Advanced-Calculus-prove-that-0-1-ln-1-x-1-x-2-dx-pi-8-ln-2-G-where-G-is-catalan-number-

Question Number 137973 by mnjuly1970 last updated on 08/Apr/21

\dots \dots . A d v a n c e d \dots \dots \dots \dots C a l c u l u s \dots \dots .

p r o v e t h a t :::

ϕ = \int_{0}^{1} \frac{l n (1 - x)}{1 + x^{2}} d x = \frac{π}{8} l n (2) - G \dots ✓

w h e r e G i s c a t a l a n n u m b e r \dots

Answered by EnterUsername last updated on 08/Apr/21

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

= \int_{0}^{\frac{π}{4}} l n (c o s x - s i n x) d x - \int_{0}^{\frac{π}{4}} l n c o s x d x

= \int_{0}^{\frac{π}{4}} l n [\sqrt{2} c o s (x + \frac{π}{4})] d x - (\frac{G}{2} - \frac{π l n 2}{4})

= \frac{π}{8} l n 2 + \int_{0}^{\frac{π}{4}} l n (s i n x) d x - (\frac{G}{2} - \frac{π l n 2}{4})

= \frac{π}{8} l n 2 - \frac{G}{2} - \frac{π l n 2}{4} - (\frac{G}{2} - \frac{π l n 2}{4}) = \frac{π}{8} l n 2 - G

Commented by mnjuly1970 last updated on 08/Apr/21

g r a t e f u l \dots

Answered by mathmax by abdo last updated on 09/Apr/21

f (a) = \int_{0}^{1} \frac{\ln (1 + ax)}{1 + x^{2}} dx \Rightarrow f^{^{'}} (a) = \int_{0}^{1} \frac{x}{(ax + 1) (x^{2} + 1)} dx

= \frac{1}{a} \int_{0}^{1} \frac{ax + 1 - 1}{(ax + 1) (x^{2} + 1)} dx = \frac{1}{a} {[arctanx]}_{0}^{1} - \frac{1}{a} \int_{0}^{1} \frac{dx}{(ax + 1) (x^{2} + 1)}

= \frac{π}{4 a} - \frac{1}{a} \int_{0}^{1} \frac{dx}{(ax + 1) (x^{2} + 1)} let decompose F (x) = \frac{1}{(ax + 1) (x^{2} + 1)}

F (x) = \frac{α}{ax + 1} + \frac{mx + n}{x^{2} + 1}

α = \frac{1}{\frac{1}{a^{2}} + 1} = \frac{a^{2}}{1 + a^{2}}

\lim_{x \to + \infty} xF (x) = 0 = \frac{α}{a} + m \Rightarrow m = - \frac{a}{1 + a^{2}}

F (o) = 1 = α + n \Rightarrow n = 1 - \frac{a^{2}}{1 + a^{2}} = \frac{1}{1 + a^{2}} \Rightarrow F (x) = \frac{a^{2}}{(a^{2} + 1) (ax + 1)}

+ \frac{- \frac{a}{a^{2} + 1} x + \frac{1}{1 + a^{2}}}{x^{2} + 1} \Rightarrow \int_{0}^{1} F (x) dx = \frac{a^{2}}{a^{2} + 1} \int_{0}^{1} \frac{dx}{ax + 1}

- \frac{1}{a^{2} + 1} \int_{0}^{1} \frac{ax - 1}{x^{2} + 1} dx = \frac{a}{a^{2} + 1} {[\ln (ax + 1)]}_{0}^{1} - \frac{a}{2 (a^{2} + 1)} \int_{0}^{1} \frac{2 x}{x^{2} + 1} dx

+ \frac{π}{4 (a^{2} + 1)} = \frac{aln (a + 1)}{a^{2} + 1} - \frac{a}{2 (a^{2} + 1)} \ln (2) + \frac{π}{4 (a^{2} + 1)} \Rightarrow

f^{^{'}} (a) = \frac{π}{4 a} - \frac{\ln (a + 1)}{a^{2} + 1} + \frac{\ln 2}{2 (a^{2} + 1)} - \frac{π}{4 a (a^{2} + 1)}

= \frac{π}{4 a} (1 - \frac{1}{a^{2} + 1}) - \frac{\ln (a + 1)}{a^{2} + 1} + \frac{\ln 2}{2 (a^{2} + 1)} = \frac{π a}{4} - \frac{\ln (a + 1)}{a^{2} + 1} + \frac{\ln 2}{2 (a^{2} + 1)}

\Rightarrow \int_{0}^{1} f^{^{'}} (a) da = \frac{π}{4} \int_{0}^{1} ada - \int_{0}^{1} \frac{\ln (a + 1)}{a^{2} + 1} da + \frac{\ln 2}{2} \int_{0}^{1} \frac{da}{a^{2} + 1}

= \frac{π}{4} \times \frac{1}{2} - \int_{0}^{1} \frac{\ln (1 + x)}{x^{2} + 1} dx + \frac{\ln 2}{2} . \frac{π}{4}

= \frac{π}{8} - \int_{0}^{1} \frac{\ln (1 + x)}{x^{2} + 1} dx + \frac{π}{8} \ln 2 = f (1) = \int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} dx \Rightarrow

2 \int_{0}^{1} \frac{\ln (1 + x)}{x^{2} + 1} dx = \frac{π}{8} + \frac{π}{8} \ln 2 \Rightarrow \int_{0}^{1} \frac{\ln (1 + x)}{x^{2} + 1} dx = \frac{π}{16} + \frac{π}{16} \ln 2 ?

if we want \int_{0}^{1} \frac{\ln (1 - x)}{x^{2} + 1} dx we use the parametric

f (a) = \int_{0}^{1} \frac{\ln (1 - ax)}{x^{2} + 1} dx \dots . be continued \dots .