Improper integral of exponential function times the integer part of $x$

Question

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.

Answer 1

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

$$\sum_{n=1}^{\infty}e^{-n} = \sum_{n=0}^{\infty}e^{-n} - e = \frac{1}{1- 1/e} - e = \frac{e}{1- 1/e}= \frac{1}{e- 1}$$

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I'm sure there's a shorter way of doing this whole thing but I'm not thinking very clearly right now.

Answer 2

Your integral is equal to $\sum\limits_{n=0}^\infty{\int_n^{n+1}{ne^{-x}dx}}$

You can simplify from here.