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Question Number 60595 by maxmathsup by imad last updated on 22/May/19
let f(a) =∫_0 ^1   ((ln^2 (x))/((1−ax)^2 )) dx  with ∣a∣<1  1)  find a explicit form of f(a)  2) determine A(θ) =∫_0 ^1   ((ln^2 (x))/((1−(cosθ)x)^2 ))dx  with 0<θ<(π/2)
letf(a)=01ln2(x)(1ax)2dxwitha∣<11)findaexplicitformoff(a)2)determineA(θ)=01ln2(x)(1(cosθ)x)2dxwith0<θ<π2
Commented by maxmathsup by imad last updated on 23/May/19
1) we have for ∣x∣<1       Σ_(n=0) ^∞  x^n  =(1/(1−x))  and  Σ_(n=1) ^∞ nx^(n−1)  =(1/((1−x)^2 )) ⇒  (1/((1−ax)^2 )) =Σ_(n=1) ^∞  n(ax)^(n−1)  =Σ_(n=1) ^∞ n a^(n−1) x^(n−1)  ⇒  f(a) =∫_0 ^1 (Σ_(n=1) ^∞ na^(n−1) x^(n−1) )ln^2 (x)dx =Σ_(n=1) ^∞ na^(n−1) ∫_0 ^1  x^(n−1) ln^2 (x)dx  =Σ_(n=1) ^∞  na^(n−1) w_n      with  w_n =∫_0 ^1  x^(n−1) ln^2 (x)dx  by parts u^′ =x^(n−1)  and v=ln^2 x  w_n =[(1/n)x^n ln^2 x]_0 ^1  −∫_0 ^1 (1/n)x^n  ((2lnx)/x)dx =−(2/n) ∫_0 ^1  x^(n−1) ln(x)    =_(byparts)     −(2/n){  [(1/n)x^n lnx]_0 ^1  −∫_0 ^1 (1/n)x^n  (dx/x)}=−(2/n){−(1/n) ∫_0 ^1  x^(n−1) dx}  =(2/n^3 ) ⇒f(a) =Σ_(n=1) ^∞  ((2na^(n−1) )/n^3 ) =2 Σ_(n=1) ^∞   (a^(n−1) /n^2 ) ⇒af(a) =2Σ_(n=1) ^∞  (a^n /n^2 )  let try to find  s(x) =Σ_(n=1) ^∞  (x^n /n^2 )   if ∣x∣<1   ....
1)wehaveforx∣<1n=0xn=11xandn=1nxn1=1(1x)21(1ax)2=n=1n(ax)n1=n=1nan1xn1f(a)=01(n=1nan1xn1)ln2(x)dx=n=1nan101xn1ln2(x)dx=n=1nan1wnwithwn=01xn1ln2(x)dxbypartsu=xn1andv=ln2xwn=[1nxnln2x]01011nxn2lnxxdx=2n01xn1ln(x)=byparts2n{[1nxnlnx]01011nxndxx}=2n{1n01xn1dx}=2n3f(a)=n=12nan1n3=2n=1an1n2af(a)=2n=1ann2lettrytofinds(x)=n=1xnn2ifx∣<1.
Commented by maxmathsup by imad last updated on 23/May/19
we have S^′ (x) =Σ_(n=1) ^∞  (x^(n−1) /n) ⇒x S^((1)) (x) =Σ_(n=1) ^∞  (x^n /n) ⇒  (x S^((1)) (x))^′  =Σ_(n=1) ^∞  x^(n−1)  =Σ_(n=0) ^∞  x^n  =(1/(1−x)) ⇒S^((1)) (x) +xS^((2)) (x) =(1/(1−x)) ⇒  S is solution of (de)      xy^(′′)  +y^′  =(1/(1−x))    let y^′  =z ⇒  xz^′  +z   =(1/(1−x))   (e)  (he) ⇒xz^′  +z =0 ⇒xz^′  =−z ⇒(z^′ /z) =−(1/x) ⇒ln∣z∣ =−ln∣x∣ +c ⇒  z =(k/(∣x∣))    let determine the solution on]0,+∞[ ⇒z =(k/x) ⇒ mvc method give  z^′  =(k^′ /x) −(k/x^2 )       (e) ⇒k^′  −(k/x) +(k/x) =(1/(1−x)) ⇒k^′  =(1/(1−x)) ⇒k(x) =−ln(1−x)+c_0   ⇒z(x) =−((ln(1−x))/x) +(c_0 /x)  y^′ =z ⇒y^′  =−((ln(1−x))/x) +(c_0 /x) ⇒ y(x) =−∫_0 ^x   ((ln(1−t))/t) dt +c_0 ln(x) +λ ⇒  S(x) = c_0 ln(x)−∫_0 ^x  ((ln(1−t))/t) dt    (  x>0)  S(e) =c_0  −∫_0 ^e    ((ln(1−t))/t) dt =Σ_(n=1) ^∞  (e^n /n^2 )  ⇒c_0 =Σ_(n=1) ^∞  (e^n /n^2 ) +∫_0 ^e  ((ln(1−t))/t) dt ⇒  S(x) =(Σ_(n=1) ^∞  (e^n /n^2 ) +∫_0 ^e  ((ln(1−t))/t) dt)ln(x)−∫_0 ^x  ((ln(1−t))/t) dt ....be continued...
wehaveS(x)=n=1xn1nxS(1)(x)=n=1xnn(xS(1)(x))=n=1xn1=n=0xn=11xS(1)(x)+xS(2)(x)=11xSissolutionof(de)xy+y=11xlety=zxz+z=11x(e)(he)xz+z=0xz=zzz=1xlnz=lnx+cz=kxletdeterminethesolutionon]0,+[z=kxmvcmethodgivez=kxkx2(e)kkx+kx=11xk=11xk(x)=ln(1x)+c0z(x)=ln(1x)x+c0xy=zy=ln(1x)x+c0xy(x)=0xln(1t)tdt+c0ln(x)+λS(x)=c0ln(x)0xln(1t)tdt(x>0)S(e)=c00eln(1t)tdt=n=1enn2c0=n=1enn2+0eln(1t)tdtS(x)=(n=1enn2+0eln(1t)tdt)ln(x)0xln(1t)tdt.becontinued

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