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Question Number 28073 by abdo imad last updated on 20/Jan/18

$$find{\int }_0^1{e}^{-2x}ln(1+t{e}^{-x})dxwith0<t<1.$$

Commented by abdo imad last updated on 26/Jan/18

$$letputf\left(t\right)={\int }_0^1{e}^{-2x}ln(1+t{e}^{-x})dx$$$$f^{{}^{\prime }}\left(t\right)={\int }_0^1{e}^{-2x}\frac{e^{-x}}{1+t{e}^{-x}}dx={\int }_0^1{e}^{-3x}\left(\sum _{n=0}^{\propto }{(-1)}^n{t}^n{e}^{-nx}\right)dx$$$$=\sum _{n=0}^{\propto }{(-1)}^n{t}^n{\int }_0^1{e}^{-(n+3)x}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{t}^n}{-(n+3)}{\left[e^{-(n+3)x}\right]}_0^1$$$$=-\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{t}^n}{n+3}(e^{-(n+3)}-1)$$$$=\sum _{n=0}^{+\mathrm{\infty }}\frac{{(-1)}^n}{n+3}{t}^n-\sum _{n=0}^{+\mathrm{\infty }}\frac{{(-1)}^n{t}^n{e}^{-(n+3)}}{n+3}$$$$\Rightarrow f\left(t\right)={\int }_0^t(....)du+\lambda$$$$=\sum _{n=0}^{\propto }\frac{{(-1)}^n}{(n+1)(n+3)}{t}^{n+1}-\sum _{n.0}^{\propto }\frac{{(-1)}^n{t}^{n+1}{e}^{-(n+3)}}{(n+1)(n+3)}+\lambda$$$$\lambda =f\left(0\right)=0and$$$$f\left(x\right)=\frac{1}{2}\sum _{n=0}^{\propto }{(-1)}^n(\frac{1}{n+1}-\frac{1}{n+3})t^{n+1}$$$$-\frac{e^{-2}}{2}\sum _{n=0}^{\propto }{(-1)}^n(\frac{1}{n+1}-\frac{1}{n+3}){\left({te}^{-1}\right)}^{n+1}$$$$=\frac{1}{2}\sum _{n=0}^{\propto }\frac{{(-1)}^n}{n+1}{t}^{n+1}-\frac{1}{2}\sum _{n=0}^{\propto }\frac{{(-1)}^n{t}^{n+1}}{n+3}$$$$-\frac{e^{-2}}{2}\sum _{n=0}^{\propto }\frac{{(-1)}^n{\left({te}^{-1}\right)}^{n+1}}{n+1}+\frac{e^{-2}}{2}\sum _{n=0}^{\propto }\frac{{(-1)}^n{\left(t{e}^{-1}\right)}^{n+1}}{n+3}$$$$andallthosesumarecalculable...becontiued.$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

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