let f(t) =∫_0 ^1 ln(1 −tx^3 )dx with 0<t≤1 find a simple form of f(t) 2)calculate ∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 36690 by prof Abdo imad last updated on 04/Jun/18

$l e t f (t) = \int_{0}^{1} l n (1 - {t x}^{3}) d x w i t h 0 < t ⩽ 1$ $f i n d a s i m p l e f o r m o f f (t)$ $2) c a l c u l a t e \int_{0}^{1} l n (2 - x^{3}) d x .$
Commented bymaxmathsup by imad last updated on 03/Aug/18
$we have f^′ (t) =−∫_0 ^1 (x^3 /(1−tx^3 ))dx =−∫_0 ^1 x^3 (Σ_(n=0) ^∞ t^n x^(3n) dx =−Σ_(n=0) ^∞ t^n ∫_0 ^1 x^(3(n+1)) dx =−Σ_(n=0) ^∞ (t^n /(3n+4)) =−(1/((^3 (√t))^4 ))Σ_(n=0) ^∞ (((^3 (√t))^(3n +4) )/(3n+4)) =−(1/((^3 (√t))^4 )) ϕ(^3 (√t)) with ϕ(x) =Σ_(n=0) ^∞ (x^(3n+4) /(3n+4)) ⇒ϕ^′ (x)= Σ_(n=0) ^∞ x^(3n+3) =x^3 Σ_(n=0) ^∞ (x^3 )^n =x^3 (1/(1−x^3 )) =(x^3 /(1−x^3 )) ⇒ϕ(x) = ∫_0 ^x (t^3 /(1−t^3 )) dt +c withc=ϕ(0)=0 ⇒ ϕ(x)=−∫_0 ^x (t^3 /(t^3 −1))dt =−∫_0 ^x ((t^3 −1 +1)/(t^3 −1))dt =−x −∫_0 ^x (dt/(t^3 −1)) let decompose F(t) = (1/(t^3 −1)) F(t)= (1/((t−1)(t^2 +t+1))) =(a/(t−1)) +((bt+c)/(t^2 +t +1)) a =lim_(t→1) (t−1)F(t) =(1/3) lim_(t→+∞) tF(t) =0 =a +b ⇒b =−a ⇒ F(t) = (1/(3(t−1))) −(1/3) ((t −3c)/(t^2 +t+1)) F(0) =−1 =−(1/3) +c ⇒c=(1/3)−1 =−(2/3) ⇒F(t)=(1/(3(t−1))) −(1/3) ((t +2)/(t^2 +t+1)) ⇒ ∫ F(t)dt = (1/3)ln∣t−1∣ −(1/6) ∫ ((2t +1+3)/(t^2 +t +1))dt =(1/3)ln∣t−1∣ −(1/6)ln∣t^2 +t+1∣ −(1/2) ∫ (dt/((t+(1/2))^2 +(3/4))) but ∫ (dt/((t+(1/2))^2 +(3/4))) =_(t+(1/2)=((√3)/2)u) ∫ (1/((3/4)(1+u^2 ))) ((√3)/2) du=(4/3) ((√3)/2) arctan(((2t+1)/(√3))) =((2(√3))/3) arctan(((2t+1)/(√3))) ⇒ ϕ(x)=−x − [ (1/3)ln∣t−1∣−(1/6)ln(t^2 +t+1)−((√3)/3) arctan(((2t+1)/(√3)))]_0 ^x =−x −{(1/3)ln∣x−1∣−(1/6)ln(x^2 +x+1)−((√3)/3) arctan(((2x+1)/(√3)))+((√3)/3) arctan((1/(√3)))} we have f^′ (t)=−(1/((^3 (√t))^4 )) ϕ(^3 (√t)) so the value of f^′ (x) is known.. be continued...$
$w e h a v e f^{^{'}} (t) = - \int_{0}^{1} \frac{x^{3}}{1 - {t x}^{3}} d x = - \int_{0}^{1} x^{3} (\sum_{n = 0}^{\infty} t^{n} x^{3 n} d x$ $= - \sum_{n = 0}^{\infty} t^{n} \int_{0}^{1} x^{3 (n + 1)} d x = - \sum_{n = 0}^{\infty} \frac{t^{n}}{3 n + 4}$ $= - \frac{1}{{(^{3} \sqrt{t})}^{4}} \sum_{n = 0}^{\infty} \frac{{(^{3} \sqrt{t})}^{3 n + 4}}{3 n + 