Menu Close

find-0-1-ln-2-x-1-x-2-2-dx-




Question Number 60586 by Mr X pcx last updated on 22/May/19
find ∫_0 ^1   ((ln^2 (x))/((1−x^2 )^2 ))dx
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 23/May/19
let  A =∫_0 ^1   ((ln^2 x)/((1−x^2 )^2 ))dx   we have for ∣x∣<1   Σ_(n=0) ^∞  x^n  =(1/(1−x))  and  Σ_(n=1) ^∞  nx^(n−1)  = (1/((1−x)^2 )) ⇒Σ_(n=1) ^∞  n x^(2n−2)  =(1/((1−x^2 )^2 )) ⇒  A =∫_0 ^1 (Σ_(n=1) ^∞  nx^(n−2) )ln^2 x dx =Σ_(n=1) ^∞ n ∫_0 ^1  x^(n−2) ln^2 x dx =Σ_(n=1) ^∞ nw_n   w_n =∫_0 ^1  x^(n−2) ln^2 x dx    by parts u^′  =x^(n−2)    and v =ln^2 x ⇒  w_n =[(1/(n−1))x^(n−1) ln^2 x]_0 ^1  −∫_0 ^1 (1/(n−1))x^(n−1)  ((2lnx)/x)dx  =−(2/(n−1)) ∫_0 ^1   x^(n−2)  ln(x)dx    by parts again  u^′  =x^(n−2)  and v =lnx ⇒  ∫_0 ^1  x^(n−2) ln(x)dx =[(1/(n−1))x^(n−1)  lnx]_0 ^1  −∫_0 ^1  (1/(n−1)) x^(n−1)  (dx/x)  =−(1/(n−1)) ∫_0 ^1 x^(n−2) dx =−(1/(n−1))[(1/(n−1)) x^(n−1) ]_0 ^1  =−(1/((n−1)^2 )) ⇒w_n =(2/((n−1)^3 ))  A =∫_0 ^1 (1+Σ_(n=2) ^∞  nx^(n−2) )ln^2 x dx =∫_0 ^1  ln^2 x dx +Σ_(n=2) ^∞  n ∫_0 ^1 x^(n−2) ln^2 xdx  =∫_0 ^1 ln^2 x dx +Σ_(n=2) ^∞ n(2/((n−1)^3 )) =∫_0 ^1  ln^2 x dx  + 2 Σ_(n=2) ^∞  (n/((n−1)^3 ))  Σ_(n=2) ^∞  (n/((n−1)^3 )) =Σ_(n=1) ^∞  ((n+1)/n^3 ) =Σ_(n=1) ^∞  (1/n^2 ) +Σ_(n=1) ^∞  (1/n^3 ) =(π^2 /6) +ξ(3)  ∫_0 ^1  ln^2 x dx =_(ln(x)=−t)        ∫_(+∞) ^0  t^2  (−e^(−t) )dt =∫_0 ^∞  t^2 e^(−t)  dt  =[−t^2 e^(−t) ]_0 ^(+∞)  −∫_0 ^(+∞)  2t (−e^(−t) )dt = 2 ∫_0 ^∞   t e^(−t)  dt  =2{ [−t e^(−t) ]_0 ^∞ −∫_0 ^∞   (−e^(−t) )dt} =2 ∫_0 ^∞  e^(−t)  dt =2[−e^(−t) ]_0 ^(+∞)  =2 ⇒   A = 2 +(π^2 /3) +2ξ(3)
$${let}\:\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}^{\mathrm{2}} {x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\:\:{we}\:{have}\:{for}\:\mid{x}\mid<\mathrm{1}\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:{and} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\:{x}^{\mathrm{2}{n}−\mathrm{2}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{2}} \right){ln}^{\mathrm{2}} {x}\:{dx}\:=\sum_{{n}=\mathrm{1}} ^{\infty} {n}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{2}} {ln}^{\mathrm{2}} {x}\:{dx}\:=\sum_{{n}=\mathrm{1}} ^{\infty} {nw}_{{n}} \\ $$$${w}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{2}} {ln}^{\mathrm{2}} {x}\:{dx}\:\:\:\:{by}\:{parts}\:{u}^{'} \:={x}^{{n}−\mathrm{2}} \:\:\:{and}\:{v}\:={ln}^{\mathrm{2}} {x}\:\Rightarrow \\ $$$${w}_{{n}} =\left[\frac{\mathrm{1}}{{n}−\mathrm{1}}{x}^{{n}−\mathrm{1}} {ln}^{\mathrm{2}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{n}−\mathrm{1}}{x}^{{n}−\mathrm{1}} \:\frac{\mathrm{2}{lnx}}{{x}}{dx} \\ $$$$=−\frac{\mathrm{2}}{{n}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{{n}−\mathrm{2}} \:{ln}\left({x}\right){dx}\:\:\:\:{by}\:{parts}\:{again}\:\:{u}^{'} \:={x}^{{n}−\mathrm{2}} \:{and}\:{v}\:={lnx}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{2}} {ln}\left({x}\right){dx}\:=\left[\frac{\mathrm{1}}{{n}−\mathrm{1}}{x}^{{n}−\mathrm{1}} \:{lnx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{n}−\mathrm{1}}\:{x}^{{n}−\mathrm{1}} \:\frac{{dx}}{{x}} \\ $$$$=−\frac{\mathrm{1}}{{n}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{2}} {dx}\:=−\frac{\mathrm{1}}{{n}−\mathrm{1}}\left[\frac{\mathrm{1}}{{n}−\mathrm{1}}\:{x}^{{n}−\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=−\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{w}_{{n}} =\frac{\mathrm{2}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\sum_{{n}=\mathrm{2}} ^{\infty} \:{nx}^{{n}−\mathrm{2}} \right){ln}^{\mathrm{2}} {x}\:{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}^{\mathrm{2}} {x}\:{dx}\:+\sum_{{n}=\mathrm{2}} ^{\infty} \:{n}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{2}} {ln}^{\mathrm{2}} {xdx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}} {x}\:{dx}\:+\sum_{{n}=\mathrm{2}} ^{\infty} {n}\frac{\mathrm{2}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}^{\mathrm{2}} {x}\:{dx}\:\:+\:\mathrm{2}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}+\mathrm{1}}{{n}^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\xi\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}^{\mathrm{2}} {x}\:{dx}\:=_{{ln}\left({x}\right)=−{t}} \:\:\:\:\:\:\:\int_{+\infty} ^{\mathrm{0}} \:{t}^{\mathrm{2}} \:\left(−{e}^{−{t}} \right){dt}\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{\mathrm{2}} {e}^{−{t}} \:{dt} \\ $$$$=\left[−{t}^{\mathrm{2}} {e}^{−{t}} \right]_{\mathrm{0}} ^{+\infty} \:−\int_{\mathrm{0}} ^{+\infty} \:\mathrm{2}{t}\:\left(−{e}^{−{t}} \right){dt}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:{t}\:{e}^{−{t}} \:{dt} \\ $$$$=\mathrm{2}\left\{\:\left[−{t}\:{e}^{−{t}} \right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \:\:\left(−{e}^{−{t}} \right){dt}\right\}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \:{dt}\:=\mathrm{2}\left[−{e}^{−{t}} \right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{2}\:\Rightarrow\: \\ $$$${A}\:=\:\mathrm{2}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:+\mathrm{2}\xi\left(\mathrm{3}\right)\: \\ $$

Leave a Reply

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