GRE question: Evaluate $\int_0^\infty \left \lfloor x \right \rfloor e^{-x} \,\mathrm{d}x$

Question

I have the following GRE question that I have no idea how to solve.

Let $\left \lfloor x \right \rfloor$ denote the greatest integer not exceeding $x$. Evaluate $\int_0^\infty \left \lfloor x \right \rfloor e^{-x} \,\mathrm{d}x$.

The answer says it should be $\frac{1}{e-1}$, and a hint that they give is

$$ \int\limits_0^\infty \left \lfloor x \right \rfloor e^{-x} \,\mathrm{d}x = \sum_{n=1}^\infty \int\limits_n^{n+1} ne^{-x} \,\mathrm{d}x\,. $$

I don't really see how we go from the integral to the summation.

Answer 1

First note that $\int_0^{\infty}\lfloor x\rfloor e^{-x}\;dx=\sum_{n=0}^{\infty}\int_{n}^{n+1}\lfloor x\rfloor e^{-x}\;dx$ by the additivity of the integral.

This decomposition is useful because $\lfloor x\rfloor$ is equal to $n$ on the interval $[n,n+1)$. So we can rewrite the integral as $$ \sum_{n=0}^{\infty}\int_{n}^{n+1}\lfloor x\rfloor e^{-x}\;dx=\sum_{n=0}^{\infty}\int_{n}^{n+1}n e^{-x}\;dx$$ Finally, the sum can be changed to start at $n=1$ if we want, since the $n=0$ term is zero anyway.

Answer 2

Note that $$ \int_0^\infty \lfloor x \rfloor e^{-x}dx = \int_0^1 \lfloor x \rfloor e^{-x}dx + \int_1^2 \lfloor x \rfloor e^{-x}dx + \int_2^3 \lfloor x \rfloor e^{-x}dx + \cdots =\\ \sum_{n=0}^\infty \int_{n}^{n+1} \lfloor x \rfloor e^{-x}dx $$ and from here it should be clear.

Answer 3

Probabilistically, this is asking about $\mathbf{E}[\lfloor X\rfloor]$ where $X\sim \mathrm{Exp}(1)$.

It is well known that if $X\sim \mathrm{Exp}(\lambda)$, then $\lfloor X \rfloor$ is geometrically distributed which value is starting from $0$. Let $n\geq 0$ be a given nonnegative integer. Then we have $$ P(\lfloor X \rfloor \geq n) = P(X\geq n) = e^{-\lambda n}. $$

Thus, it follows that $\lfloor X \rfloor$ is geometrically distributed with parameter $p=1-e^{-\lambda}$ which starts with value $0$. This has expectation $\frac 1p -1$.

In the problem we have $\lambda =1$. Thus, the answer is $$ \mathbf{E}[ \lfloor X \rfloor ] = \frac { e^{-1} }{1-e^{-1} } = \frac 1{e-1}. $$