find f(x)= ∫_0 ^1 ln(1+xt^3 )dt with ∣x∣<1 . 2) calculate ∫_0 ^1 ln(1+4t^3 )dt and ∫


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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
$1) we have f(x)= ∫_0 ^1 (Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n t^(3n) )dt = Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n (1/(3n+1)) = Σ_(n=1) ^∞ (((−1)^(n−1) )/(n(3n+1)))x^n ((f(x))/3) =Σ_(n=1) ^∞ (((−1)^(n−1) )/(3n(3n+1))) x^n = Σ_(n=1) ^∞ (−1)^(n−1) { (1/(3n)) −(1/(3n+1))}x^n =(1/3) Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n −Σ_(n=1) ^∞ (((−1)^(n−1) )/(3n+1)) x^n but Σ_(n=1) ^∞ (((−1)^n )/n) x^n =ln(1+x) let w(x)=Σ_(n=1) ^∞ (((−1)^(n−1) )/(3n+1)) x^n if 0<x<1 w(x)= (1/x^(1/3) ) Σ_(n=1) ^∞ (((−1)^(n−1) )/(3n+1)) (^3 (√x))^(3n+1) =^3 ((√x))^(−1) ϕ(^3 (√x)) with ϕ(x)=Σ_(n=1) ^∞ (((−1)^(n−1) )/(3n+1)) x^(3n+1) ⇒ϕ^′ (x)=Σ_(n=1) ^∞ (−1)^(n−1) x^(3n) =Σ_(n=0) ^∞ (−1)^n x^(3n+3) =x^3 (1/(1+x^3 )) =(x^3 /(1+x^3 )) ⇒ϕ(x)=∫_0 ^x (t^3 /(1+t^3 ))dt +c c=ϕ(o)=0 ⇒ ϕ(x)= ∫_0 ^x (t^3 /(1+t^3 ))dt =∫_0 ^x ((1+t^3 −1)/(1+t^3 ))dt = x −∫_0 ^x (dt/(1+t^3 )) let decompose F(t)= (1/(1+t^3 )) F(t)=(1/((t+1)(t^2 −t+1))) = (a/(t+1)) +((bt +c)/(t^2 −t +1)) a=lim_(t→−1) (t+1)ϕ(t)=(1/3) lim_(t→+∞) t ϕ(t) = 0 =a+b ⇒b=−(1/3) ⇒F(t)= (1/(3(t+1))) +((−(1/3)t +c)/(t^2 −t+1)) F(0) =1 = (1/3) +c ⇒c=(2/3) ⇒F(x)= (1/(3(t+1))) +(1/3) ((−t +1)/(t^2 −t +1)) ∫_0 ^x F(t)dt =(1/3) ∫_0 ^x (dt/(t+1)) −(1/6) ∫_0 ^x ((2t−1−1)/(t^2 −t +1))dt = [ (1/3)ln∣t+1∣−(1/6)ln(∣t^2 −t+1∣]_0 ^x +(1/6) ∫_0 ^x (dt/(t^2 −t +1)) =[ (1/6)ln((((t+1)^2 )/(t^2 −t+1)))]_0 ^x +(1/6) ∫_0 ^x (dt/(t^2 −t +1)) =(1/6)ln(((x^2 +2x+1)/(x^2 −x+1))) +I I =(1/6) ∫_0 ^x (dt/(t^2 −2 (t/2) +(1/4) +(3/4))) ⇒6I = ∫_0 ^x (dt/((t−(1/2))^2 +(3/4))) =_(t−(1/2) =((√3)/2)u) (4/3)∫_(−(1/(√3))) ^((2x−1)/(√3)) (1/(1+u^2 )) ((√3)/2) du = (2/(√3)) [arctan (((2x−1)/(√3))) +arctan((1/(√3)))} ⇒ ∫_0 ^x F(t)dt = (1/6)ln(((x^2 +2x+1)/(x^2 −x+1))) +(1/(3(√3))){ arctan(((2x−1)/(√3))) +arctan((1/(√3)))} ϕ(x) =x −(1/6)ln(((x^2 +2x+1)/(x^2 −x+1))) −(1/(3(√3))){ arctan(((2x−1)/(√3)))+(π/6)} w(x)= (1/((^3 (√x)))) ϕ(^3 (√x)) ⇒ ((f(x))/3) =(1/3)ln(1+x) −w(x) ⇒ f(x) =ln(1+x)−3w(x) .$
$1) w e h a v e f (x) = \int_{0}^{1} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} t^{3 n}) d t$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} \frac{1}{3 n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n (3 n + 1)} x^{n}$ $\frac{f (x)}{3} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3 n (3 n + 1)} x^{n} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} {\frac{1}{3 n} - \frac{1}{3 n + 1}} x^{n}$ $= \frac{1}{3} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3 n + 1} x^{n} b u t$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} x^{n} = l n (1 + x) l e t w (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3 n + 1} x^{n}$ $i f 0 < x < 1 w (x) = \frac{1}{x^{\frac{1}{3}}} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3 n + 1} {(^{3} \sqrt{x})}^{3 n + 1} =^{3} {(\sqrt{x})}^{- 