Pages

News: Currently the LaTeX and hidden solutions on this blog do not work on Google Reader.
Email me if you have suggestions on how to improve this blog!

Monday 15 August 2011

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

2 comments:

Post a Comment

This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.