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


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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 byabdo.msup.com last updated on 01/Jul/18
$we have ln^′ (1−u)=((−1)/(1−u)) =−Σ_(n=0) ^∞ u^n ⇒ln(1−u) =−Σ_(n=0) ^∞ (u_(n+1) /(n+1)) =−Σ_(n=1) ^∞ (u^n /n) ⇒ln(1−x^2 t^2 )=−Σ_(n=1) ^∞ ((x^(2n) t^(2n) )/n) ⇒((ln(1−x^2 t^2 ))/t^2 ) =−Σ_(n=1) ^∞ ((x^(2n) t^(2n−2) )/n) ⇒ f(x)=−∫_0 ^1 {Σ_(n=1) ^∞ ((x^(2n) t^(2n−2) )/n)}dt =−Σ_(n=1) ^∞ (x^(2n) /n) ∫_0 ^1 t^(2n−2) dt =−Σ_(n=1) ^∞ (x^(2n) /(n(2n−1))) ⇒ ((f(x))/2) =−Σ_(n=1) ^∞ (x^(2n) /(2n(2n−1))) =−Σ_(n=1) ^∞ ( (1/(2n−1)) −(1/(2n)))x^(2n) =Σ_(n=1) ^∞ (x^(2n) /(2n)) −Σ_(n=1) ^∞ (x^(2n) /(2n−1)) but Σ_(n=1) ^∞ (x^(2n) /(2n)) =−(1/2)ln(1−x^2 ) let w(x)=Σ_(n=1) ^∞ (x^(2n) /(2n−1)) w(x)=x Σ_(n=1) ^∞ (x^(2n−1) /(2n−1)) =xϕ(x) ϕ^′ (x) =Σ_(n=1) ^∞ x^(2n−2) =Σ_(n=1) ^∞ (x^2 )^(n−1) =Σ_(n=0) ^∞ (x^2 )^n =(1/(1−x^2 )) ⇒ϕ(x)=∫ (dx/(1−x^2 )) +c =(1/2)∫ ( (1/(1+x)) +(1/(1−x)))dx +c =(1/2)ln∣((1+x)/(1−x))∣ +c but c=ϕ(0) ⇒ w(x)=(x/2)ln∣((1+x)/(1−x))∣⇒ ((f(x))/2) =−(1/2)ln(1−x^2 )−(x/2)ln∣((1+x)/(1−x))∣ ⇒ f(x)=−ln(1−x^2 )−(x/2)ln∣((1+x)/(1−x))∣ .$
$w e h a v e {l n}^{^{'}} (1 - u) = \frac{- 1}{1 - u} = - \sum_{n = 0}^{\infty} u^{n}$ $\Rightarrow l n (1 - u) = - \sum_{n = 0}^{\infty} \frac{u_{n + 1}}{n + 1} = - \sum_{n = 1}^{\infty} \frac{u^{n}}{n}$ $\Rightarrow l n (1 - x^{2} t^{2}) = - \sum_{n = 1}^{\infty} \frac{x^{2 n} t^{2 n}}{n}$ $\Rightarrow \frac{l n (1 - x^{2} t^{2})}{t^{2}} = - \sum_{n = 1}^{\infty} \frac{x^{2 n} t^{2 n - 2}}{n} \Rightarrow$ $f (x) = - \int_{0}^{1} {\sum_{n = 1}^{\infty} \frac{x^{2 n} t^{2 n - 2}}{n}} d t$ $= - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{n} \int_{0}^{1} t^{2 n - 2} d t$ $= - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{n (2 n - 1)} \Rightarrow$ $\frac{f (x)}{2} = - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n (2 n - 1)}$ $= - \sum_{n = 1}^{\infty} (\frac{1}{2 n - 1} - \frac{1}{2 n}) x^{2 n}$ $= \sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n} - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n - 1} b u t$ $\sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n} = - \frac{1}{2} l n (1 - x^{2})$ $l e t w (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n - 1}$ $w (x) = x \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{2 n - 1} = x φ (x)$ $φ^{^{'}} (x) = \sum_{n = 1}^{\infty} x^{2 n - 2} = \sum_{n = 1}^{\infty} {(x^{2})}^{n - 1}$ $= \sum_{n = 0}^{\infty} {(x^{2})}^{n} = \frac{1}{1 - x^{2}} \Rightarrow φ (x) = \int \frac{d x}{1 - x^{2}} + c$ $= \frac{1}{2} \int (\frac{1}{1 + x} + \frac{1}{1 - x}) d x + c$ $= \frac{1}{2} l n ∣ \frac{1 + x}{1 - x} ∣ + c b u t c = φ (0) \Rightarrow$ $w (x) = \frac{x}{2} l n ∣ \frac{1 + x}{1 - x} ∣\Rightarrow$ $\frac{f (x)}{2} = - \frac{1}{2} l n (1 - x^{2}) - \frac{x}{2} l n ∣ \frac{1 + x}{1 - x} ∣ \Rightarrow$ $f (x) = - l n (1 - x^{2}) - \frac{x}{2} l n ∣ \frac{1 + x}{1 - x} ∣ .$
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