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let-f-x-0-1-ln-1-x-2-t-2-t-2-dt-with-x-lt-1-find-f-x-at-a-simple-form-

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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 by abdo.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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