ways? Any help would be greatly appreciated. Thank you.

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Edit :

Using the integral representation of polygamma function \begin{equation} \Psi_m(z)=(-1)^m\int_0^1\frac{x^{z-1}}{1-x}\ln^m x\,dx \end{equation} then we have \begin{align} \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)&=-\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\int_0^1\frac{x^{n}}{1-x}\ln^3 x\,dx\\ &=-\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n+1}x^{n}}{n}\cdot\frac{\ln^3 x}{1-x}\,dx\\ &=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx\\ \end{align} I am looking for an approach to evaluate the above integral without using residue method or double summation.

Answer 1

Edited: I have changed the approach as I realised that the use of summation is quite redundant (since the resulting sums have to be converted back to integrals). I feel that this new method is slightly cleaner and more systematic.

We can break up the integral into \begin{align} -&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\rm d}x\\ =&\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\int^1_0\frac{(1+x)\ln^3{x}\ln(1-x^2)}{(1+x)(1-x)}{\rm d}x\\ =&\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\int^1_0\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}{\rm d}x-\int^1_0\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}{\rm d}x\\ =&\frac{15}{16}\int^1_0\frac{\ln^3{x}\ln(1-x)}{1-x}{\rm d}x-\frac{1}{16}\int^1_0\frac{x^{-1/2}\ln^3{x}\ln(1-x)}{1-x}{\rm d}x\\ =&\frac{15}{16}\frac{\partial^4\beta}{\partial a^3 \partial b}(1,0^{+})-\frac{1}{16}\frac{\partial^4\beta}{\partial a^3 \partial b}(0.5,0^{+}) \end{align} After differentiating and expanding at $b=0$ (with the help of Mathematica), \begin{align} &\frac{\partial^4\beta}{\partial a^3 \partial b}(a,0^{+})\\ =&\left[\frac{\Gamma(a)}{\Gamma(a+b)}\left(\frac{1}{b}+\mathcal{O}(1)\right)\left(\left(-\frac{\psi_4(a)}{2}+(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a)\right)b+\mathcal{O}(b^2)\right)\right]_{b=0}\\ =&-\frac{1}{2}\psi_4(a)+(\gamma+\psi_0(a))\psi_3(a)+3\psi_1(a)\psi_2(a) \end{align} Therefore, \begin{align} -&\int^1_0\frac{\ln^3{x}\ln(1+x)}{1-x}{\rm d}x\\ =&-\frac{15}{32}\psi_4(1)+\frac{45}{16}\psi_1(1)\psi_2(1)+\frac{1}{32}\psi_4(0.5)+\frac{1}{8}\psi_3(0.5)\ln{2}-\frac{3}{16}\psi_1(0.5)\psi_2(0.5)\\ =&-12\zeta(5)+\frac{3\pi^2}{8}\zeta(3)+\frac{\pi^4}{8}\ln{2} \end{align} The relation between $\psi_{m}(1)$, $\psi_m(0.5)$ and $\zeta(m+1)$ is established easily using the series representation of the polygamma function.

Answer 2

\begin{align} \sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n+1)}{n} &=-12\zeta(5)+\frac{45}{4}\zeta(4)\ln{2}+\frac{9}{4}\zeta(2)\zeta(3) \end{align}

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Let $\displaystyle f(z)=\frac{\pi\csc(\pi z)\psi_3(-z)}{z}$. Then at the positive integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,n) &=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{6(-1)^n}{z(z-n)^5}+\frac{6(-1)^n\zeta(2)}{z(z-n)^3}+(-1)^n\frac{(33/2)\zeta(4)+6H_n^{(4)}}{z(z-n)}\right]\\ &=6\sum^\infty_{n=1}\frac{(-1)^n}{n^5}+6\zeta(2)\sum^\infty_{n=1}\frac{(-1)^n}{n^3}+\frac{33}{2}\zeta(4)\sum^\infty_{n=1}\frac{(-1)^n}{n}+6\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n}\\ &=-\frac{45}{8}\zeta(5)-\frac{9}{2}\zeta(2)\zeta(3)-\frac{33}{2}\zeta(4)\ln{2}+6\sum^\infty_{n=1}\frac{(-1)^nH_n^{(4)}}{n} \end{align} At zero, $${\rm Res}(f,0)=24\zeta(5)$$ At the negative integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,-n) &=\sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n)}{n}\\ &=6\zeta(4)\ln{2}-6\sum^\infty_{n=1}\frac{(-1)^{n-1}H_{n-1}^{(4)}}{n}\\ &=\frac{45}{8}\zeta(5)+6\zeta(4)\ln{2}+6\sum^\infty_{n=1}\frac{(-1)^{n}H_{n}^{(4)}}{n}\\ \end{align} Since the sum of residues is zero, \begin{align} 12\sum^\infty_{n=1}\frac{(-1)^{n}H_{n}^{(4)}}{n}=-24\zeta(5)+\frac{21}{2}\zeta(4)\ln{2}+\frac{9}{2}\zeta(2)\zeta(3)\\ \end{align} This implies that \begin{align} \sum^\infty_{n=1}\frac{(-1)^{n-1}\psi_3(n+1)}{n} &=-12\zeta(5)+\frac{45}{4}\zeta(4)\ln{2}+\frac{9}{4}\zeta(2)\zeta(3) \end{align} Refer to this paper if you have any doubts.

