let f(x) =e^(−x^2 ) ln(1−x) developp f at integr serie.


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Question Number 65285 by mathmax by abdo last updated on 27/Jul/19

$l e t f (x) = e^{- x^{2}} l n (1 - x)$ $d e v e l o p p f a t i n t e g r s e r i e .$
Commented by mathmax by abdo last updated on 30/Jul/19
$we have e^(−x^2 ) =Σ_(n=0) ^∞ (((−x^2 )^n )/(n!)) =Σ_(n=0) ^∞ (((−1)^n x^(2n) )/(n!)) ln^′ (1−x) =−(1/(1−x)) =−Σ_(n=0) ^∞ x^n ⇒ln(1−x) =−Σ_(n=0) ^∞ (x^(n+1) /(n+1)) +c (c=0)⇒ln(1−x) =−Σ_(n=1) ^∞ (x^n /n) ⇒ f(x) =−(Σ_(n=0) ^∞ (((−1)^n x^(2n) )/(n!)))(Σ_(n=1) ^∞ (x^n /n)) =−(1+Σ_(n=1) ^∞ (((−1)^n x^(2n) )/(n!)))(Σ_(n=1) ^∞ (x^n /n)) −Σ_(n=1) ^∞ (x^n /n) −(Σ_(n=1) ^∞ (((−1)^n x^(2n) )/(n!)))(Σ_(n=1) ^∞ (x^n /n)) (Σ_(n=1) ^∞ a_n )(Σ_(n=1) ^∞ b_n ) =Σ c_n with c_n =Σ_(i+j=n) a_i b_j =Σ_(i=1) ^(n−1) a_i b_(n−i) =Σ_(i=1) ^(n−1) (((−1)^i x^(2i) )/(i!)) (x^(n−i) /((n−i))) =Σ_(i=1) ^(n−1) (((−1)^i )/((n−i)i!)) x^(n+i) ⇒ f(x) =Σ_(n=1) ^∞ {Σ_(i=1) ^(n−1) (((−1)^i )/((n−i)i!))x^(n+i) }−Σ_(n=1) ^∞ (x^n /n) .$
$w e h a v e e^{- x^{2}} = \sum_{n = 0}^{\infty} \frac{{(- x^{2})}^{n}}{n!} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{n!}$ ${l n}^{^{'}} (1 - x) = - \frac{1}{1 - x} = - \sum_{n = 0}^{\infty} x^{n} \Rightarrow l n (1 - x) = - \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} + c$ $(c = 0) \Rightarrow l n (1 - x) = - \sum_{n = 1}^{\infty} \frac{x^{n}}{n} \Rightarrow$ $f (x) = - (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{n!}) (\sum_{n = 1}^{\infty} \frac{x^{n}}{n})$ $= - (1 + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{n!}) (\sum_{n = 1}^{\infty} \frac{x^{n}}{n})$ $- \sum_{n = 1}^{\infty} \frac{x^{n}}{n} - (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{n!}) (\sum_{n = 1}^{\infty} \frac{x^{n}}{n})$ $(\sum_{n = 1}^{\infty} a_{n}) (\sum_{n = 1}^{\infty} b_{n}) = Σ c_{n} w i t h c_{n} = \sum_{i + j = n} a_{i} b_{j} = \sum_{i = 1}^{n - 1} a_{i} b_{n - i}$ $= \sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i} x^{2 i}}{i!} \frac{x^{n - i}}{(n - i)} = \sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i}}{(n - i) i!} x^{n + i} \Rightarrow$ $f (x) = \sum_{n = 1}^{\infty} {\sum_{i = 1}^{n - 1} \frac{{(- 1)}^{i}}{(n - i) i!} x^{n + i}} - \sum_{n = 1}^{\infty} \frac{x^{n}}{n} .$
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