calculate-I-0-1-ln-1-t-2-4-t-2-dt-

Question Number 38463 by maxmathsup by imad last updated on 25/Jun/18

c a l c u l a t e I = \int_{0}^{1} \frac{l n (1 - \frac{t^{2}}{4})}{t^{2}} d t

Commented by abdo mathsup 649 cc last updated on 05/Jul/18

w e h a v e p r o v e d ? t h a t

\int_{0}^{1} \frac{l n (1 - x^{2} t^{2})}{t^{2}} d t = - l n (1 - x^{2}) - \frac{x}{2} l n (\frac{1 + x}{1 - x})

w i t h ∣ x ∣< 1 \Rightarrow

\int_{0}^{1} \frac{l n (1 - \frac{t^{2}}{4})}{t^{2}} d t = - l n (1 - {(\frac{1}{2})}^{2}) - \frac{1}{4} l n (\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}})

= - l n (\frac{3}{4}) - \frac{1}{4} l n (3) = l n (\frac{4}{3}) - \frac{l n (3)}{4}

= 2 l n (2) - l n (3) - \frac{1}{4} l n (3)

= 2 l n (2) - \frac{5}{4} l n (3) .