Answer 3

Different approach using only series manipulations.

By using the identity

$$\frac{\ln(1+x)}{1-x}=\sum_{n=1}^\infty \overline{H}_n x^n$$ which can be easily proved by series-expanding the numerator and denominator.

Multiply both sides by $\ln^3x$ then $\int_0^1$ we get

$$I=\int_0^1\frac{\ln(1+x)\ln^3x}{1-x}\ dx=\sum_{n=1}^\infty \overline{H}_n\int_0^1 x^n \ln^3x\ dx=-6\sum_{n=1}^\infty\frac{\overline{H}_n}{(n+1)^4}=-6\sum_{n=1}^\infty\frac{\overline{H}_{n-1}}{n^4}$$

Now use $\overline{H}_{n-1}=\overline{H}_n+\frac{(-1)^n}{n}$

$$ \Longrightarrow I=-6\sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}-6\sum_{n=1}^\infty\frac{(-1)^n}{n^5}=\frac{45}{8}\zeta(5)-6\sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}\tag1$$

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$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}=1+\sum_{n=2}^\infty\frac{\overline{H}_n}{n^4}=1+\sum_{n=1}^\infty\frac{\overline{H}_{2n}}{(2n)^4}+\sum_{n=1}^\infty\frac{\overline{H}_{2n+1}}{(2n+1)^4}$$

By writing $\overline{H}_{2n}=H_{2n}-H_n$ and $\overline{H}_{2n+1}=H_{2n+1}-H_n$ we have

$$\sum_{n=1}^\infty\frac{\overline{H}_{2n}}{(2n)^4}=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}-\sum_{n=1}^\infty\frac{H_{n}}{(2n)^4}=\frac12\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^4}+\frac7{16}\sum_{n=1}^\infty\frac{H_{n}}{n^4}$$

and

$$\sum_{n=1}^\infty\frac{\overline{H}_{2n+1}}{(2n+1)^4}=\color{blue}{\sum_{n=1}^\infty\frac{H_{2n+1}}{(2n+1)^4}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}$$

$$=\color{blue}{-1+\sum_{n=0}^\infty\frac{H_{2n+1}}{(2n+1)^4}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}$$

$$=\color{blue}{-1+\frac12\sum_{n=0}^\infty\frac{(-1)^nH_{n+1}}{(n+1)^4}+\frac12\sum_{n=0}^\infty\frac{H_{n+1}}{(n+1)^4}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}$$

$$=\color{blue}{-1-\frac12\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^4}+\frac12\sum_{n=1}^\infty\frac{H_{n}}{n^4}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}$$

Combine the two sums,

$$\Longrightarrow \sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}=\frac{15}{16}\sum_{n=1}^\infty\frac{H_n}{n^4}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}$$

From here we have

$$\sum_{n=1}^{\infty} \frac{H_{n}}{(n+a)^{2}}=\left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2}$$

Differentiate with respect to $a$ twice then set $a=1/2$ we get

$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}=\frac{31}{8}\zeta(5)-\frac{15}{8}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)$$

Substituting this result along with $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ gives

$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}=-\frac{17}{16}\zeta(5)+\frac{15}{8}\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)\tag2$$

Finally plug $(2)$ in $(1)$ we get

$$I=12\zeta(5)-\frac{45}{4}\ln2\zeta(4)-\frac94\zeta(2)\zeta(3)$$

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Edit

Another way to calculate $\displaystyle \sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}$ is to use the generalization

$$\sum_{k = 1}^\infty \frac{\overline H_k}{k^m} = \zeta (m) \log 2 - \frac{1}{2} m \zeta (m + 1) + \eta (m + 1) + \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1).$$

where $\eta (s) = \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^s} = (1 - 2^{1 - s}) \zeta (s)$ is the Dirichlet eta function and $\zeta (s) = \sum_{n = 1}^\infty \frac{1}{n^s}$ is the Riemann zeta function.

With $m=4$ we have

$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^4}=-\frac{17}{16}\zeta(5)+\frac{15}{8}\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)$$

The generalization can be found here (see Theorem 3.5 on page 9).

