Question Number 60496 by maxmathsup by imad last updated on 21/May/19
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{lnx}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by Mr X pcx last updated on 22/May/19
![let A =∫_0 ^1 ((ln(x))/((1−x)^2 ))dx we have Σ_(n=0) ^∞ x^n =(1/(1−x)) with ∣x∣<1 ⇒ Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒ A =∫_0 ^1 (Σ_(n=1) ^∞ nx^(n−1) )ln(x)dx =Σ_(n=1) ^∞ n ∫_0 ^1 x^(n−1) ln(x)dx=Σ_(n=1) ^∞ nw_n by parts u^′ =x^(n−1) and v=ln(x) ⇒ w_n =[(1/n)x^n ln(x)]_0 ^1 −∫_0 ^1 (1/n)x^n (dx/x) =−(1/n) ∫_0 ^1 x^(n−1) dx =−(1/n)[(1/n) x^n ]_0 ^1 =−(1/n^2 ) ⇒ A =Σ_(n=1) ^∞ n(−(1/n^2 )) =−Σ_(n=1) ^∞ (1/n) =−∞ so this integral diverges....!](https://www.tinkutara.com/question/Q60585.png)
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{dx}\:\:{we}\:{have} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:{with}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} {nx}^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \right){ln}\left({x}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} {n}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {ln}\left({x}\right){dx}=\sum_{{n}=\mathrm{1}} ^{\infty} {nw}_{{n}} \\ $$$${by}\:{parts}\:\:{u}^{'} ={x}^{{n}−\mathrm{1}} \:\:{and}\:{v}={ln}\left({x}\right)\:\Rightarrow \\ $$$${w}_{{n}} =\left[\frac{\mathrm{1}}{{n}}{x}^{{n}} {ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{n}}{x}^{{n}} \:\frac{{dx}}{{x}} \\ $$$$=−\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{1}} {dx}\:=−\frac{\mathrm{1}}{{n}}\left[\frac{\mathrm{1}}{{n}}\:{x}^{{n}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\Rightarrow\:{A}\:=\sum_{{n}=\mathrm{1}} ^{\infty} {n}\left(−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\:=−\infty\:\:{so}\:{this}\:{integral} \\ $$$${diverges}….! \\ $$