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nice-calculus-evaluation-0-1-log-1-x-log-1-x-m-n-july-197o-




Question Number 115055 by mnjuly1970 last updated on 23/Sep/20
           ...  nice  calculus...          evaluation :                            χ=∫_0 ^( 1) log(1−x).log(1+x) =???                             ...m.n.july.197o...
$$\:\:\:\:\:\:\:\:\:\:\:…\:\:{nice}\:\:{calculus}… \\ $$$$\:\:\:\:\:\:\:\:{evaluation}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\chi=\int_{\mathrm{0}} ^{\:\mathrm{1}} {log}\left(\mathrm{1}−{x}\right).{log}\left(\mathrm{1}+{x}\right)\:=??? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{m}.{n}.{july}.\mathrm{197}{o}… \\ $$$$\: \\ $$
Answered by mathmax by abdo last updated on 23/Sep/20
I =∫_0 ^1  ln(1−x).ln(1+x)dx   we have (d/dx)ln(1−x) =((−1)/(1−x))  =−Σ_(n=0) ^∞  x^n  ⇒ln(1−x) =−Σ_(n=0) ^∞  (x^(n+1) /(n+1)) =−Σ_(n=1) ^∞  (x^n /n)  also ln(1+x) =−Σ_(n=1) ^∞ (((−1)^n  x^n )/n) ⇒  ln(1−x).ln(1+x) =(Σ_(n=1) ^∞  (x^n /n)).(Σ_(n=1) ^∞  (((−1)^n )/n) x^n )  =Σ_(n=1) ^∞  c_n x^n  with c_n =Σ_(i+j=n)      a_i b_j     =Σ_(i+j =n)    (1/i)×(((−1)^j )/j)  =Σ_(i=1) ^(n−1)    (1/i)×(((−1)^(n−i) )/(n−i)) ⇒ln(1−x).ln(1+x) =Σ_(n=1) ^∞ (Σ_(i=1) ^(n−1)  (((−1)^(n−i) )/(i(n−i))))x^n   ⇒∫_0 ^1 ln(1−x)ln(1+x)dx =Σ_(n=1) ^∞ (Σ_(i=1) ^n  (((−1)^(n−i) )/(i(n−i))))×(1/(n+1))  =Σ_(n=1) ^∞    (((−1)^n )/(n+1))(Σ_(i=1) ^(n−1)  (((−1)^i )/(i(n−i))))  but  Σ_(i=1) ^(n−1)  (((−1)^i )/(i(n−i))) =(1/n)Σ_(i=1) ^(n−1)  (−1)^i ((1/i)−(1/(n−i))) =(1/n)Σ_(i=1) ^(n−1) (((−1)^i )/i)  −(1/n)Σ_(i=1) ^(n−1)  (((−1)^i )/(n−i))(n−i=p) =(1/n)Σ_(i=1) ^(n−1)  (((−1)^i )/i)−(1/n)Σ_1 ^(n−1)  (((−1)^(n−p) )/p)  =(1/n)Σ_(i=1) ^(n−1)  (((−1)^i )/i)−(((−1)^n )/n) Σ_(i=1) ^(n−1)  (((−1)^i )/i)  =(1/n)(1−(−1)^n )Σ_(i=1) ^(n−1)  (((−1)^i )/i)  =(1/n) Σ_(i=1) ^(2n)  (((−1)^i )/i)  ....be continued....
