Question Number 111965 by mnjuly1970 last updated on 05/Sep/20
$$\:\:\:\:\:\:....{advanced}\:\:{calculus}... \\ $$$${evaluate}: \\ $$$$ \\ $$$$\:\:\:\:{i}::\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {xH}_{{x}} {dx}\:=???\:\: \\ $$$$\:\:\:\:\:{ii}::\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}^{\mathrm{2}} \mathrm{2}^{{n}\:} }\:=??? \\ $$$$\:\:\:\:\:{iii}::\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}=??? \\ $$$$\:\:\:\:\:\:\:\:\:{m}.{n}.\:{july}\:\mathrm{1970}...# \\ $$
Answered by mindispower last updated on 05/Sep/20
![∫_0 ^1 ((ln(x+1))/((x+1)(x+2)))dx ∫_0 ^1 ((ln(x+1))/(x+1))dx_(=A) −∫_0 ^1 ((ln(x+1))/(x+2))dx_(=B) A=(1/2)[ln^2 (x+1)]_0 ^1 =((ln^2 (2))/2) B=−[ln(x+1)ln(x+2)]_0 ^1 +∫_0 ^1 ((ln(x+2))/(x+1))dx By part and x+1=u⇒ B=−ln(2)ln(3)+∫_1 ^2 ((ln(u+1))/u)du u=−y⇒B=−ln(2)ln(3)+∫_(−1) ^(−2) ((ln(1−y))/y)dy B=−ln(2)ln(3)−[−∫_(−1) ^(−2) ((ln(1−y))/y)dy] =−ln(2)ln(3)−Li_2 (−2)+Li_2 (−1) Li_2 (z)=Σ(z^k /k^2 ) Li_2 (−1)=Σ_(k≥1) (1/(4k^2 ))−Σ_(k≥0) (1/((2k+1)^2 ))=((ζ(2))/4)−{ζ(2)−((ζ(2))/4)} =−(1/2)=−(1/2)ζ(2)=−(π^2 /(12)) ∫_0 ^1 ((ln(x+1))/((x+1)(x+2)))dx=A+B=((ln^2 (2))/2)−ln(2)ln(3)−Li_2 (−2)−(π^2 /(12)) =−Li_2 (−2)−(1/2)ln(2)ln((2/9))−(π^2 /(12))](Q112049.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}}{dx}_{={A}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{2}}{d}\underset{={B}} {{x}} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}^{\mathrm{2}} \left({x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$${B}=−\left[{ln}\left({x}+\mathrm{1}\right){ln}\left({x}+\mathrm{2}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\mathrm{2}\right)}{{x}+\mathrm{1}}{dx} \\ $$$${By}\:{part}\:{and}\:{x}+\mathrm{1}={u}\Rightarrow \\ $$$${B}=−{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{3}\right)+\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{ln}\left({u}+\mathrm{1}\right)}{{u}}{du} \\ $$$${u}=−{y}\Rightarrow{B}=−{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{3}\right)+\int_{−\mathrm{1}} ^{−\mathrm{2}} \frac{{ln}\left(\mathrm{1}−{y}\right)}{{y}}{dy} \\ $$$${B}=−{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{3}\right)−\left[−\int_{−\mathrm{1}} ^{−\mathrm{2}} \frac{{ln}\left(\mathrm{1}−{y}\right)}{{y}}{dy}\right] \\ $$$$=−{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{3}\right)−{Li}_{\mathrm{2}} \left(−\mathrm{2}\right)+{Li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$$${Li}_{\mathrm{2}} \left({z}\right)=\Sigma\frac{{z}^{{k}} }{{k}^{\mathrm{2}} } \\ $$$${Li}_{\mathrm{2}} \left(−\mathrm{1}\right)=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{2}} }−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{4}}−\left\{\zeta\left(\mathrm{2}\right)−\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{4}}\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}}\zeta\left(\mathrm{2}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx}={A}+{B}=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}−{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{3}\right)−{Li}_{\mathrm{2}} \left(−\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$=−{Li}_{\mathrm{2}} \left(−\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right){ln}\left(\frac{\mathrm{2}}{\mathrm{9}}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mathdave last updated on 06/Sep/20
$${mistake}\:{from}\:{the}\:{first}\:{step} \\ $$
Commented by mathdave last updated on 06/Sep/20
$${it}\:{supposed}\:{to}\:{be}\:{something}\:{like}\:{this} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{\left(\mathrm{1}+{x}\right)\left(\mathrm{2}+{x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{x}\right)}{dx}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{2}+{x}\right)}{dx} \\ $$
Commented by mindispower last updated on 06/Sep/20
$${i}\:{have}\:{starte}\:{withe}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\mathrm{1}\right)}{\left(\mathrm{1}+{x}\right)\left(\mathrm{2}+{x}\right)}{dx} \\ $$$${just}\:{because}\:{i}\:{dont}\:{want}\:{write}\:{evrey}\:{steps}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$
Commented by mathdave last updated on 06/Sep/20
$${aside}\:{that}\:{u}\:{dont}\:{wamna}\:{used} \\ $$$${everything}\:{when}\:{getting}\:{ur}\:{final} \\ $$$${answer}\:\frac{\mathrm{1}}{\mathrm{3}}\:{wasnt}\:{reflecting}\:{there} \\ $$
Commented by mnjuly1970 last updated on 06/Sep/20
$${thank}\:{you}\:{so}\:{much}\:\:{for}\:{you} \\ $$$${carefulness}... \\ $$
Commented by mnjuly1970 last updated on 06/Sep/20
$${thank}\:{you}\:{so}\:{much}\:{for}\:{your}\: \\ $$$${effort}\:.{grateful}.... \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{grest}\:\mathrm{sir} \\ $$