find the value of h(t)=∫_0 ^1 ln(1+tx^2 ) with ∣t∣≤1 2) calculate ∫_0 ^1 ln(1+x^2 )dx 3) calculate ∫


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Question Number 36435 by prof Abdo imad last updated on 02/Jun/18

$f i n d t h e v a l u e o f h (t) = \int_{0}^{1} l n (1 + {t x}^{2}) w i t h ∣ t ∣⩽ 1$ $2) c a l c u l a t e \int_{0}^{1} l n (1 + x^{2}) d x$ $3) c a l c u l a t e \int_{0}^{1} l n (1 - x^{2}) d x$
Commented by abdo.msup.com last updated on 03/Jun/18

$a n o t h e r m e t h o d b u t e a s y$ $\int_{0}^{1} l n (1 - x^{2}) d x = \int_{0}^{1} l n (1 - x) d x +$ $\int_{0}^{1} l n (1 + x) d x b u t$ $\int_{0}^{1} l n (1 - x) d x =_{1 - x = t} - \int_{1}^{0} l n t d t$ $= \int_{0}^{1} l n (t) d t = {[t l n (t)]}_{0}^{1} = 0$ $\int_{0}^{1} l n (1 + x) d x =_{1 + x = t} \int_{1}^{2} l n (t) d t$ $= {[t l n (t)]}_{1}^{2} = 2 l n (2) s o$ $\int_{0}^{1} l n (1 - x^{2}) d x = 2 l n (2) .$
Commented by abdo.msup.com last updated on 03/Jun/18
$1)we have ∣tx^2 ∣≤1 ⇒ln(1+tx^2 ) =ln(1−(−tx^2 )) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)(tx^2 )^n =Σ_(n=1) ^∞ (((−1)^n )/n) t^(n−1) x^(2n) ⇒ h(t) = Σ_(n=1) ^∞ (((−1)^(n−1) t^n )/n) ∫_0 ^1 x^(2n) dx =Σ_(n=1) ^∞ (((−1)^(n−1) t^n )/(n(2n+1))) (1/2)h(t) =Σ_(n=1) ^∞ { (1/(2n))−(1/(2n+1))}(−1)^(n−1) t^n =(1/2) Σ_(n=1) ^∞ (((−1)^(n−1) )/n) t^n − Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n+1)) t^n but Σ_(n=1) ^∞ (((−1)^(n−1) )/n) t^n =ln(1+t) Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n+1)) t^n =(1/(√t)) Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n+1))((√t))^(2n+1) =(1/(√t)) w((√t)) with w(x) =Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n+1))x^(2n+1) w^′ (x) = Σ_(n=1) ^∞ (−1)^(n−1) x^(2n) = Σ_(n=0) ^∞ (−1)^n x^(2n +2) = x^2 (1/(1+x^2 )) ⇒ w(x) = ∫ (x^2 /(1+x^2 )) dx +λ =x −arctanx +λ but λ =w(0)=0 ⇒ w(x)=x −arctan(x) and Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n+1)) t^n = (1/(√t))w((√t)) =(1/(√t)){ (√t) −arctan((√t)) so ((h(t))/2) = (1/2)ln(1+t) −1 +((arctan((√t)))/(√t)) ⇒ h(t) =ln(1+t) +((2arctan((√t)))/(√t)) −2$
$1) w e h a v e ∣ {t x}^{2} ∣⩽ 1 \Rightarrow l n (1 + {t x}^{2})$ $= l n (1 - (- {t x}^{2})) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {({t x}^{2})}^{n}$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} t^{n - 1} x^{2 n} \Rightarrow$ $h (t) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} t^{n}}{n} \int_{0}^{1} x^{2 n} d x$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} t^{n}}{n (2 n + 1)}$ $\frac{1}{2} h (t) = \sum_{n = 1}^{\infty} {\frac{1}{2 n} - \frac{1}{2 n + 1}} {(- 1)}^{n - 1} t^{n}$ $= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} t^{n} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{2 n + 1} t^{n}$ $b u t \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} t^{n} = l n (1 + t)$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{2 n + 1} t^{n} = \frac{1}{\sqrt{t}} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{2 n + 1} {(\sqrt{t})}^{2 n + 1}$ $= \frac{1}{\sqrt{t}} w (\sqrt{t}) w i t h w (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{2 n + 1} x^{2 n + 1}$ $w^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} x^{2 n}$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n + 2} = x^{2} \frac{1}{1 + x^{2}} \Rightarrow$ $w (x) = \int \frac{x^{2}}{1 + x^{2}} d x + λ$ $= x - a r c t a n x + λ b u t λ = w (0) = 0 \Rightarrow$ $w (x) = x - a r c t a n (x) a n d$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{2 n + 1} t^{n} = \frac{1}{\sqrt{t}} w (\sqrt{t})$ $= \frac{1}{\sqrt{t}} {\sqrt{t} - a r c t a n (\sqrt{t}) s o$ $\frac{h (t)}{2} = \frac{1}{2} l n (1 + t) - 1 + \frac{a r c t a n (\sqrt{t})}{\sqrt{t}} \Rightarrow$ $h (t) = l n (1 + t) + \frac{2 a r c t a n (\sqrt{t})}{\sqrt{t}} - 2$
Commented by abdo.msup.com last updated on 03/Jun/18