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right).\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}\:\:\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\:=\frac{−\mathrm{1}}{\mathrm{1}−\mathrm{x}} \\ $$$$=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{n}} \:\Rightarrow\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\:=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\:=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}} \\ $$$$\mathrm{also}\:\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}\:\Rightarrow \\ $$$$\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right).\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:=\left(\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}\right).\left(\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\:\mathrm{x}^{\mathrm{n}} \right) \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{c}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \:\mathrm{with}\:\mathrm{c}_{\mathrm{n}} =\sum_{\mathrm{i}+\mathrm{j}=\mathrm{n}} \:\:\:\:\:\mathrm{a}_{\mathrm{i}} \mathrm{b}_{\mathrm{j}} \:\:\:\:=\sum_{\mathrm{i}+\mathrm{j}\:=\mathrm{n}} \:\:\:\frac{\mathrm{1}}{\mathrm{i}}×\frac{\left(−\mathrm{1}\right)^{\mathrm{j}} }{\mathrm{j}} \\ $$$$=\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\mathrm{i}}×\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{i}} }{\mathrm{n}−\mathrm{i}}\:\Rightarrow\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right).\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{i}} }{\mathrm{i}\left(\mathrm{n}−\mathrm{i}\right)}\right)\mathrm{x}^{\mathrm{n}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{i}} }{\mathrm{i}\left(\mathrm{n}−\mathrm{i}\right)}\right)×\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}+\mathrm{1}}\left(\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{i}} }{\mathrm{i}\left(\mathrm{n}−\mathrm{i}\right)}\right)\:\:\mathrm{but} \\ $$$$\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{i}} }{\mathrm{i}\left(\mathrm{n}−\mathrm{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\left(−\mathrm{1}\right)^{\mathrm{i}} \left(\frac{\mathrm{1}}{\mathrm{i}}−\frac{\mathrm{1}}{\mathrm{n}−\mathrm{i}}\right)\:=\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{\mathrm{i}} }{\mathrm{i}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{i}} }{\mathrm{n}−\mathrm{i}}\left(\mathrm{n}−\mathrm{i}=\mathrm{p}\right)\:=\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{i}} }{\mathrm{i}}−\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{p}} }{\mathrm{p}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{i}} }{\mathrm{i}}−\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\:\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{i}} }{\mathrm{i}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{n}}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{n}} \right)\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{i}} }{\mathrm{i}}\:\:=\frac{\mathrm{1}}{\mathrm{n}}\:\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{2n}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{i}} }{\mathrm{i}}\:\:….\mathrm{be}\:\mathrm{continued}…. \\ $$
Commented by mnjuly1970 last updated on 23/Sep/20
thank you so much sir  max ...
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$$${max}\:… \\ $$
Commented by mathmax by abdo last updated on 23/Sep/20
you are welcome sir
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$
Answered by mathdave last updated on 23/Sep/20
solution  if let I_1 =∫_0 ^1 ln(1−x)ln(1+x)dx.....(1)  put x=−x  I_2 =∫_(−1) ^0 ln(1+x)ln(1−x)dx........(2) since the integrand are unchanged then we add them up  which mean that I_1 =I_2   I=I_1 +I_2  =2I_1 , I_1 =(1/2)I  I_1 =∫_0 ^1 ln(1−x)ln(1+x)dx=(1/2)∫_0 ^1 ln(1−x)ln(1+x)dx  put x=2y−1,dx=2dy  I_1 =(1/2)∫_0 ^1 ln(2−2y)ln(2y)2dy=∫_0 ^1 [(ln2+ln(1−y))(ln2+lny)]dy  I=∫_0 ^1 (ln^2 2+ln2(ln(1−y)+lny)+lnyln(1−y))dx  I=ln^2 (2)∫_0 ^1 dy+ln2∫_0 ^1 (lny+ln(1−y))dy+∫_0 ^1 lnyln(1−y)dy  let A=ln^2 (2)∫_0 ^1 dy=ln^2 (2)......(1)  let  B=ln2∫_0 ^1 (lny+ln(1−y))dy=2∫_0 ^1 lnydy  B=2ln2(ylny−y)_0 ^1 =−2ln2.....(2)  let  C=∫_0 ^1 lnyln(1−y)dy=−Σ_(n=1) ^∞ (1/n)∫_0 ^1 y^n lnydy  C=−Σ_(n=1) ^∞ (1/n)(∂/∂a)∣_(a=1) ∫_0 ^1 y^n .y^(a−1) dy=−Σ_(n=1) ^∞ (1/n)(∂/∂a)∣_(a=1) ((1/(n+a)))  C=Σ_(n=1) ^∞ (1/(n(n+1)^2 ))=Σ_(n=1) ^∞ ((1/n)−(1/(n+1))−(1/((n+1)^2 )))  C=Σ_(n=1) ^∞ (1/n)−Σ_(n=2) ^∞ (1/n)−Σ_(n=2) ^∞ (1/n^2 )=Σ_(n=1) ^∞ ((1/n))−Σ_(n=1) ^∞ ((1/n)−1)−Σ_(n=1) ^∞ ((π^2 /6)−1)  C=1−(π^2 /6)+1=(2−(π^2 /6)).........(3)  but  I=A+B+C=ln^2 (2)−2ln2+2−(π^2 /6)=2−(π^2 /6)+ln^2 (2)−ln4  ∵∫_0 ^1 ln(1−x)ln(1+x)dx=2−(π^2 /6)+ln^2 (2)−ln4  by mathdave(23/09/2020)
