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Question Number 44471 by abdo.msup.com last updated on 29/Sep/18
calculste ∫_0 ^∞ ((ln(x))/((1+x)^2 ))dx
$${calculste}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 30/Sep/18
let I =∫_0 ^∞ ((ln(x))/((1+x)^2 ))dx changement x=(1/t) give I =−∫_0 ^∞ ((−ln(t))/((1+(1/t))^2 )) (−(1/t^2 ))dt =−∫_0 ^∞ ((ln(t))/(t^2 (((t+1)/t))^2 ))dt =−∫_0 ^∞ ((ln(t))/((1+t)^2 )) =−I ⇒ 2I =0 ⇒ I =0 .
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}\:\:{changement}\:{x}=\frac{\mathrm{1}}{{t}}\:{give}\: \\ $$$${I}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{−{ln}\left({t}\right)}{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} }\:\left(−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} \left(\frac{{t}+\mathrm{1}}{{t}}\right)^{\mathrm{2}} }{dt}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\:=−{I}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\mathrm{0}\:\Rightarrow\:{I}\:=\mathrm{0}\:. \\ $$
Commented by maxmathsup by imad last updated on 30/Sep/18
another way I =∫_0 ^1 ((ln(x))/((1+x)^2 ))dx +∫_1 ^(+∞) ((ln(x))/((1+x)^2 ))dx but changement x=(1/t) give ∫_1 ^(+∞) ((ln(x))/((1+x)^2 ))dx =−∫_0 ^1 ((−ln(t))/((1+(1/t))^2 ))(−(dt/t^2 )) =−∫_0 ^1 ((ln(t))/((1+t)^2 ))dt ⇒ I =∫_0 ^1 ((ln(x))/((1+x)^2 ))dx−∫_0 ^1 ((ln(t))/((1+t)^2 ))dt =0
$${another}\:{way}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}\:+\int_{\mathrm{1}} ^{+\infty} \:\frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}\:\:{but}\:{changement}\: \\ $$$${x}=\frac{\mathrm{1}}{{t}}\:\:\:{give}\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{−{ln}\left({t}\right)}{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} }\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right) \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}\:\Rightarrow\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}\:=\mathrm{0} \\ $$

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