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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}
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Did you recently update your age? Happy birthday! And thanks for the questions.
@6237915540123150851.0 Yes I did! Haha. Very observant of you. My birthday was a couple days ago. Thanks!
2 comments:
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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.
Yes I did! Haha. Very observant of you. My birthday was a couple days ago. Thanks!
Did you recently update your age? Happy birthday! And thanks for the questions.
Yes I did! Haha. Very observant of you. My birthday was a couple days ago. Thanks!
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