let f(x)= ∫_0 ^1 ((ln(1−x^2 t^2 ))/t^2 )dt with ∣x∣<1 find f(x) at a simple form .


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 38464 by maxmathsup by imad last updated on 25/Jun/18

$l e t f (x) = \int_{0}^{1} \frac{l n (1 - x^{2} t^{2})}{t^{2}} d t w i t h ∣ x ∣< 1$ $f i n d f (x) a t a s i m p l e f o r m .$
Commented by abdo.msup.com last updated on 01/Jul/18
$we have ln^′ (1−u)=((−1)/(1−u)) =−Σ_(n=0) ^∞ u^n ⇒ln(1−u) =−Σ_(n=0) ^∞ (u_(n+1) /(n+1)) =−Σ_(n=1) ^∞ (u^n /n) ⇒ln(1−x^2 t^2 )=−Σ_(n=1) ^∞ ((x^(2n) t^(2n) )/n) ⇒((ln(1−x^2 t^2 ))/t^2 ) =−Σ_(n=1) ^∞ ((x^(2n) t^(2n−2) )/n) ⇒ f(x)=−∫_0 ^1 {Σ_(n=1) ^∞ ((x^(2n) t^(2n−2) )/n)}dt =−Σ_(n=1) ^∞ (x^(2n) /n) ∫_0 ^1 t^(2n−2) dt =−Σ_(n=1) ^∞ (x^(2n) /(n(2n−1))) ⇒ ((f(x))/2) =−Σ_(n=1) ^∞ (x^(2n) /(2n(2n−1))) =−Σ_(n=1) ^∞ ( (1/(2n−1)) −(1/(2n)))x^(2n) =Σ_(n=1) ^∞ (x^(2n) /(2n)) −Σ_(n=1) ^∞ (x^(2n) /(2n−1)) but Σ_(n=1) ^∞ (x^(2n) /(2n)) =−(1/2)ln(1−x^2 ) let w(x)=Σ_(n=1) ^∞ (x^(2n) /(2n−1)) w(x)=x Σ_(n=1) ^∞ (x^(2n−1) /(2n−1)) =xϕ(x) ϕ^′ (x) =Σ_(n=1) ^∞ x^(2n−2) =Σ_(n=1) ^∞ (x^2 )^(n−1) =Σ_(n=0) ^∞ (x^2 )^n =(1/(1−x^2 )) ⇒ϕ(x)=∫ (dx/(1−x^2 )) +c =(1/2)∫ ( (1/(1+x)) +(1/(1−x)))dx +c =(1/2)ln∣((1+x)/(1−x))∣ +c but c=ϕ(0) ⇒ w(x)=(x/2)ln∣((1+x)/(1−x))∣⇒ ((f(x))/2) =−(1/2)ln(1−x^2 )−(x/2)ln∣((1+x)/(1−x))∣ ⇒ f(x)=−ln(1−x^2 )−(x/2)ln∣((1+x)/(1−x))∣ .$
$w e h a v e {l n}^{^{'}} (1 - u) = \frac{- 1}{1 - u} = - \sum_{n = 0}^{\infty} u^{n}$ $\Rightarrow l n (1 - u) = - \sum_{n = 0}^{\infty} \frac{u_{n + 1}}{n + 1} = - \sum_{n = 1}^{\infty} \frac{u^{n}}{n}$ $\Rightarrow l n (1 - x^{2} t^{2}) = - \sum_{n = 1}^{\infty} \frac{x^{2 n} t^{2 n}}{n}$ $\Rightarrow \frac{l n (1 - x^{2} t^{2})}{t^{2}} = - \sum_{n = 1}^{\infty} \frac{x^{2 n} t^{2 n - 2}}{n} \Rightarrow$ $f (x) = - \int_{0}^{1} {\sum_{n = 1}^{\infty} \frac{x^{2 n} t^{2 n - 2}}{n}} d t$ $= - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{n} \int_{0}^{1} t^{2 n - 2} d t$ $= - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{n (2 n - 1)} \Rightarrow$ $\frac{f (x)}{2} = - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n (2 n - 1)}$ $= - \sum_{n = 1}^{\infty} (\frac{1}{2 n - 1} - \frac{1}{2 n}) x^{2 n}$ $= \sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n} - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n - 1} b u t$ $\sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n} = - \frac{1}{2} l n (1 - x^{2})$ $l e t w (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n - 1}$ $w (x) = x \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{2 n - 1} = x φ (x)$ $φ^{^{'}} (x) = \sum_{n = 1}^{\infty} x^{2 n - 2} = \sum_{n = 1}^{\infty} {(x^{2})}^{n - 1}$ $= \sum_{n = 0}^{\infty} {(x^{2})}^{n} = \frac{1}{1 - x^{2}} \Rightarrow φ (x) = \int \frac{d x}{1 - x^{2}} + c$ $= \frac{1}{2} \int (\frac{1}{1 + x} + \frac{1}{1 - x}) d x + c$ $= \frac{1}{2} l n ∣ \frac{1 + x}{1 - x} ∣ + c b u t c = φ (0) \Rightarrow$ $w (x) = \frac{x}{2} l n ∣ \frac{1 + x}{1 - x} ∣\Rightarrow$ $\frac{f (x)}{2} = - \frac{1}{2} l n (1 - x^{2}) - \frac{x}{2} l n ∣ \frac{1 + x}{1 - x} ∣ \Rightarrow$ $f (x) = - l n (1 - x^{2}) - \frac{x}{2} l n ∣ \frac{1 + x}{1 - x} ∣ .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum