Menu Close

calculate-0-1-lnx-1-x-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 60496 by maxmathsup by imad last updated on 21/May/19
calculate ∫_0 ^1 ((lnx)/((1−x)^2 ))dx
calculate∫01lnx(1−x)2dx
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....!
letA=∫01ln(x)(1−x)2dxwehave∑n=0∞xn=11−xwith∣x∣<1⇒∑n=1∞nxn−1=1(1−x)2⇒A=∫01(∑n=1∞nxn−1)ln(x)dx=∑n=1∞n∫01xn−1ln(x)dx=∑n=1∞nwnbypartsu′=xn−1andv=ln(x)⇒wn=[1nxnln(x)]01−∫011nxndxx=−1n∫01xn−1dx=−1n[1nxn]01=−1n2⇒A=∑n=1∞n(−1n2)=−∑n=1∞1n=−∞sothisintegraldiverges….!

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *