Question Number 189091 by mnjuly1970 last updated on 12/Mar/23
$$ \\ $$$$\:\:\:\:\mathrm{1}\::\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\:\mathrm{1}\:\right)^{\:{n}} \mathrm{H}_{\:{n}} }{{n}^{\:\mathrm{2}} }\:=\:? \\ $$$$\:\:\:\:\mathrm{2}\::\:\:\:\:\:\eta\:\left(−\mathrm{1}\:\right)=\:? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by qaz last updated on 15/Mar/23
![A=Σ(−1)^n H_n ∫_0 ^∞ xe^(−nx) dx =−∫_0 ^∞ ((xln(1+e^(−x) ))/(1+e^(−x) ))dx=∫_0 ^1 ((lntln(1+t))/(t(1+t)))dt =∫_0 ^1 (((lntln(1+t))/t)−((lntln(1+t))/(1+t)))dt =∫_0 ^1 lnt(Li_2 (−t))′dt−∫_0 ^1 (((ln(t/(1+t))+ln(1+t))ln(1+t))/(1+t))dt =−∫_0 ^1 ((Li_2 (−t))/t)dt−(1/3)ln^3 2−∫_0 ^1 ((ln(1−(1/(1+t)))ln(1+t))/(1+t))dt =Li_3 (−1)−(1/3)ln^3 2−∫_0 ^1 (Li_2 ((1/(1+t))))′ln(1+t)dt =Li_3 (−1)−(1/3)ln^3 2−Li_2 ((1/2))ln2+∫_0 ^1 ((Li_2 ((1/(1+t))))/(1+t))dt =Li_3 (−1)−(1/3)ln^3 2−Li_2 ((1/2))ln2−∫_0 ^1 (Li_3 ((1/(1+t))))′dt =Li_3 (−1)−(1/3)ln^3 2−Li_2 ((1/2))ln2−Li_3 ((1/2))+Li_3 (1) =Li_3 (−1)−(1/3)ln^3 2−((π^2 /(12))−(1/2)ln^2 2)ln2−[(7/8)Li_3 (1)−(π^2 /(12))ln2+(1/6)ln^3 2]+Li_3 (1) =Li_3 (−1)+(1/8)Li_3 (1)=(2^(1−3) −1)ζ(3)+(1/8)ζ(3)=−(5/8)ζ(3) ...](https://www.tinkutara.com/question/Q189349.png)
$${A}=\Sigma\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} \int_{\mathrm{0}} ^{\infty} {xe}^{−{nx}} {dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \frac{{xln}\left(\mathrm{1}+{e}^{−{x}} \right)}{\mathrm{1}+{e}^{−{x}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lntln}\left(\mathrm{1}+{t}\right)}{{t}\left(\mathrm{1}+{t}\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{lntln}\left(\mathrm{1}+{t}\right)}{{t}}−\frac{{lntln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {lnt}\left({Li}_{\mathrm{2}} \left(−{t}\right)\right)'{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({ln}\frac{{t}}{\mathrm{1}+{t}}+{ln}\left(\mathrm{1}+{t}\right)\right){ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}{dt} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left(−{t}\right)}{{t}}{dt}−\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}{dt} \\ $$$$={Li}_{\mathrm{3}} \left(−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}−\int_{\mathrm{0}} ^{\mathrm{1}} \left({Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}\right)\right)'{ln}\left(\mathrm{1}+{t}\right){dt} \\ $$$$={Li}_{\mathrm{3}} \left(−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}−{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right){ln}\mathrm{2}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}\right)}{\mathrm{1}+{t}}{dt} \\ $$$$={Li}_{\mathrm{3}} \left(−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}−{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right){ln}\mathrm{2}−\int_{\mathrm{0}} ^{\mathrm{1}} \left({Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}\right)\right)'{dt} \\ $$$$={Li}_{\mathrm{3}} \left(−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}−{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right){ln}\mathrm{2}−{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{Li}_{\mathrm{3}} \left(\mathrm{1}\right) \\ $$$$={Li}_{\mathrm{3}} \left(−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}−\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \mathrm{2}\right){ln}\mathrm{2}−\left[\frac{\mathrm{7}}{\mathrm{8}}{Li}_{\mathrm{3}} \left(\mathrm{1}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}{ln}\mathrm{2}+\frac{\mathrm{1}}{\mathrm{6}}{ln}^{\mathrm{3}} \mathrm{2}\right]+{Li}_{\mathrm{3}} \left(\mathrm{1}\right) \\ $$$$={Li}_{\mathrm{3}} \left(−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{8}}{Li}_{\mathrm{3}} \left(\mathrm{1}\right)=\left(\mathrm{2}^{\mathrm{1}−\mathrm{3}} −\mathrm{1}\right)\zeta\left(\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)=−\frac{\mathrm{5}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$$$… \\ $$
Commented by senestro last updated on 02/Apr/23
