An Alternating Euler Sum

In today´s post we will compute the amazing alternate infinite sum

$\sum_{n=1}^\infty\frac{(-1)^{n}\text{H}_n}{n^3}=-\frac{11 \pi^4}{360}+2\text{Li}_4\left(\frac12\right)+\frac{7\ln(2)\zeta(3)}{4}+\frac{\ln^4(2)}{12} -\frac{\ln^2(2)\pi^2}{12}$

$\begin{aligned} \sum_{n=1}^\infty\frac{(-1)^{n}\text{H}_n}{n^3}&=\sum_{n=1}^\infty (-1)^{n}\text{H}_n \left( \frac12 \int_0^1x^{n-1}\ln^2(x)\,dx\right)\\ &= \frac12 \int_0^1 \left(\sum_{n=1}^\infty (-1)^{n}\text{H}_nx^{n-1} \right)\ln^2(x)\,dx\\ &= \frac12 \int_0^1 \left(\sum_{n=1}^\infty (-1)^{n}\text{H}_nx^{n} \right)\frac{\ln^2(x)}{x}\,dx\\ &= -\frac12 \int_0^1 \left(\frac{\ln(1-(-x))}{1-(-x)} \right)\frac{\ln^2(x)}{x}\,dx \qquad (*)\\ &= - \frac12 \int_0^1 \frac{\ln^2(x)\ln(1+x)}{x(1+x)} \,dx\\ &=\frac12 \int_0^1 \frac{\ln^2(x)\ln(1+x)}{1+x} \,dx- \frac12 \int_0^1 \frac{\ln^2(x)\ln(1+x)}{x} \,dx\\ &=-\frac12 \int_0^1 \frac{\ln^2(x)\ln(1+x)}{x} \,dx+ \frac12\left( \int_0^{1/2} \frac{\ln^2\left(\frac{x}{1-x}\right)\ln\left(\frac{1}{1-x}\right)}{\frac{1}{1-x}} \,\frac{dx}{(1-x)^2}\right) \qquad \left(x \to \frac{x}{1-x} \right)\\ &=-\frac12 \int_0^1 \frac{\ln^2(x)\ln(1+x)}{x} \,dx+ \frac12\left(\int_0^{1/2} \frac{\left(\ln(x)-\ln(1-x) \right)^2\left(-\ln(1-x) \right)}{1-x} \,dx\right)\\ &=-\frac12 \int_0^1 \frac{\ln^2(x)\ln(1+x)}{x} \,dx+ \frac12\left(-\int_0^{1/2} \frac{\ln^2(x)\ln(1-x) }{1-x} \,dx +2\int_0^{1/2} \frac{\ln(x)\ln^2(1-x) }{1-x} \,dx-\int_0^{1/2} \frac{\ln^3(1-x) }{1-x} \,dx\right)\\ &=-\frac12 I_1+ \frac12\left(-I_2 +2I_3-I_4\right)\\ &=-\frac12\left(\frac74\zeta(4) \right)+ \frac12\left( \frac{\ln^4(2)}{4}+\frac14\zeta(4)+2\left(2\text{Li}_4\left(\frac12\right)-\frac{\ln^2(2)\zeta(2)}{2}+\frac{7\ln(2) \zeta(3)}{4}-2\zeta(4)-\frac{\ln^4(2)}{6} \right)+\frac{\ln^4(2)}{4}\right)\\ &=-\frac78\zeta(4) + \frac12\left(4\text{Li}_4\left(\frac12\right)-\frac{15}{4}\zeta(4)+\frac{7\ln(2)\zeta(3)}{2}-\ln^2(2)\zeta(2)+\frac{\ln^4(2)}{6}\right)\\ &=-\frac{11 \pi^4}{360}+2\text{Li}_4\left(\frac12\right)+\frac{7\ln(2)\zeta(3)}{4}+\frac{\ln^4(2)}{12} -\frac{\ln^2(2)\pi^2}{12}\qquad \blacksquare \end{aligned}$

Where in (*) we used the generating function previously proved here

$\sum_{n=1}^\infty\text{H}_nx^n=-\frac{\ln(1-x)}{1-x}$

Evaluation of

$\begin{aligned} I_1&=\int_0^1 \frac{\ln^2(x)\ln(1+x)}{x} \,dx\\ &=\int_0^1 \frac{\ln^2(x)}{x}\left(\sum_{k=1}^\infty(-1)^{k-1}\frac{x^k}{k} \right)\,dx\\ &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\int_0^1 x^{k-1}\ln^2(x)\,dx\\ &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\left(\frac{(-1)^22!}{(k-1+1)^{2+1}} \right)\\ &=2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^4}\\ &=2\eta(4)\\ &=2(1-2^{1-4})\zeta(4)\\ &=2\,\frac78\zeta(4)\\ &=\frac74\zeta(4) \qquad \blacksquare \end{aligned}$

Evaluation of

$\begin{aligned} I_2&=\int_0^{1/2}\frac{\ln^2(x)\ln(1-x)}{1-x}\,dx =uv-\int v du \\&=-\frac{\ln^2(x)\ln(1-x)}{2}\Big|_0^{1/2}+\int_0^{1/2}\frac{\ln(x)\ln^2(1-x)}{x}\,dx\\ &=-\frac{\ln^4(2)}{2}+\int_{1/2}^{1}\frac{\ln^2(x)\ln(1-x)}{1-x}\,dx \qquad \left(x \to 1-x \right)\\ &=-\frac{\ln^4(2)}{4}+\frac12\int_{0}^{1}\frac{\ln^2(x)\ln(1-x)}{1-x}\,dx \qquad \left( \text{by adding } \int_0^{1/2} \text{to both sides} \right)\\ &=-\frac{\ln^4(2)}{4}-\frac14\zeta(4)\qquad \blacksquare\\ \end{aligned}$

