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Question Number 38464 by maxmathsup by imad last updated on 25/Jun/18

let f(x)= ∫_0 ^1   ((ln(1−x^2 t^2 ))/t^2 )dt  with ∣x∣<1  find  f(x) at a simple form .

$${let}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt}\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$ $${find}\:\:{f}\left({x}\right)\:{at}\:{a}\:{simple}\:{form}\:. \\ $$

Commented byabdo.msup.com last updated on 01/Jul/18

we have ln^′ (1−u)=((−1)/(1−u)) =−Σ_(n=0) ^∞ u^n   ⇒ln(1−u) =−Σ_(n=0) ^∞ (u_(n+1) /(n+1)) =−Σ_(n=1) ^∞ (u^n /n)  ⇒ln(1−x^2 t^2 )=−Σ_(n=1) ^∞  ((x^(2n) t^(2n) )/n)  ⇒((ln(1−x^2 t^2 ))/t^2 ) =−Σ_(n=1) ^∞  ((x^(2n) t^(2n−2) )/n) ⇒  f(x)=−∫_0 ^1 {Σ_(n=1) ^∞  ((x^(2n) t^(2n−2) )/n)}dt  =−Σ_(n=1) ^∞  (x^(2n) /n) ∫_0 ^1   t^(2n−2) dt  =−Σ_(n=1) ^∞    (x^(2n) /(n(2n−1))) ⇒  ((f(x))/2) =−Σ_(n=1) ^∞  (x^(2n) /(2n(2n−1)))  =−Σ_(n=1) ^∞  ( (1/(2n−1)) −(1/(2n)))x^(2n)   =Σ_(n=1) ^∞   (x^(2n) /(2n)) −Σ_(n=1) ^∞  (x^(2n) /(2n−1)) but  Σ_(n=1) ^∞  (x^(2n) /(2n)) =−(1/2)ln(1−x^2 )  let w(x)=Σ_(n=1) ^∞  (x^(2n) /(2n−1))  w(x)=x Σ_(n=1) ^∞  (x^(2n−1) /(2n−1)) =xϕ(x)  ϕ^′ (x) =Σ_(n=1) ^∞   x^(2n−2) =Σ_(n=1) ^∞ (x^2 )^(n−1)   =Σ_(n=0) ^∞  (x^2 )^n  =(1/(1−x^2 )) ⇒ϕ(x)=∫ (dx/(1−x^2 )) +c  =(1/2)∫  ( (1/(1+x)) +(1/(1−x)))dx +c  =(1/2)ln∣((1+x)/(1−x))∣ +c  but c=ϕ(0) ⇒  w(x)=(x/2)ln∣((1+x)/(1−x))∣⇒  ((f(x))/2) =−(1/2)ln(1−x^2 )−(x/2)ln∣((1+x)/(1−x))∣ ⇒  f(x)=−ln(1−x^2 )−(x/2)ln∣((1+x)/(1−x))∣ .

$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}−{u}\right)=\frac{−\mathrm{1}}{\mathrm{1}−{u}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} {u}^{{n}} \\ $$ $$\Rightarrow{ln}\left(\mathrm{1}−{u}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{u}_{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{u}^{{n}} }{{n}} \\ $$ $$\Rightarrow{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} {t}^{\mathrm{2}{n}} }{{n}} \\ $$ $$\Rightarrow\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} {t}^{\mathrm{2}{n}−\mathrm{2}} }{{n}}\:\Rightarrow \\ $$ $${f}\left({x}\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} {t}^{\mathrm{2}{n}−\mathrm{2}} }{{n}}\right\}{dt} \\ $$ $$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{\mathrm{2}{n}−\mathrm{2}} {dt} \\ $$ $$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}{n}} }{{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}\:\Rightarrow \\ $$ $$\frac{{f}\left({x}\right)}{\mathrm{2}}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$ $$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(\:\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}{n}}\right){x}^{\mathrm{2}{n}} \\ $$ $$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}−\mathrm{1}}\:{but} \\ $$ $$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}}\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$ $${let}\:{w}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}−\mathrm{1}} \\ $$ $${w}\left({x}\right)={x}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}−\mathrm{1}} }{\mathrm{2}{n}−\mathrm{1}}\:={x}\varphi\left({x}\right) \\ $$ $$\varphi^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{x}^{\mathrm{2}{n}−\mathrm{2}} =\sum_{{n}=\mathrm{1}} ^{\infty} \left({x}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} \\ $$ $$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({x}^{\mathrm{2}} \right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow\varphi\left({x}\right)=\int\:\frac{{dx}}{\mathrm{1}−{x}^{\mathrm{2}} }\:+{c} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\left(\:\frac{\mathrm{1}}{\mathrm{1}+{x}}\:+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right){dx}\:+{c} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid\:+{c}\:\:{but}\:{c}=\varphi\left(\mathrm{0}\right)\:\Rightarrow \\ $$ $${w}\left({x}\right)=\frac{{x}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid\Rightarrow \\ $$ $$\frac{{f}\left({x}\right)}{\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−\frac{{x}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid\:\Rightarrow \\ $$ $${f}\left({x}\right)=−{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−\frac{{x}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid\:. \\ $$ $$ \\ $$

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