# Subject GRE Exam 0568 Q.51

The question and its answer is given in the following pictures:

The first line in the solution is not clear for me, my questions are:

1- where is the interval from 0 to 1 ?

2- why we changed the x inside the floor function to n and did not change the x in the power of e ?

Could anyone illustrate this for me please?

• Please don't post images on the site. They don't load on some devices, take a long time to load on others, and are not searchable on the site. See this guide for writing Maths nicely on the site Jul 6 '17 at 11:45
• As long as I have time I will do what u said ..... thank u so much for your advice@lioness99a
– user426277
Jul 6 '17 at 12:04
• I see that this is a good resource,are those questions available nline? BTW what is this source you posted the image? Jul 6 '17 at 12:04
• @BAYMAX this test is answered by Charles Rambo and it is available online.
– user426277
Jul 6 '17 at 12:07

The solution is too complicated according to me.

A way to simplify,

\begin{align} J&=\sum_{n=1}^{\infty} \int_n^{n+1} ne^{-x}\\ &=\sum_{n=1}^{\infty} \Big[-ne^{-x}\Big]_n^{n+1}\\ &=\sum_{n=1}^{\infty}n\left(e^{-n}-e^{-(n+1)}\right)\\ &=\sum_{n=1}^{\infty}ne^{-n}-\sum_{n=1}^{\infty}ne^{-(n+1)}\\ &=\sum_{n=1}^{\infty}ne^{-n}-\sum_{n=1}^{\infty}(n+1)e^{-(n+1)}+\sum_{n=1}^{\infty}e^{-(n+1)}\\ &=\sum_{n=1}^{\infty}ne^{-n}-\sum_{n=2}^{\infty}ne^{-n}+\sum_{n=1}^{\infty}e^{-(n+1)}\\ &=e^{-1}+\sum_{n=1}^{\infty}e^{-(n+1)}\\ &=\sum_{n=0}^{\infty}e^{-(n+1)}\\ &=\frac{1}{\text{e}}\sum_{n=0}^{\infty}\left(e^{-1}\right)^n\\ &=\dfrac{1}{\text{e}}\frac{1}{1-\dfrac{1}{\text{e}}}\\ &=\frac{1}{\text{e}-1} \end{align}

For your first question: The first summand $\int_0^1 \lfloor x \rfloor e^{-x}$ gives 0, because $\lfloor x \rfloor = 0$ for all $x\in [0,1)$. That's why the whole integral vanishes.

For your second question: In the interval $[n,n+1)$ the function $\lfloor x \rfloor e^{-x} = n e^{-x}$, that's why only the $x$ that is 'floored' changed to $n$.

Our task here is nothing but calculating Laplace Transform of $\lfloor t\rfloor$ at $s=1$

\begin{align}F(s)&=\displaystyle\int_0^\infty \lfloor t \rfloor e^{-st}dt\\&=\displaystyle\int_0^1 \lfloor t \rfloor e^{-st}dt +\displaystyle\int_1^2 \lfloor t \rfloor e^{-st}dt +\displaystyle\int_2^3 \lfloor t \rfloor e^{-st}dt + \cdots\\&=\displaystyle\sum_{n=0}^\infty n\displaystyle\int_{n}^{n+1} e^{-st}dt\\&=\dfrac{1}{s}\displaystyle\sum_{n=0}^\infty n \left[ e^{-sn}-e^{-s(n+1)} \right]\\&=\dfrac{1}{s}\left(1-e^{-s}\right)\displaystyle\sum_{n=0}^\infty n e^{-sn}\\&=\dfrac{1}{s}\left(1-e^{-s}\right)\dfrac{e^{-s}}{\left(1-e^{-s}\right)^{2}}\\&=\dfrac{1}{s\left(e^s-1\right)}\\&=\dfrac{\coth\left(\dfrac{s}{2}\right)-1}{2s}\end{align}\tag*{}

Now just put $s=1$

Note that:

$$\displaystyle\sum_{n=0}^{\infty} e^{-n x} = \dfrac{1}{1-e^{-x}}$$ $$\displaystyle\sum_{n=0}^{\infty}n e^{-n x}=- \dfrac{d}{dx}\left(\dfrac{1}{1-e^{-x}}\right)$$

• What if I did not study Laplace transform?
– user426277
Jul 6 '17 at 18:42
• that actually does not matter much , you can consider Laplace transform as a mere improper integral , consider this integral $$\int_{0}^{\infty}e^{-st} \sin t \,dt$$ Formally this is known as LT of $\sin t$ , but for a guy who have never heard of LT it's just a improper integral over the positive real number line Jul 6 '17 at 18:49

In answer to your first question, on that interval the integrand vanishes, so there is no contribution to the integral. In answer to your second question, on each of the length-$1$ intervals considered the floor factor is constant but $e^{-x}$ is not.

$$\forall x \in [n, n+1[, \lfloor x \rfloor = n$$

and $$\int_0^1{\lfloor x \rfloor e^{-x}dx} = \int_0^1{0 e^{-x}dx} =0$$