let-f-x-0-1-ln-1-xt-3-dt-with-x-lt-1-1-find-a-explicit-form-of-f-x-2-calculate-0-1-ln-1-1-2-t-3-dt-3-calculate-A-0-1-ln-1-sin-t-3-dt-with-0-lt-lt-pi-2-

Question Number 61978 by maxmathsup by imad last updated on 13/Jun/19

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

1) f i n d a e x p l i c i t f o r m o f f (x)

2) c a l c u l a t e \int_{0}^{1} l n (1 - \frac{1}{\sqrt{2}} t^{3}) d t

3) c a l c u l a t e A (θ) = \int_{0}^{1} l n (1 - s i n θ t^{3}) d t w i t h 0 < θ < \frac{π}{2}

Commented by maxmathsup by imad last updated on 14/Jun/19

1) w e h a v e f^{^{'}} (x) = - \int_{0}^{1} \frac{t^{3}}{1 - {x t}^{3}} d t = \frac{1}{x} \int_{0}^{1} \frac{1 - {x t}^{3} - 1}{1 - {x t}^{3}} d t

= \frac{1}{x} - \frac{1}{x} \int_{0}^{1} \frac{d t}{1 - {x t}^{3}} l e t α =^{3} \sqrt{x} \Rightarrow \int_{0}^{1} \frac{d t}{1 - {x t}^{3}} = \int_{0}^{1} \frac{d t}{1 - {(α t)}^{3}}

=_{α t = u} \int_{0}^{α} \frac{d u}{α (1 - u^{3})} = - \frac{1}{α} \int_{0}^{α} \frac{d u}{u^{3} - 1} l e t d e c o m p o s e

F (u) = \frac{1}{u^{3} - 1} = \frac{1}{(u - 1) (u^{2} + u + 1)} = \frac{a}{u - 1} + \frac{b u + c}{u^{2} + u + 1}

a = {l i m}_{u \to 1} (u - 1) F (u) = \frac{1}{3}

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

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

\int_{0}^{α} \frac{d u}{u^{3} - 1} = \int_{0}^{α} \frac{d u}{3 (u - 1)} - \frac{1}{6} \int_{0}^{α} \frac{2 u + 1 + 3}{u^{2} + u + 1}

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

\int_{0}^{α} \frac{d u}{u^{2} + u + 1} = \int_{0}^{α} \frac{d u}{{(u + \frac{1}{2})}^{2} + \frac{3}{4}} =_{u + \frac{1}{2} = \frac{\sqrt{3}}{2} z} \frac{4}{3} \int_{\frac{1}{\sqrt{3}}}^{\frac{2 α + 1}{\sqrt{3}}} \frac{1}{1 + z^{2}} \frac{\sqrt{3}}{2} d z

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

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

f^{^{'}} (x) = \frac{1}{x} + \frac{1}{x α} {\frac{1}{6} l n (\frac{α^{2} - 2 α + 1}{α^{2} + α + 1}) - \frac{1}{\sqrt{3}} a r c t a n (\frac{2 α + 1}{\sqrt{3}}) + \frac{1}{2} a r c t a n (\frac{1}{\sqrt{3}})} = A (x)

a n d α =^{3} \sqrt{x} \Rightarrow f (x) = \int A (x) d x + C \dots . b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 14/Jun/19

l e t u s e a n o t h e r w a y l e t α =^{3} \sqrt{x} \Rightarrow

f (x) = \int_{0}^{1} l n (1 - {(α t)}^{3}) d t = \int_{0}^{1} l n {(1 - α t) (1 + α t + α^{2} t^{2})} d t

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

c h a n g e m e n t 1 - α t = u g i v e \int_{0}^{1} l n (1 - α t) d t = \frac{- 1}{α} \int_{1}^{1 - α} l n (u) d u

= - \frac{1}{α} {[u l n u - u]}_{1}^{1 - α} = - \frac{1}{α} {(1 - α) l n (1 - α) - (1 - α) + 1}

= - \frac{1}{α} {(1 - α) l n (1 - α) + α} = \frac{α - 1}{α} l n (1 - α) - 1 . b y p a r t s

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

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

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

= 2 - \int_{0}^{1} \frac{α t + 2}{α^{2} t^{2} + α t + 1} d t = 2 - \frac{1}{2 α} \int_{0}^{1} \frac{2 α^{2} t + α + 3 α}{α^{2} t^{2} + α t + 1} d t

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

\int_{0}^{1} \frac{d t}{α^{2} t^{2} + α t + 1} = \frac{1}{α^{2}} \int_{0}^{1} \frac{d t}{t^{2} + \frac{t}{α} + \frac{1}{α^{2}}} = \frac{1}{α^{2}} \int_{0}^{1} \frac{d t}{t^{2} + + \frac{2 t}{2 α} + \frac{1}{4 α^{2}} + \frac{1}{α^{2}} - \frac{1}{4 α^{2}}}

= \frac{1}{α^{2}} \int_{0}^{1} \frac{d t}{{(t + \frac{1}{2 α})}^{2} + \frac{3}{α^{2}}} =_{t + \frac{1}{2 α} = \frac{\sqrt{3}}{α} x} \frac{1}{α^{2}} \int_{\frac{1}{2 \sqrt{3}}}^{\frac{2 α + 1}{2 \sqrt{3}}} \frac{α^{2}}{3} \frac{1}{1 + x^{2}} \frac{\sqrt{3}}{α} d x (w e t a k e x > 0)

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

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

f (x) = \frac{α - 1}{α} l n (1 - α) - 1 + l n (α^{2} + α + 1) - 2 + \frac{1}{2 α} l n (α^{2} + α + 1)

+ \frac{\sqrt{3}}{2 α} {a r c t a n (\frac{2 α + 1}{2 \sqrt{3}}) - a r c t a n (\frac{1}{2 \sqrt{3}})} w i t h α =^{3} \sqrt{x}

Commented by maxmathsup by imad last updated on 14/Jun/19

f (x) = \frac{(α - 1) l n (1 - α)}{α} - 3 + (1 + \frac{1}{2 α}) l n (α^{2} + α + 1)

+ \frac{\sqrt{3}}{2 α} {a r c t a n (\frac{2 α + 1}{2 \sqrt{3}}) - a r c t a n (\frac{1}{2 \sqrt{3}})} .

Commented by maxmathsup by imad last updated on 14/Jun/19

2) \int_{0}^{1} l n (1 - \frac{1}{\sqrt{2}} t^{3}) = f (\frac{1}{\sqrt{2}}) w e h a v e α = x^{\frac{1}{3}} \Rightarrow α = {(2^{- \frac{1}{2}})}^{\frac{1}{3}} = 2^{- \frac{1}{6}} = \frac{1}{(^{6} \sqrt{2})} \Rightarrow

f (x) = \frac{(α - 1) l n (1 - α)}{α} - 3 + (1 + \frac{1}{2 α}) l n (α^{2} + α + 1)

+ \frac{\sqrt{3}}{2 α} {a r c t a n (\frac{2 α + 1}{2 \sqrt{3}}) - a r c t a n (\frac{1}{2 \sqrt{3}})} .

Commented by maxmathsup by imad last updated on 14/Jun/19

3) \int_{0}^{1} l n (1 - s i n θ t^{3}) d θ = f (s i n θ) w e α =^{3} \sqrt{x} \Rightarrow α =^{3} \sqrt{s i n θ}

s o t h e v a l u e o f t h i s i n t e g r a l i s k n o w n .