Menu Close

Question-157665

Tinku Tara 0 Comments
Pin




Question Number 157665 by mnjuly1970 last updated on 26/Oct/21
Answered by Raxreedoroid last updated on 26/Oct/21
I=∫_0 ^( 1) x^(n−1) ln (1−x) dw=x^(n−1) ,w=(x^n /n) t=ln (1−x),dt=((−1)/(1−x))dx I=[((x^n ln (1−x))/n)+(1/n)∫(x^n /(1−x))dx]_0 ^1 u=1−x,x=1−u,dx=−du I=[(((1−u)^n ln (u))/n)−(1/n)∫(((1−u)^n )/u)du]_1 ^0 I=[(((1−u)^n ln (u))/n)+(1/n)∫(1−u)^(n−1) +(1/u)du]_1 ^0 I=[(((1−u)^n ln (u))/n)−(1/n)((((1−u)^n )/n)+ln u)]_1 ^0 I=[(((1−u)^n nln (u)−(1−u)^n −nln u)/n^2 )]_1 ^0 I=lim_(u→0^+ ) (((1−u)^n nln (u)−(1−u)^n −nln u)/n^2 ) I=lim_(u→0^+ ) ((nln u((1−u)^n −1)−(1−u)^n )/n^2 ) I=lim_(u→0^+ ) (((ln u((1−u)^n −1))/n))−(1/n^2 ) I=−(1/n^2 ) ϕ(n)=−(1/n^2 ) S=−Σ_(n=1) ^∞ (((−1)^n )/n^3 )=Σ_(n=1) ^∞ (1/n^3 )−2Σ_(n=1) ^∞ (1/((2n)^3 )) S=ζ(3)−2Σ_(n=1) ^∞ (1/(8n^3 ))=ζ(3)−(1/4)ζ(3) S=(3/4)ζ(3)
$${I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}−\mathrm{1}} \mathrm{ln}\:\left(\mathrm{1}−{x}\right) \\ $$$${dw}={x}^{{n}−\mathrm{1}} ,{w}=\frac{{x}^{{n}} }{{n}} \\ $$$${t}=\mathrm{ln}\:\left(\mathrm{1}−{x}\right),{dt}=\frac{−\mathrm{1}}{\mathrm{1}−{x}}{dx} \\ $$$${I}=\left[\frac{{x}^{{n}} \mathrm{ln}\:\left(\mathrm{1}−{x}\right)}{{n}}+\frac{\mathrm{1}}{{n}}\int\frac{{x}^{{n}} }{\mathrm{1}−{x}}{dx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${u}=\mathrm{1}−{x},{x}=\mathrm{1}−{u},{dx}=−{du} \\ $$$${I}=\left[\frac{\left(\mathrm{1}−{u}\right)^{{n}} \mathrm{ln}\:\left({u}\right)}{{n}}−\frac{\mathrm{1}}{{n}}\int\frac{\left(\mathrm{1}−{u}\right)^{{n}} }{{u}}{du}\right]_{\mathrm{1}} ^{\mathrm{0}} \\ $$$${I}=\left[\frac{\left(\mathrm{1}−{u}\right)^{{n}} \mathrm{ln}\:\left({u}\right)}{{n}}+\frac{\mathrm{1}}{{n}}\int\left(\mathrm{1}−{u}\right)^{{n}−\mathrm{1}} +\frac{\mathrm{1}}{{u}}{du}\right]_{\mathrm{1}} ^{\mathrm{0}} \\ $$$${I}=\left[\frac{\left(\mathrm{1}−{u}\right)^{{n}} \mathrm{ln}\:\left({u}\right)}{{n}}−\frac{\mathrm{1}}{{n}}\left(\frac{\left(\mathrm{1}−{u}\right)^{{n}} }{{n}}+\mathrm{ln}\:{u}\right)\right]_{\mathrm{1}} ^{\mathrm{0}} \\ $$$${I}=\left[\frac{\left(\mathrm{1}−{u}\right)^{{n}} {n}\mathrm{ln}\:\left({u}\right)−\left(\mathrm{1}−{u}\right)^{{n}} −{n}\mathrm{ln}\:{u}}{{n}^{\mathrm{2}} }\right]_{\mathrm{1}} ^{\mathrm{0}} \\ $$$${I}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\left(\mathrm{1}−{u}\right)^{{n}} {n}\mathrm{ln}\:\left({u}\right)−\left(\mathrm{1}−{u}\right)^{{n}} −{n}\mathrm{ln}\:{u}}{{n}^{\mathrm{2}} } \\ $$$${I}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{{n}\mathrm{ln}\:{u}\left(\left(\mathrm{1}−{u}\right)^{{n}} −\mathrm{1}\right)−\left(\mathrm{1}−{u}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$${I}=\underset{{u}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\frac{\mathrm{ln}\:{u}\left(\left(\mathrm{1}−{u}\right)^{{n}} −\mathrm{1}\right)}{{n}}\right)−\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$${I}=−\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\varphi\left({n}\right)=−\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$${S}=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{3}} } \\ $$$${S}=\zeta\left(\mathrm{3}\right)−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{8}{n}^{\mathrm{3}} }=\zeta\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{3}\right) \\ $$$${S}=\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$
Answered by qaz last updated on 26/Oct/21
S=Σ_(n=1) ^∞ (((−1)^n )/n)∫_0 ^1 x^(n−1) ln(1−x)dx =−∫_0 ^1 ((ln(1−x)ln(1+x))/x)dx =(1/4)∫_0 ^1 ((ln^2 ((1−x)/(1+x)))/x)dx−(1/4)∫_0 ^1 ((ln^2 (1−x^2 ))/x)dx =(1/2)∫_0 ^1 ((ln^2 x)/(1−x^2 ))dx−(1/8)∫_0 ^1 ((ln^2 x)/( 1−x))dx................((1−x)/(1+x))→x =(1/2)Σ_(n=0) ^∞ ∫_0 ^1 x^(2n) ln^2 xdx−(1/8)Σ_(n=0) ^∞ ∫_0 ^1 x^n ln^2 xdx =(1/2)Σ_(n=0) ^∞ (((−1)^2 2!)/((2n+1)^3 ))−(1/8)Σ_(n=0) ^∞ (((−1)^2 2!)/((n+1)^3 )) =(1−2^(−3) )ζ(3)−(1/4)ζ(3) =(5/8)ζ(3)
$$\mathrm{S}=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}}{\mathrm{x}}\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \mathrm{x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \mathrm{x}}{\:\mathrm{1}−\mathrm{x}}\mathrm{dx}…………….\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}\rightarrow\mathrm{x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2n}} \mathrm{ln}^{\mathrm{2}} \mathrm{xdx}−\frac{\mathrm{1}}{\mathrm{8}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{n}} \mathrm{ln}^{\mathrm{2}} \mathrm{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}} \mathrm{2}!}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{8}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}} \mathrm{2}!}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\left(\mathrm{1}−\mathrm{2}^{−\mathrm{3}} \right)\zeta\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{3}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *