find-f-x-0-1-ln-1-xt-3-dt-with-x-lt-1-2-calculate-0-1-ln-1-4t-3-dt-and-0-1-ln-2-t-3-dt-

Question Number 38465 by maxmathsup by imad last updated on 25/Jun/18

f i n d f (x) = \int_{0}^{1} l n (1 + {x t}^{3}) d t w i t h ∣ x ∣< 1 .

2) c a l c u l a t e \int_{0}^{1} l n (1 + 4 t^{3}) d t a n d \int_{0}^{1} l n (2 + t^{3}) d t .

Commented by maxmathsup by imad last updated on 30/Jun/18

1) w e h a v e f (x) = \int_{0}^{1} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} t^{3 n}) d t

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} \frac{1}{3 n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n (3 n + 1)} x^{n}

\frac{f (x)}{3} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3 n (3 n + 1)} x^{n} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} {\frac{1}{3 n} - \frac{1}{3 n + 1}} x^{n}

= \frac{1}{3} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3 n + 1} x^{n} b u t

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} x^{n} = l n (1 + x) l e t w (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3 n + 1} x^{n}

i f 0 < x < 1 w (x) = \frac{1}{x^{\frac{1}{3}}} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3 n + 1} {(^{3} \sqrt{x})}^{3 n + 1} =^{3} {(\sqrt{x})}^{- 1} φ (^{3} \sqrt{x}) w i t h

φ (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3 n + 1} x^{3 n + 1} \Rightarrow φ^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} x^{3 n}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{3 n + 3} = x^{3} \frac{1}{1 + x^{3}} = \frac{x^{3}}{1 + x^{3}} \Rightarrow φ (x) = \int_{0}^{x} \frac{t^{3}}{1 + t^{3}} d t + c

c = φ (o) = 0 \Rightarrow φ (x) = \int_{0}^{x} \frac{t^{3}}{1 + t^{3}} d t = \int_{0}^{x} \frac{1 + t^{3} - 1}{1 + t^{3}} d t

= x - \int_{0}^{x} \frac{d t}{1 + t^{3}} l e t d e c o m p o s e F (t) = \frac{1}{1 + t^{3}}

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

a = {l i m}_{t \to - 1} (t + 1) φ (t) = \frac{1}{3}

{l i m}_{t \to + \infty} t φ (t) = 0 = a + b \Rightarrow b = - \frac{1}{3} \Rightarrow F (t) = \frac{1}{3 (t + 1)} + \frac{- \frac{1}{3} t + c}{t^{2} - t + 1}

F (0) = 1 = \frac{1}{3} + c \Rightarrow c = \frac{2}{3} \Rightarrow F (x) = \frac{1}{3 (t + 1)} + \frac{1}{3} \frac{- t + 1}{t^{2} - t + 1}

\int_{0}^{x} F (t) d t = \frac{1}{3} \int_{0}^{x} \frac{d t}{t + 1} - \frac{1}{6} \int_{0}^{x} \frac{2 t - 1 - 1}{t^{2} - t + 1} d t

= [\frac{1}{3} l n ∣ t + 1 ∣ - \frac{1}{6} l n {(∣ t^{2} - t + 1 ∣]}_{0}^{x} + \frac{1}{6} \int_{0}^{x} \frac{d t}{t^{2} - t + 1}

= {[\frac{1}{6} l n (\frac{{(t + 1)}^{2}}{t^{2} - t + 1})]}_{0}^{x} + \frac{1}{6} \int_{0}^{x} \frac{d t}{t^{2} - t + 1} = \frac{1}{6} l n (\frac{x^{2} + 2 x + 1}{x^{2} - x + 1}) + I

I = \frac{1}{6} \int_{0}^{x} \frac{d t}{t^{2} - 2 \frac{t}{2} + \frac{1}{4} + \frac{3}{4}} \Rightarrow 6 I = \int_{0}^{x} \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{3}{4}}

=_{t - \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int_{- \frac{1}{\sqrt{3}}}^{\frac{2 x - 1}{\sqrt{3}}} \frac{1}{1 + u^{2}} \frac{\sqrt{3}}{2} d u

= \frac{2}{\sqrt{3}} [a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + a r c t a n (\frac{1}{\sqrt{3}})} \Rightarrow

\int_{0}^{x} F (t) d t = \frac{1}{6} l n (\frac{x^{2} + 2 x + 1}{x^{2} - x + 1}) + \frac{1}{3 \sqrt{3}} {a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + a r c t a n (\frac{1}{\sqrt{3}})}

φ (x) = x - \frac{1}{6} l n (\frac{x^{2} + 2 x + 1}{x^{2} - x + 1}) - \frac{1}{3 \sqrt{3}} {a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + \frac{π}{6}}

w (x) = \frac{1}{(^{3} \sqrt{x})} φ (^{3} \sqrt{x}) \Rightarrow \frac{f (x)}{3} = \frac{1}{3} l n (1 + x) - w (x) \Rightarrow

f (x) = l n (1 + x) - 3 w (x) .

Commented by maxmathsup by imad last updated on 30/Jun/18

2) l e t I = \int_{0}^{1} l n (2 + t^{3}) d t

I = \int_{0}^{1} l n (2 (1 + \frac{1}{2} t^{3})) d t = l n (2) + \int_{0}^{1} l n (1 + \frac{1}{2} t^{3}) d t

= l n (2) + f (\frac{1}{2}) = l n (2) + l n (\frac{3}{2}) - 3 w (\frac{1}{2})

= l n (2) + l n (\frac{3}{2}) - 3^{3} \sqrt{2} φ (\frac{1}{(^{3} \sqrt{2})}) t h e v a l u e o f φ (x) i s k n o w n .