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Question Number 27613 by abdo imad last updated on 10/Jan/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}{e}^{-\left[x\right]-x}dx.$$

Commented by abdo imad last updated on 12/Jan/18

$$I_n=\sum _{k=0}^n{\int }_k^{k+1}{e}^{-\left[x\right]-x}dxI=lim{I}_n$$$$I_n=\sum _{k=}^n{e}^{-k}{\left[e^{-x}\right]}_k^{k+1}=\sum _{k=0}^n{e}^{-k}(e^{-k}-e^{-k-1})$$$$I_n=\sum _{k=0}^n{e}^{-2k}-e^{-1}\sum _{k=0}^n{e}^{-2k}=(1-e^{-1})\sum _{k=0}^n{\left(e^{-2}\right)}^k$$$${\int }_0^{\mathrm{\infty }}{e}^{-\left[x\right]-x}dx={lim}_{n->\propto }{I}_n=(1-e^{-1}).\frac{1}{1-e^{-2}}$$$$=\frac{e^2(1-e^{-1})}{e^2-1}=\frac{e^2-e}{e^2-1}=\frac{e(e-1)}{(e+1)(e-1)}=\frac{e}{e+1}.$$

Answered by prakash jain last updated on 12/Jan/18

$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}{e}^{-i}{\int }_i^{i+1}{e}^{-x}dx$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}{e}^{-i}{[-e^{-x}]}_i^{i+1}$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{e^{-i}}{e^i}-\frac{e^{-i}}{e^{i+1}}$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{1}{e^{2i}}-\frac{1}{e^{2i+1}}$$$$=\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{1}{e^{2i}}-\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{1}{e^{2i+1}}$$$$=\frac{1}{1-\frac{1}{e^2}}-\frac{1/e}{1-\frac{1}{e^2}}$$$$=\frac{e(e-1)}{e^2-1}=\frac{e}{e+1}$$

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