$w e h a v e p r o v e d t h a t$ $\int_{0}^{1} l n (1 + {t x}^{2}) d x = l n (1 + t) + \frac{2 a r c t a n (\sqrt{t})}{\sqrt{t}} - 2$ $f o r t = 1 w e g e t$ $\int_{0}^{1} l n (1 + x^{2}) d x = l n (2) + 2 \frac{π}{4} - 2$ $= l n (2) + \frac{π}{2} - 2 .$
Commented by abdo.msup.com last updated on 03/Jun/18

$3) w e h a v e {l n}^{^{'}} (1 - u) = \frac{- 1}{1 - u}$ $= - \sum_{n = 0}^{\infty} u^{n} \Rightarrow l n (1 - u) = - \sum_{n = 0}^{\infty} \frac{u^{n + 1}}{n + 1} + c$ $= - \sum_{n = 1}^{\infty} \frac{u^{n}}{n} = \sum_{n = 1}^{\infty} \frac{u^{n - 1}}{n} (c = 0) \Rightarrow$ $l n (1 - x^{2}) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 2}}{n} \Rightarrow$ $\int_{0}^{1} l n (1 - x^{2}) d x = \sum_{n = 1}^{\infty} \frac{1}{n (2 n - 1)} = A$ $\frac{A}{2} = \sum_{n = 1}^{\infty} \frac{1}{2 n (2 n - 1)}$ $= \sum_{n = 1}^{\infty} (\frac{1}{2 n - 1} - \frac{1}{2 n}) =$ ${l i m}_{n \to + \infty} \sum_{k = 1}^{n} (\frac{1}{2 k - 1} - \frac{1}{2 k}) b u t$ $\sum_{k = 1}^{n} \frac{1}{2 k - 1} = 1 + \frac{1}{3} + \frac{1}{5} + . . . + \frac{1}{2 n - 1}$ $= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . . . . + \frac{1}{2 n - 1} + \frac{1}{2 n}$ $- \frac{1}{2} - \frac{1}{4} - . . . . - \frac{1}{2 n} = H_{2 n} - \frac{1}{2} H_{n}$ $\sum_{k = 1}^{n} \frac{1}{2 k - 1} - \frac{1}{2} \sum_{k = 1}^{\infty} \frac{1}{k}$ $= H_{2 n} - \frac{1}{2} H_{n} - \frac{1}{2} H_{n} = H_{2 n} - H_{n}$ $= l n (2 n) + γ + o (\frac{1}{n}) - l n (n) - δ - o (\frac{1}{n})$ $= l n (\frac{2 n}{n}) + o (\frac{1}{n}) \to l n (2) (n \to + \infty) s o$ $s o \int_{0}^{1} l n (1 - x^{2}) d x = 2 l n 2 .$
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