4} = - \frac{1}{{(^{3} \sqrt{t})}^{4}} φ (^{3} \sqrt{t}) w i t h$ $φ (x) = \sum_{n = 0}^{\infty} \frac{x^{3 n + 4}}{3 n + 4} \Rightarrow φ^{^{'}} (x) = \sum_{n = 0}^{\infty} x^{3 n + 3} = x^{3} \sum_{n = 0}^{\infty} {(x^{3})}^{n}$ $= x^{3} \frac{1}{1 - x^{3}} = \frac{x^{3}}{1 - x^{3}} \Rightarrow φ (x) = \int_{0}^{x} \frac{t^{3}}{1 - t^{3}} d t + c w i t h c = φ (0) = 0 \Rightarrow$ $φ (x) = - \int_{0}^{x} \frac{t^{3}}{t^{3} - 1} d t = - \int_{0}^{x} \frac{t^{3} - 1 + 1}{t^{3} - 1} d t = - x - \int_{0}^{x} \frac{d t}{t^{3} - 1} l e t d e c o m p o s e$ $F (t) = \frac{1}{t^{3} - 1}$ $F (t) = \frac{1}{(t - 1) (t^{2} + t + 1)} = \frac{a}{t - 1} + \frac{b t + c}{t^{2} + t + 1}$ $a = {l i m}_{t \to 1} (t - 1) F (t) = \frac{1}{3}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b \Rightarrow b = - a \Rightarrow$ $F (t) = \frac{1}{3 (t - 1)} - \frac{1}{3} \frac{t - 3 c}{t^{2} + t + 1}$ $F (0) = - 1 = - \frac{1}{3} + c \Rightarrow c = \frac{1}{3} - 1 = - \frac{2}{3} \Rightarrow F (t) = \frac{1}{3 (t - 1)} - \frac{1}{3} \frac{t + 2}{t^{2} + t + 1} \Rightarrow$ $\int F (t) d t = \frac{1}{3} l n ∣ t - 1 ∣ - \frac{1}{6} \int \frac{2 t + 1 + 3}{t^{2} + t + 1} d t$ $= \frac{1}{3} l n ∣ t - 1 ∣ - \frac{1}{6} l n ∣ t^{2} + t + 1 ∣ - \frac{1}{2} \int \frac{d t}{{(t + \frac{1}{2})}^{2} + \frac{3}{4}} b u t$ $\int \frac{d t}{{(t + \frac{1}{2})}^{2} + \frac{3}{4}} =_{t + \frac{1}{2} = \frac{\sqrt{3}}{2} u} \int \frac{1}{\frac{3}{4} (1 + u^{2})} \frac{\sqrt{3}}{2} d u = \frac{4}{3} \frac{\sqrt{3}}{2} a r c t a n (\frac{2 t + 1}{\sqrt{3}})$ $= \frac{2 \sqrt{3}}{3} a r c t a n (\frac{2 t + 1}{\sqrt{3}}) \Rightarrow$ $φ (x) = - x - {[\frac{1}{3} l n ∣ t - 1 ∣ - \frac{1}{6} l n (t^{2} + t + 1) - \frac{\sqrt{3}}{3} a r c t a n (\frac{2 t + 1}{\sqrt{3}})]}_{0}^{x}$ $= - x - {\frac{1}{3} l n ∣ x - 1 ∣ - \frac{1}{6} l n (x^{2} + x + 1) - \frac{\sqrt{3}}{3} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + \frac{\sqrt{3}}{3} a r c t a n (\frac{1}{\sqrt{3}})}$ $w e h a v e f^{^{'}} (t) = - \frac{1}{{(^{3} \sqrt{t})}^{4}} φ (^{3} \sqrt{t}) s o t h e v a l u e o f f^{^{'}} (x) i s k n o w n . .$ $b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum