$\int_{0}^{\infty} [x]e^{-x}dx$ could any one tell me how to solve this integration? It was a question from a prev year of my institute qual: Thank you for help.
$$\int_{0}^{\infty} [x]e^{-x}dx$$

\begin{align}
\int_0^{\infty} \lfloor x \rfloor e^{-x} dx & = \sum_{n=0}^{\infty} \int_n^{n+1} \lfloor x \rfloor e^{-x} dx = \sum_{n=0}^{\infty} \int_n^{n+1} n e^{-x} dx = \sum_{n=0}^{\infty} n \left(e^{-n} - e^{-(n+1)}\right)\\
& = (e^{-1} - e^{-2}) + (2e^{-2} - 2e^{-3}) + (3e^{-3} - 4e^{-4}) + \cdots\\
& = \dfrac1e + \dfrac1{e^2} + \dfrac1{e^3} + \cdots = \dfrac{\dfrac1e}{1-\dfrac1e} = \dfrac1{e-1}
\end{align}
