Question Number 206522 by MrGHK last updated on 17/Apr/24
Answered by Berbere last updated on 17/Apr/24
![u′=xln^2 (x)⇒u=(1/2)(x^2 ln^2 (x)−x^2 ln(x)+(x^2 /2)) v=Li_2 (((1+x)/x));v′=−(1/x^2 ).−((ln(−(1/x)))/((1+x)/x))=−((ln(−x))/(x(1+x))) IBP =(1/2)[(x^2 ln^2 (x)−(x^2 /2)ln(x)+(x^2 /2))Li_2 (((1+x)/x))]_0 ^1 +(1/2)∫_0 ^1 (x/(1+x))ln(−x)(xln^2 (x)−xln(x)+(x/2))dx “Li_2 (z)+Li_2 (1−(1/z))=−(((ln(z))^2 )/2) ” ⇒Li_2 (((1+z)/z))=Li_2 (−z)−(1/2)(ln(−z))^2 ⇒lim_(x→0) xLi_2 (((1+x)/x))=0 A=(1/4)Li_2 (2)+(1/2)∫_0 ^1 ((x^2 ln^2 (x)ln(−x))/(1+x))−(1/2)∫_0 ^1 ((x^2 ln(x)ln(−x))/(1+x))dx +(1/4)∫_0 ^1 (x^2 /(1+x))ln(−x)dx ln(−x)=ln(x)+iπ ∫_0 ^1 ((x^2 ln^n (x)ln(−x))/(1+x))dx =∫_0 ^1 ((x^2 ln^(n+1) (x))/(1+x))dx+iπ∫_0 ^1 ((x^2 ln^n (x)dx)/(1+x))= ∫_0 ^1 ((x^2 ln^m (x))/(1+x))=Σ_(n≥0) (−1)^n x^(n+2) ln^m (x)dx=H(m) =Σ_(n≥0) (−1)^n ∫_0 ^1 x^(n+2) ln^m (x)=Σ_(n≥0) (−1)^n ∫_0 ^∞ (−t)^m e^(−(n+2)t) dt =(−1)^m Σ_(n≥0) (((−1)^n Γ(m+1))/((n+2)^(m+1) ))=(−1)^m m!(Σ_(n≥0) (((−1)^(n+2) )/((n+2)^(m+1) ))) =(−1)^m m!(Σ_(n≥1) (((−1)^n )/n^(m+1) )+1) Σ_(n≥1) (((−1)^n )/n^(m+1) )=(1−(1/2^m ))ζ(m+1);∀m≥1 H=(−1)^m m!((1−(1/2^m ))ζ(m+1)+1) A=((Li_2 (2))/4)+(1/2)∫_0 ^1 ((x^2 ln^3 (x))/(1+x))+((iπ)/2)∫_0 ^1 ((x^2 ln^2 (x))/(1+x))dx−(1/2)∫_0 ^1 ((x^2 ln^2 (x))/(1+x))+(1/2)∫_0 ^1 x^2 ((iπln(x))/(1+x))dx +(1/4)∫_0 ^1 ((x^2 ln(x))/(1+x))dx+((iπ)/4)∫_0 ^1 (x^2 /(1+x)) dx =((Li_2 (2))/4)+(1/2)H(3)−((H(2))/2)+(1/4)H(1)+((iπ)/2)(H(2)+H(1)) +i(π/2)∫_0 ^1 (x^2 /(1+x))dx easy To continue](https://www.tinkutara.com/question/Q206529.png)
$${u}'={xln}^{\mathrm{2}} \left({x}\right)\Rightarrow{u}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} {ln}^{\mathrm{2}} \left({x}\right)−{x}^{\mathrm{2}} {ln}\left({x}\right)+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$${v}={Li}_{\mathrm{2}} \left(\frac{\mathrm{1}+{x}}{{x}}\right);{v}'=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }.−\frac{{ln}\left(−\frac{\mathrm{1}}{{x}}\right)}{\frac{\mathrm{1}+{x}}{{x}}}=−\frac{{ln}\left(−{x}\right)}{{x}\left(\mathrm{1}+{x}\right)} \\ $$$${IBP} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left({x}^{\mathrm{2}} {ln}^{\mathrm{2}} \left({x}\right)−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({x}\right)+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right){Li}_{\mathrm{2}} \left(\frac{\mathrm{1}+{x}}{{x}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}}{ln}\left(−{x}\right)\left({xln}^{\mathrm{2}} \left({x}\right)−{xln}\left({x}\right)+\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$“{Li}_{\mathrm{2}} \left({z}\right)+{Li}_{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{z}}\right)=−\frac{\left({ln}\left({z}\right)\right)^{\mathrm{2}} }{\mathrm{2}}\:'' \\ $$$$\Rightarrow{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}+{z}}{{z}}\right)={Li}_{\mathrm{2}} \left(−{z}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(−{z}\right)\right)^{\mathrm{2}} \Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{xLi}_{\mathrm{2}} \left(\frac{\mathrm{1}+{x}}{{x}}\right)=\mathrm{0} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{4}}{Li}_{\mathrm{2}} \left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}^{\mathrm{2}} \left({x}\right){ln}\left(−{x}\right)}{\mathrm{1}+{x}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}\left({x}\right){ln}\left(−{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$+\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}}{ln}\left(−{x}\right){dx} \\ $$$${ln}\left(−{x}\right)={ln}\left({x}\right)+{i}\pi \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}^{{n}} \left({x}\right){ln}\left(−{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}^{{n}+\mathrm{1}} \left({x}\right)}{\mathrm{1}+{x}}{dx}+{i}\pi\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}^{{n}} \left({x}\right){dx}}{\mathrm{1}+{x}}= \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}^{{m}} \left({x}\right)}{\mathrm{1}+{x}}=\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} {x}^{{n}+\mathrm{2}} {ln}^{{m}} \left({x}\right){dx}={H}\left({m}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{2}} {ln}^{{m}} \left({x}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} \left(−{t}\right)^{{m}} {e}^{−\left({n}+\mathrm{2}\right){t}} {dt} \\ $$$$=\left(−\mathrm{1}\right)^{{m}} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \Gamma\left({m}+\mathrm{1}\right)}{\left({n}+\mathrm{2}\right)^{{m}+\mathrm{1}} }=\left(−\mathrm{1}\right)^{{m}} {m}!\left(\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{2}} }{\left({n}+\mathrm{2}\right)^{{m}+\mathrm{1}} }\right) \\ $$$$=\left(−\mathrm{1}\right)^{{m}} {m}!\left(\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{m}+\mathrm{1}} }+\mathrm{1}\right) \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{m}+\mathrm{1}} }=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{m}} }\right)\zeta\left({m}+\mathrm{1}\right);\forall{m}\geqslant\mathrm{1} \\ $$$${H}=\left(−\mathrm{1}\right)^{{m}} {m}!\left(\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{m}} }\right)\zeta\left({m}+\mathrm{1}\right)+\mathrm{1}\right) \\ $$$${A}=\frac{{Li}_{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}^{\mathrm{3}} \left({x}\right)}{\mathrm{1}+{x}}+\frac{{i}\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \frac{{i}\pi{ln}\left({x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$+\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {ln}\left({x}\right)}{\mathrm{1}+{x}}{dx}+\frac{{i}\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}}\:{dx}\: \\ $$$$=\frac{{Li}_{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}{H}\left(\mathrm{3}\right)−\frac{{H}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}{H}\left(\mathrm{1}\right)+\frac{{i}\pi}{\mathrm{2}}\left({H}\left(\mathrm{2}\right)+{H}\left(\mathrm{1}\right)\right) \\ $$$$+{i}\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}}{dx}\:{easy}\:{To}\:{continue} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by MrGHK last updated on 18/Apr/24
$$\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{Nice}}\:\boldsymbol{\mathrm{solution}} \\ $$
Commented by Berbere last updated on 18/Apr/24
$${withe}\:{Pleasur} \\ $$