1} φ (^{3} \sqrt{x}) w i t h$ $φ (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{3 n + 1} x^{3 n + 1} \Rightarrow φ^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} x^{3 n}$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{3 n + 3} = x^{3} \frac{1}{1 + x^{3}} = \frac{x^{3}}{1 + x^{3}} \Rightarrow φ (x) = \int_{0}^{x} \frac{t^{3}}{1 + t^{3}} d t + c$ $c = φ (o) = 0 \Rightarrow φ (x) = \int_{0}^{x} \frac{t^{3}}{1 + t^{3}} d t = \int_{0}^{x} \frac{1 + t^{3} - 1}{1 + t^{3}} d t$ $= x - \int_{0}^{x} \frac{d t}{1 + t^{3}} l e t d e c o m p o s e F (t) = \frac{1}{1 + t^{3}}$ $F (t) = \frac{1}{(t + 1) (t^{2} - t + 1)} = \frac{a}{t + 1} + \frac{b t + c}{t^{2} - t + 1}$ $a = {l i m}_{t \to - 1} (t + 1) φ (t) = \frac{1}{3}$ ${l i m}_{t \to + \infty} t φ (t) = 0 = a + b \Rightarrow b = - \frac{1}{3} \Rightarrow F (t) = \frac{1}{3 (t + 1)} + \frac{- \frac{1}{3} t + c}{t^{2} - t + 1}$ $F (0) = 1 = \frac{1}{3} + c \Rightarrow c = \frac{2}{3} \Rightarrow F (x) = \frac{1}{3 (t + 1)} + \frac{1}{3} \frac{- t + 1}{t^{2} - t + 1}$ $\int_{0}^{x} F (t) d t = \frac{1}{3} \int_{0}^{x} \frac{d t}{t + 1} - \frac{1}{6} \int_{0}^{x} \frac{2 t - 1 - 1}{t^{2} - t + 1} d t$ $= [\frac{1}{3} l n ∣ t + 1 ∣ - \frac{1}{6} l n {(∣ t^{2} - t + 1 ∣]}_{0}^{x} + \frac{1}{6} \int_{0}^{x} \frac{d t}{t^{2} - t + 1}$ $= {[\frac{1}{6} l n (\frac{{(t + 1)}^{2}}{t^{2} - t + 1})]}_{0}^{x} + \frac{1}{6} \int_{0}^{x} \frac{d t}{t^{2} - t + 1} = \frac{1}{6} l n (\frac{x^{2} + 2 x + 1}{x^{2} - x + 1}) + I$ $I = \frac{1}{6} \int_{0}^{x} \frac{d t}{t^{2} - 2 \frac{t}{2} + \frac{1}{4} + \frac{3}{4}} \Rightarrow 6 I = \int_{0}^{x} \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{3}{4}}$ $=_{t - \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int_{- \frac{1}{\sqrt{3}}}^{\frac{2 x - 1}{\sqrt{3}}} \frac{1}{1 + u^{2}} \frac{\sqrt{3}}{2} d u$ $= \frac{2}{\sqrt{3}} [a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + a r c t a n (\frac{1}{\sqrt{3}})} \Rightarrow$ $\int_{0}^{x} F (t) d t = \frac{1}{6} l n (\frac{x^{2} + 2 x + 1}{x^{2} - x + 1}) + \frac{1}{3 \sqrt{3}} {a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + a r c t a n (\frac{1}{\sqrt{3}})}$ $φ (x) = x - \frac{1}{6} l n (\frac{x^{2} + 2 x + 1}{x^{2} - x + 1}) - \frac{1}{3 \sqrt{3}} {a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + \frac{π}{6}}$ $w (x) = \frac{1}{(^{3} \sqrt{x})} φ (^{3} \sqrt{x}) \Rightarrow \frac{f (x)}{3} = \frac{1}{3} l n (1 + x) - w (x) \Rightarrow$ $f (x) = l n (1 + x) - 3 w (x) .$
Commented by maxmathsup by imad last updated on 30/Jun/18

$2) l e t I = \int_{0}^{1} l n (2 + t^{3}) d t$ $I = \int_{0}^{1} l n (2 (1 + \frac{1}{2} t^{3})) d t = l n (2) + \int_{0}^{1} l n (1 + \frac{1}{2} t^{3}) d t$ $= l n (2) + f (\frac{1}{2}) = l n (2) + l n (\frac{3}{2}) - 3 w (\frac{1}{2})$ $= l n (2) + l n (\frac{3}{2}) - 3^{3} \sqrt{2} φ (\frac{1}{(^{3} \sqrt{2})}) t h e v a l u e o f φ (x) i s k n o w n .$
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