Where we used the following result

$\begin{aligned} \int_0^1\frac{\ln^2(x)\ln(1-x)}{1-x}\,dx&=-\int_0^1\ln^2(x)\left(\sum_{n=1}^\infty \text{H}_{n-1}x^{n-1} \right) \,dx\\ &=-\sum_{n=1}^\infty \text{H}_{n-1} \int_0^1 x^{n-1} \ln^2(x)\,dx\\ &=-\sum_{n=1}^\infty \left(\text{H}_{n}-\frac1n \right)\left( \frac{2}{n^3}\right)\\ &=-2\left(\sum_{n=1}^\infty \frac{\text{H}_{n}}{n^3}-\sum_{n=1}^\infty\frac1{n^4} \right)\\ &=-2\left(\frac{\zeta^2(2)}{2}-\zeta(4) \right)\\ &=2\zeta(4)-\zeta^2(2) \\ &= 2\frac{\pi^4}{90}-\frac{\pi^4}{36}\\ &=-\frac{\pi^4}{180}\\ &=-\frac12\frac{}{}\zeta(4)\qquad \blacksquare\\ \end{aligned}$

Evaluation of

$\begin{aligned} I_3&=\int_0^{1/2}\frac{\ln(x)\ln^2(1-x)}{1-x}\,dx\\&=uv-\int v\,du\\ &=-\frac{\ln(x)\ln^3(1-x)}{3}\Big|_0^{1/2}+\frac13\int_0^{1/2}\frac{\ln^3(1-x)}{x}\,dx \qquad \left(u=\ln(x) \,\,dv=\frac{\ln^2(1-x)}{1-x}\,dx \right)\\ &=-\frac{\ln^4(2)}{3}+\frac13 \int_{1/2}^{1}\frac{\ln^3(x)}{1-x}\,dx \qquad \left( 1-x \to x\right)\\ &=-\frac{\ln^4(2)}{3}+\frac13 \int_{0}^{1}\frac{\ln^3(x)}{1-x}\,dx-\frac13 \int_{0}^{1/2}\frac{\ln^3(x)}{1-x}\,dx\\ &=-\frac{\ln^4(2)}{3}+\frac13 \left(-\frac{\pi^4}{15} \right)-\frac13 \left(-6\text{Li}_4\left(\frac12\right)-\frac{21\ln(2) \zeta(3)}{4}+\frac{\pi^2\ln^2(2)}{4}-\frac{\ln^4(2)}{2} \right)\\ &=2\text{Li}_4\left(\frac12\right)-\frac{\ln^2(2)\zeta(2)}{2}+\frac{7\ln(2) \zeta(3)}{4}-2\zeta(4)-\frac{\ln^4(2)}{6} \qquad \blacksquare \end{aligned}$

Where We used the previously proved results

$\begin{aligned} &\int_0^{1}\frac{\ln^3(x)}{1-x}\,dx=-\frac{\pi^4}{15}\\ & \\ &\int_0^{1/2}\frac{\ln^3(x)}{1-x}\,dx=-6\text{Li}_4\left(\frac12\right)-\frac{21\ln(2) \zeta(3)}{4}+\frac{\pi^2\ln^2(2)}{4}-\frac{\ln^4(2)}{2} \end{aligned}$

Evaluation of

$\begin{aligned} \int\frac{\ln^3(1-x)}{1-x}\,dx&=-\int\frac{\ln^3(w)}{w}\,dw \qquad \left( \text{by} \,\,1-x=w\right)\\ &=-\left(\ln^4(w)- \int\frac{\ln^3(w)}{w}\,dw\right)\\ &=-\frac{\ln^4(w)}{4}\\ &=-\frac{\ln^4(1-x)}{4}\\ & \\ I_4&=\int_0^{1/2}\frac{\ln^3(1-x)}{1-x}\,dx\\ &=-\frac{\ln^4(1-x)}{4}\Big|_0^{1/2}\\ &=-\frac{\ln^4(2)}{4} \qquad \blacksquare \end{aligned}$

Appendix

$\frac{(-1)^{k-1}}{\Gamma(k)}\int_0^1x^{n-1}\ln^{k-1}(x)\,dx=\frac{1}{n^k}$

Proof:

We start from

$\begin{aligned} \frac{1}{\Gamma(k)}\int_0^\infty e^{-nx}x^{k-1}\,dx&=\frac{1}{\Gamma(k)}\int_0^\infty e^{-x}\left(\frac{x}{n}\right)^{k-1}\,\frac{dx}{n}\\ &=\frac{1}{\Gamma(k) n^k}\int_0^\infty e^{-x}x^{k-1}\,dx\\ &=\frac{\Gamma(k)}{\Gamma(k) n^k}\\ &=\frac{1}{ n^k} \end{aligned}$

On the other hand

$\begin{aligned} \frac{1}{\Gamma(n)}\int_0^\infty e^{-nx}x^{k-1}\,dx&=\frac{1}{\Gamma(n)}\int_1^0 e^{n\ln(x)}\left(-\ln(x)\right)^{k-1}\,\frac{(-dx)}{x} \qquad \left(x \to -\ln(x) \right)\\ &=\frac{(-1)^{k-1}}{\Gamma(n)}\int_0^1x^{n-1}\ln^{k-1}(x)\,dx \end{aligned}$

Equating both equations we obtain the desired result

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