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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:

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[hide] Raoul Golan said...

Did you recently update your age? Happy birthday! And thanks for the questions.

on 16 August 2011 at 00:15
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[hide] Paul Liu said...

Yes I did! Haha. Very observant of you. My birthday was a couple days ago. Thanks!

on 16 August 2011 at 07:38
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