$${solution} \\ $$$${if}\:{let}\:{I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−{x}\right)\mathrm{ln}\left(\mathrm{1}+{x}\right){dx}…..\left(\mathrm{1}\right) \\ $$$${put}\:{x}=−{x} \\ $$$${I}_{\mathrm{2}} =\int_{−\mathrm{1}} ^{\mathrm{0}} \mathrm{ln}\left(\mathrm{1}+{x}\right)\mathrm{ln}\left(\mathrm{1}−{x}\right){dx}……..\left(\mathrm{2}\right)\:{since}\:{the}\:{integrand}\:{are}\:{unchanged}\:{then}\:{we}\:{add}\:{them}\:{up} \\ $$$${which}\:{mean}\:{that}\:{I}_{\mathrm{1}} ={I}_{\mathrm{2}} \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \:=\mathrm{2}{I}_{\mathrm{1}} ,\:{I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{I} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−{x}\right)\mathrm{ln}\left(\mathrm{1}+{x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−{x}\right)\mathrm{ln}\left(\mathrm{1}+{x}\right){dx} \\ $$$${put}\:{x}=\mathrm{2}{y}−\mathrm{1},{dx}=\mathrm{2}{dy} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{2}−\mathrm{2}{y}\right)\mathrm{ln}\left(\mathrm{2}{y}\right)\mathrm{2}{dy}=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\left(\mathrm{ln2}+\mathrm{ln}\left(\mathrm{1}−{y}\right)\right)\left(\mathrm{ln2}+\mathrm{ln}{y}\right)\right]{dy} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{ln}^{\mathrm{2}} \mathrm{2}+\mathrm{ln2}\left(\mathrm{ln}\left(\mathrm{1}−{y}\right)+\mathrm{ln}{y}\right)+\mathrm{ln}{y}\mathrm{ln}\left(\mathrm{1}−{y}\right)\right){dx} \\ $$$${I}=\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {dy}+\mathrm{ln2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{ln}{y}+\mathrm{ln}\left(\mathrm{1}−{y}\right)\right){dy}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}{y}\mathrm{ln}\left(\mathrm{1}−{y}\right){dy} \\ $$$${let}\:{A}=\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {dy}=\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)……\left(\mathrm{1}\right) \\ $$$${let} \\ $$$${B}=\mathrm{ln2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{ln}{y}+\mathrm{ln}\left(\mathrm{1}−{y}\right)\right){dy}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}{ydy} \\ $$$${B}=\mathrm{2ln2}\left({y}\mathrm{ln}{y}−{y}\right)_{\mathrm{0}} ^{\mathrm{1}} =−\mathrm{2ln2}…..\left(\mathrm{2}\right) \\ $$$${let} \\ $$$${C}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}{y}\mathrm{ln}\left(\mathrm{1}−{y}\right){dy}=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{{n}} \mathrm{ln}{ydy} \\ $$$${C}=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {y}^{{n}} .{y}^{{a}−\mathrm{1}} {dy}=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{1}} \left(\frac{\mathrm{1}}{{n}+{a}}\right) \\ $$$${C}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$${C}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\mathrm{1}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\right) \\ $$$${C}=\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\mathrm{1}=\left(\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)………\left(\mathrm{3}\right) \\ $$$${but} \\ $$$${I}={A}+{B}+{C}=\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{2ln2}+\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{ln4} \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−{x}\right)\mathrm{ln}\left(\mathrm{1}+{x}\right){dx}=\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{ln4} \\ $$$${by}\:{mathdave}\left(\mathrm{23}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 23/Sep/20
thank you mr dave   nice solution that was i  wanted...
$${thank}\:{you}\:{mr}\:{dave}\: \\ $$$${nice}\:{solution}\:{that}\:{was}\:{i} \\ $$$${wanted}… \\ $$
Commented by mnjuly1970 last updated on 23/Sep/20
final answer:  ln^2 (2)−ln(4)+2−(π^2 /6)  thank you so much mr  dave.
$${final}\:{answer}: \\ $$$${ln}^{\mathrm{2}} \left(\mathrm{2}\right)−{ln}\left(\mathrm{4}\right)+\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${thank}\:{you}\:{so}\:{much}\:{mr} \\ $$$${dave}. \\ $$
Commented by mnjuly1970 last updated on 23/Sep/20
you are very gentlman  mr dave.thank you for   your effort.
$${you}\:{are}\:{very}\:{gentlman} \\ $$$${mr}\:{dave}.{thank}\:{you}\:{for}\: \\ $$$${your}\:{effort}. \\ $$
Commented by Tawa11 last updated on 06/Sep/21
great
$$\mathrm{great} \\ $$

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