calculate-0-ln-x-x-2-x-1-dx-

Question Number 55457 by maxmathsup by imad last updated on 24/Feb/19

c a l c u l a t e \int_{0}^{\infty} \frac{l n (x)}{x^{2} + x + 1} d x .

Commented by maxmathsup by imad last updated on 05/Mar/19

l e t I = \int_{0}^{\infty} \frac{l n (x)}{x^{2} + x + 1} d x \Rightarrow I = \int_{0}^{\infty} \frac{l n x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} d x

=_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t} \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{l n (\frac{\sqrt{3}}{2} t - \frac{1}{2})}{\frac{3}{4} (1 + t^{2})} \frac{\sqrt{3}}{2} d t = \frac{4}{3} \frac{\sqrt{3}}{2} \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{l n (\frac{\sqrt{3}}{2} t - \frac{1}{2})}{1 + t^{2}} d t

= \frac{2}{\sqrt{3}} \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{l n (\frac{\sqrt{3}}{2} t - \frac{1}{2})}{t^{2} + 1} d t \Rightarrow I = \frac{2}{\sqrt{3}} \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{l n (\frac{\sqrt{3}}{2} x - \frac{1}{2})}{x^{2} + 1} d x l e t

f (t) = \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{l n (t x - 1)}{x^{2} + 1} d x \Rightarrow f^{^{'}} (t) = \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{x}{(t x - 1) (x^{2} + 1)} d x l e t d e c o m p i s e

F (x) = \frac{x}{(t x - 1) (x^{2} + 1)} \Rightarrow F (x) = \frac{a}{t x - 1} + \frac{b x + c}{x^{2} + 1}

a = {l i m}_{x \to \frac{1}{t}} (t x - 1) F (x) = \frac{1}{t (\frac{1}{t^{2}} + 1)} = \frac{1}{t + \frac{1}{t}} = \frac{t}{t^{2} + 1}

{l i m}_{x \to + \infty} x F (x) = \frac{a}{t} + b = 0 \Rightarrow b = - \frac{a}{t} = - \frac{1}{t^{2} + 1} \Rightarrow

F (x) = \frac{t}{(t^{2} + 1) (t x - 1)} + \frac{- \frac{1}{t^{2} + 1} x + c}{x^{2} + 1}

F (0) = 0 = - \frac{t}{t^{2} + 1} + c \Rightarrow c = \frac{t}{t^{2} + 1} \Rightarrow

F (x) = \frac{t}{(t^{2} + 1) (t x - 1)} + \frac{1}{t^{2} + 1} \frac{- x + t}{x^{2} + 1} \Rightarrow

\int F (x) d x = \frac{1}{t^{2} + 1} l n ∣ t x - 1 ∣ - \frac{1}{2 (t^{2} + 1)} l n (x^{2} + 1) + \frac{t}{t^{2} + 1} a r c t a n (x) + c \Rightarrow

\int_{\frac{1}{\sqrt{3}}}^{+ \infty} F (x) d x = \frac{1}{t^{2} + 1} {[l n ∣ \frac{t x - 1}{\sqrt{x^{2} + 1}} ∣]}_{\frac{1}{\sqrt{3}}}^{+ \infty} + \frac{t}{t^{2} + 1} {[a r c t a n t]}_{\frac{1}{\sqrt{3}}}^{+ \infty}

= \frac{1}{t^{2} + 1} {l n ∣ t ∣ - l n ∣ \frac{\frac{t}{\sqrt{3}} - 1}{\frac{2}{\sqrt{3}}} ∣} + \frac{t}{t^{2} + 1} {\frac{π}{2} - \frac{π}{6}}

= \frac{1}{t^{2} + 1} {l n ∣ t ∣ - l n ∣ \frac{t - \sqrt{3}}{2} ∣} + \frac{π t}{3 (t^{2} + 1)} = f^{^{'}} (t) \Rightarrow

f (t) = \int \frac{l n ∣ t ∣}{t^{2} + 1} d t - \int \frac{l n ∣ t - \sqrt{3}) - l n (2)}{t^{2} + 1} d t + \frac{π}{6} l n (t^{2} + 1) + C

= \int \frac{l n ∣ t ∣}{t^{2} + 1} d t - \int \frac{l n ∣ t - \sqrt{3} ∣}{t^{2} + 1} d t + l n (2) a r c t a n t + C \dots . b e c o n t i n u e d \dots .

Commented by Abdo msup. last updated on 05/Mar/19

l e t t r y a n o t h e r w a y w e h a v e I = \int_{0}^{\infty} \frac{(1 - x) l n x}{1 - x^{3}} d x

= \int_{0}^{1} \frac{(1 - x) l n (x)}{1 - x^{3}} d x + \int_{1}^{+ \infty} \frac{(1 - x) l n (x)}{1 - x^{3}} d x b u t

\int_{1}^{\infty} \frac{(1 - x) l n (x)}{1 - x^{3}} d x =_{x = \frac{1}{t}} - \int_{0}^{1} \frac{(1 - \frac{1}{t}) (- l n (t))}{1 - \frac{1}{t^{3}}} (\frac{- d t}{t^{2}})

= - \int_{0}^{1} \frac{(t - 1) l n (t)}{t^{3} - 1} d t = \int_{0}^{1} \frac{(t - 1) l n (t)}{1 - t^{3}} d t

= - \int_{0}^{1} \frac{(1 - t) l n (t)}{1 - t^{3}} d t \Rightarrow I = 0