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2022-10-02

Solve:

$${\int}_{0}^{\mathrm{\infty}}[x]{e}^{-x}dx$$

$${\int}_{0}^{\mathrm{\infty}}[x]{e}^{-x}dx$$

Helena Bentley

Beginner2022-10-03Added 10 answers

Answer:

$$\begin{array}{rl}{\int}_{0}^{\mathrm{\infty}}\lfloor x\rfloor {e}^{-x}dx& =\sum _{n=0}^{\mathrm{\infty}}{\int}_{n}^{n+1}\lfloor x\rfloor {e}^{-x}dx=\sum _{n=0}^{\mathrm{\infty}}{\int}_{n}^{n+1}n{e}^{-x}dx=\sum _{n=0}^{\mathrm{\infty}}n({e}^{-n}-{e}^{-(n+1)})\\ & =({e}^{-1}-{e}^{-2})+(2{e}^{-2}-2{e}^{-3})+(3{e}^{-3}-4{e}^{-4})+\cdots \\ & ={\displaystyle \frac{1}{e}}+{\displaystyle \frac{1}{{e}^{2}}}+{\displaystyle \frac{1}{{e}^{3}}}+\cdots ={\displaystyle \frac{{\displaystyle \frac{1}{e}}}{1-{\displaystyle \frac{1}{e}}}}={\displaystyle \frac{1}{e-1}}\end{array}$$

$$\begin{array}{rl}{\int}_{0}^{\mathrm{\infty}}\lfloor x\rfloor {e}^{-x}dx& =\sum _{n=0}^{\mathrm{\infty}}{\int}_{n}^{n+1}\lfloor x\rfloor {e}^{-x}dx=\sum _{n=0}^{\mathrm{\infty}}{\int}_{n}^{n+1}n{e}^{-x}dx=\sum _{n=0}^{\mathrm{\infty}}n({e}^{-n}-{e}^{-(n+1)})\\ & =({e}^{-1}-{e}^{-2})+(2{e}^{-2}-2{e}^{-3})+(3{e}^{-3}-4{e}^{-4})+\cdots \\ & ={\displaystyle \frac{1}{e}}+{\displaystyle \frac{1}{{e}^{2}}}+{\displaystyle \frac{1}{{e}^{3}}}+\cdots ={\displaystyle \frac{{\displaystyle \frac{1}{e}}}{1-{\displaystyle \frac{1}{e}}}}={\displaystyle \frac{1}{e-1}}\end{array}$$

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