I would like to solve $\int_{0}^{\infty} \lfloor x\rfloor e^{-x} dx$ for its exact solution. This was on a previous GRE Mathematics exam.
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1No, it's $\frac{1}{e-1}$ – Adam Staples May 23 '14 at 07:01
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(And this is a definite integral) – davidlowryduda May 23 '14 at 07:04
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Fixed, I meant improper I suppose. – Adam Staples May 23 '14 at 07:05
2 Answers
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Your integral is equal to $\sum\limits_{n=0}^\infty{\int_n^{n+1}{ne^{-x}dx}}$
You can simplify from here.
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Then $\left.\sum_{n=0}^{\infty} -ne^{-x}\right|n^{n+1}$ = $\sum{n=0}^{\infty} ne^{-n} (1-\frac{1}{e})$ – Adam Staples May 23 '14 at 07:10
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http://math.stackexchange.com/a/525040/99024 This gives a possible way of computing the sum above. – Adam Staples May 23 '14 at 07:31
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Break the integral up into a sum of integrals over intervals of length 1. This way the floor of $x$ becomes constant for that interval. You get
$$\int_0^\infty = \sum_{k=0}^{\infty}k\left(\int_k^{k+1}e^{-x}dx\right) = \sum_{k=0}^{\infty}k\left(-e^{-(k+1)} + e^{-k}\right)$$
Breaking up this sum into two we get
$$\sum_{k=0}^{\infty}k\left(-e^{-(k+1)} + e^{-k}\right) = -\sum_{l=1}^{\infty}(l-1)e^{-l} + \sum_{m=0}^{\infty}me^{-m}$$
Since the first term of the second sum is clearly zero, we can start the summation from 1 instead of zero to get
$$-\sum_{l=1}^{\infty}(l-1)e^{-l} + \sum_{m=0}^{\infty}me^{-m} = -\sum_{l=1}^{\infty}(l-1)e^{-l} + \sum_{m=1}^{\infty}me^{-m}$$
Now we're able to combine these back into one sum to get
$$\sum_{n=1}^{\infty}\left( (1-n)e^{-n} + ne^{-n} \right) = \sum_{n=1}^{\infty}e^{-n},$$
which is a power series. This finally gives us
$$\sum_{n=1}^{\infty}e^{-n} = \sum_{n=0}^{\infty}e^{-n} - e = \frac{1}{1- 1/e} - e = \frac{e}{1- 1/e}= \frac{1}{e- 1}$$
I'm sure there's a shorter way of doing this whole thing but I'm not thinking very clearly right now.
Phonon
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How do you get the equality when you break up the sums? I see that the second sum corresponds to the first term in the parenthesis, but what about the first sum. How come $ke^{-(k+1)}=(k-1)e^{-k}$? – DanZimm May 23 '14 at 07:21
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The right answer should be $\frac{1}{e-1}$. I'm attempting to fix the error. – Adam Staples May 23 '14 at 07:22
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math.stackexchange.com/a/525040/99024 This gives a possible way of computing the sum above. Man being this clever on a GRE timed Math exam is difficult for sure. – Adam Staples May 23 '14 at 07:32
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@AdamStaples I'm starting to study for the exam too, I feel the same way >.< – DanZimm May 23 '14 at 07:39
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Basically Once you have $\sum_{n=0}^{\infty} ne^{-n}$, we're done. This can be done by realizing $\sum_{n=0}^{\infty} ne^{-nx}$ is the derivative of $\sum_{n=0}^{\infty} e^{-nx} = \frac{1}{1-e^{-x}}$. Integrating both sides and then plugging in x = 1 gives the desired result. – Adam Staples May 23 '14 at 07:42
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@AdamStaples Got it. Power series was starting summation at 1, not 0. All correct now. – Phonon May 23 '14 at 07:53