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Question Number 60701 by maxmathsup by imad last updated on 24/May/19

$$letf\left(x\right)=e^{-x^2}ln(2-x)developpfatintegrserie.$$

Commented by maxmathsup by imad last updated on 27/May/19

$$wehavef\left(x\right)=e^{-x^2}ln\left(2(1-\frac{x}{2})\right)=ln\left(2\right)e^{-x^2}+e^{-x^2}ln(1-\frac{x}{2})$$$$if\mid \frac{x}{2}\mid <1\Rightarrow -2<x<2weget{e}^{-x^2}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-x^2)}^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n}}{n!}$$$$ln(1+u)=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{u^n}{n}\Rightarrow ln(1-u)=-\sum _{n=1}^{\mathrm{\infty }}\frac{u^n}{n}\Rightarrow$$$$ln(1-\frac{x}{2})=-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n{2}^n}\Rightarrow f\left(x\right)=ln\left(2\right)\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n}}{n!}$$$$-\left(\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n}}{n!}\right)\left(\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n{2}^n}\right)$$$$=ln\left(2\right)\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n}}{n!}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n{2}^n}+\left(\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}{x}^{2n}\right)\left(\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n{2}^n}{x}^n\right)$$$$let{a}_n=\frac{{(-1)}^n}{n!}{x}^{2n}and{b}_n=\frac{1}{n{2}^n}{x}^n\Rightarrow$$$$\left(\sum _{n=1}^{\mathrm{\infty }}(...)\right)\left(\mathrm{\Sigma }(...)\right)=\left(\sum _{n=1}^{\mathrm{\infty }}{a}_n\right)\left(\sum _{n=1}^{\mathrm{\infty }}{b}_n\right)=\sum _{n=1}^{\mathrm{\infty }}{c}_n$$$$c_n=\sum _{i+j=n}{a}_i{b}_j=\sum _{i=1}^n{a}_i{b}_{n-i}=\sum _{i=1}^n\frac{{(-1)}^i}{i!}{x}^{2i}\frac{1}{(n-i)2^{n-i}}{x}^{n-i}$$$$=\sum _{i=1}^n\frac{{(-1)}^i}{i!(n-i)2^{n-i}}{x}^{n+i}\Rightarrow$$$$f\left(x\right)=ln\left(2\right)\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n}}{n!}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n{2}^n}+\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{i=1}^n\frac{{(-1)}^i}{i!(n-i)!2^{n-i}}{x}^i\right)x^n$$$$....becontinued....$$

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