Answer 4

We can have a nice generalization,

From

$$\frac{\ln(1+x)}{1-x}=\sum_{n=1}^\infty \overline{H}_n x^n$$

We have

$$I_m=\int_0^1\frac{\ln(1+x)\ln^{m-1}x}{1-x}\ dx=\sum_{n=1}^\infty \overline{H}_n \int_0^1 x^n\ln^{m-1}x\ dx$$

$$=(-1)^{m-1}(m-1)!\sum_{n=1}^\infty \frac{\overline{H}_n}{(n+1)^m}$$

$$=(-1)^{m-1}(m-1)!\sum_{n=1}^\infty \frac{\overline{H}_{n-1}}{n^m}$$

$$=(-1)^{m-1}(m-1)!\sum_{n=1}^\infty \frac{\overline{H}_n+\frac{(-1)^n}{n}}{n^m}$$

$$=(-1)^{m-1}(m-1)!\left[\sum_{n=1}^\infty \frac{\overline{H}_n}{n^m}-\eta(m+1)\right]$$

Substitute

$$\sum_{n = 1}^\infty \frac{\overline H_n}{n^m} = \ln 2\zeta (m) - \frac{1}{2} m \zeta (m + 1) + \eta (m + 1) + \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1)$$

We get

$$I_m=(-1)^{m}(m-1)!\left[\frac{1}{2} m \zeta (m + 1)-\ln 2\zeta (m) - \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1)\right]$$

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The generalization $\displaystyle \small \sum_{n = 1}^\infty \frac{\overline H_n}{n^m}$ can be found here (see Theorem 3.5 on page 9).

Answer 5

Computation of $\displaystyle U=\int_0^1 \frac{\ln(1+x)\ln^3 x}{1-x}\,dx$

\begin{align*} U&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^3 t}{1-t}\,dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x\frac{\ln^3 t}{1-t}\,dt\right)\,dx\\ &=-6\zeta(4)\ln 2+\int_0^1\int_0^1 \left(\frac{\ln^3(tx)}{(1+t)(1+x)}-\frac{\ln^3(tx)}{(1+t)(1-tx)}\right)\,dt\,dx\\ &=-6\zeta(4)\ln 2+6\left(\int_0^1\frac{\ln^2 t}{1+t}\,dt\right)\left(\int_0^1\frac{\ln x}{1+x}\,dx\right)+\\ &2\left(\int_0^1\frac{\ln^3 t}{1+t}\,dt\right)\left(\int_0^1\frac{1}{1+x}\,dx\right)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\ &=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\ &\overset{\text{IBP}}=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\left[\ln\left(\frac{t}{1+t}\right)\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\right]_0^1+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\ &=-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\ &=-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+24\zeta(5)-U\\ U&=\boxed{-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)} \end{align*}

Answer 6

The following proof is very simple and short but it works for only the odd powers of $\ln(x):$

By integrating

$$\operatorname{Li}_2(-x)+\operatorname{Li}_2\left(-\frac1x\right)=-\frac{\ln^2(x)}{2}-2\eta(2)$$

repeatedly, we get

$$\operatorname{Li}_{2a}(-x)+\operatorname{Li}_{2a}\left(-\frac1x\right)=-2\sum_{k=0}^a \frac{\eta(2a-2k)}{(2k)!}\ln^{2k}(x)$$

Using the integral form of the polylogarithm function, we have

\begin{gather*} \operatorname{Li}_{2a}(-x)+\operatorname{Li}_{2a}\left(-\frac1x\right)\\ =\frac{-1}{(2a-1)!}\int_0^1\frac{-x\ln^{2a-1}(y)}{1+xy}\mathrm{d}y-\frac1{(2a-1)!}\int_0^1\frac{-\frac1{x}\ln^{2a-1}(y)}{1+\frac{y}{x}}\mathrm{d}y\\ =\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\frac{x}{1+xy}+\frac{1}{x+y}\right)\mathrm{d}y\\ =\frac{1}{(2a-1)!}\int_0^1\frac{(1+2xy+x^2)\ln^{2a-1}(y)}{(1+xy)(x+y)}\mathrm{d}y. \end{gather*} and so \begin{equation} \int_0^1\frac{(1+2xy+x^2)\ln^{2a-1}(y)}{(1+xy)(x+y)}\mathrm{d}y=-2(2a-1)!\sum_{k=0}^a \frac{\eta(2a-2k)}{(2k)!}\ln^{2k}(x). \end{equation}

Divide the latter identity by $1+x$ using $\int_0^1\frac{\ln^s(x)}{1+x}dx=(-1)^s s!\eta(s+1),$ we get

$$-2(2a-1)!\sum_{k=0}^a \eta(2a-2k)\eta(2k+1)=\int_0^1\ln^{2a-1}(y)\left(\int_0^1\frac{1+2xy+x^2}{(1+xy)(x+y)(1+x)}\mathrm{d}x\right)\mathrm{d}y$$

$$=\int_0^1\ln^{2a-1}(y)\left(\frac{2\ln(1+y)}{1-y}+\frac{\ln(1+y)}{y}-\frac{\ln(y)}{1-y}-\frac{2\ln(2)}{1-y}\right)\mathrm{d}y$$

$$=2\int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{1-x}\mathrm{d}x-(2a-1)!\eta(2a+1)-(2a)!\zeta(2a+1)$$ $$+2(2a-1)!\ln(2)\zeta(2a)$$

$$\Longrightarrow \int_0^1\frac{\ln^{2a-1}(x)\ln(1+x)}{1-x}\mathrm{d}x$$ $$=(2a-1)!\left(\frac12(2a-4^{-a}+1)\zeta(2a+1)-\ln(2)\zeta(2a)-\sum_{k=0}^a \eta(2a-2k)\eta(2k+1)\right).$$