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Question Number 32477 by prof Abdo imad last updated on 25/Mar/18
calcilate ∫_0 ^1 ((ln(1−x^2 ))/x^2 )dx
$${calcilate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$
Commented by prof Abdo imad last updated on 31/Mar/18
for ∣t∣<1 (1/(1−t))=Σ_(n=0) ^∞ t^n ⇒ −ln∣1−t∣=Σ_(n=0) ^∞ (t^(n+1) /(n+1)) = Σ_(n=1) ^∞ (t^n /n) ⇒ Σ_(n=1) ^∞ (t^n /n) = −ln(1−t) and ln(1−x^2 )= −Σ_(n=1) ^∞ (x^(2n) /n) ⇒ ((ln(1−x^2 ))/x^2 ) =−Σ_(n=1) ^∞ (x^(2n−2) /n) ⇒ ∫_0 ^1 ((ln(1−x^2 ))/x^2 )dx =−Σ_(n=1) ^∞ (1/(n(2n−1))) let put S_n = Σ_(k=1) ^n (1/(k(2k−1))) we have (1/2)S_n = Σ_(k=1) ^n (1/(2k(2k−1))) = Σ_(k=1) ^n ((1/(2k−1)) −(1/(2k))) = Σ_(k=1) ^n (1/(2k−1)) −(1/2) H_n but Σ_(k=1) ^n (1/(2k−1)) = 1 +(1/3) +....+ (1/(2n−1)) =1+(1/2) +(1/3) +...+ (1/(2n)) −(1/2) −(1/4) −....−(1/(2n)) = H_(2n) −(1/2) H_n (1/2) S_n = H_(2n) −H_n =ln(2n) +γ −ln(n) −γ +o(1) =ln(((2n)/n) )+γ +o(1)→ ln(2) ⇒lim_(n→∞) S_n = 2ln(2) ⇒ ∫_0 ^1 ((ln(1−x^2 ))/x^2 ) =−2ln(2) .
$${for}\:\mid{t}\mid<\mathrm{1}\:\:\:\frac{\mathrm{1}}{\mathrm{1}−{t}}=\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{{n}} \:\:\:\Rightarrow\:−{ln}\mid\mathrm{1}−{t}\mid=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{t}^{{n}} }{{n}}\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{t}^{{n}} }{{n}}\:=\:−{ln}\left(\mathrm{1}−{t}\right)\:{and} \\ $$$${ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{{n}}\:\Rightarrow\:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{x}^{\mathrm{2}{n}−\mathrm{2}} }{{n}} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}\:\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$${let}\:{put}\:\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}\left(\mathrm{2}{k}−\mathrm{1}\right)}\:\:{we}\:{have} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{S}_{{n}} \:\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{k}\left(\mathrm{2}{k}−\mathrm{1}\right)}\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}{k}}\right) \\ $$$$=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \:\:{but} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:=\:\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+….+\:\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+…+\:\:\frac{\mathrm{1}}{\mathrm{2}{n}}\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:−….−\frac{\mathrm{1}}{\mathrm{2}{n}} \\ $$$$=\:{H}_{\mathrm{2}{n}} \:−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:{S}_{{n}} =\:\:{H}_{\mathrm{2}{n}} \:−{H}_{{n}} \:\:\:={ln}\left(\mathrm{2}{n}\right)\:+\gamma\:\:−{ln}\left({n}\right)\:−\gamma\:+{o}\left(\mathrm{1}\right) \\ $$$$={ln}\left(\frac{\mathrm{2}{n}}{{n}}\:\right)+\gamma\:+{o}\left(\mathrm{1}\right)\rightarrow\:{ln}\left(\mathrm{2}\right)\:\Rightarrow{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} =\:\mathrm{2}{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:=−\mathrm{2}{ln}\left(\mathrm{2}\right)\:. \\ $$

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