Math GRE - #14

If $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$ (i.e. the floor of x), then which of the choice below is the value of \[\int_0^\infty{\lfloor x\rfloor e^{-x}dx}?\]

• $\dfrac{e}{e^2-1}$


• $\dfrac{1}{e-1}$


• $\dfrac{e-1}{e}$


• $1$


• $\infty$

Solution :

Since the floor function is discrete, we can turn this integral into a discrete sum of integrals by noting that:
\[\begin{eqnarray*}
\int_0^\infty{\lfloor x\rfloor e^{-x}dx} & = & \int_0^1{0\cdot e^{-x}dx}+\int_1^2{1\cdot e^{-x}dx}+\int_2^3{2\cdot e^{-x}dx}+\cdots \\
& = & \sum_{n=0}^\infty{\int_n^{n+1}{n\cdot e^{-x}dx}}
\end{eqnarray*}\]
The sum above turns out to be telescoping, as we can see that:
\[\begin{eqnarray*}
\sum_{n=0}^\infty{\int_n^{n+1}{n\cdot e^{-x}dx}} & = & \sum_{n=0}^\infty{n\left(e^{-n}-e^{-(n+1)}\right)} \\
& = & \sum_{n=0}^\infty{n\cdot e^{-n}}-\sum_{n=0}^\infty{n\cdot e^{-(n+1)}} \\
& = & \sum_{n=0}^\infty{n\cdot e^{-n}}-\sum_{n=0}^\infty{(n+1)\cdot e^{-(n+1)}}+\sum_{n=0}^\infty {e^{-(n+1)}} \\
& = & \sum_{n=0}^\infty {e^{-(n+1)}}
\end{eqnarray*}\]
Where the last line is due to adjusting the summation indices on the previous line. We can also see that \[\sum_{n=0}^\infty {e^{-(n+1)}}=\frac{1}{e-1}\] as the above is a geometric series.

Therefore, the answer is: \[\int_0^\infty{\lfloor x\rfloor e^{-x}dx}=\sum_{n=0}^\infty{\int_n^{n+1}{n\cdot e^{-x}dx}}=\frac{1}{e-1}\]